Unsolved Problems in Number Theory, Logic, and Cryptography Yahoo Group Pythagorean "five"tuples and "six"tuples. =============================================== sergey_beliy@ymail.com Message 1 of 9 Sep 4, 2012 ----------------------------------------------- Hi.May be it some sort of intresting. Among a^2+b^2+c^2+d^2=e^2's i knew only 1^2+1^2+1^2+1^2=2^2. Here are some more. We are looking for e's. We have odd squares 9,25,49...... 1. 9 is the 5th odd number: (9+1)/2=5 2. in its turn 5 is the 3rd odd number: (5+1)/2=3 3. in its turn 3 is the 2nd odd number: (3+1)/2=2 or we can do it in 2 steps 1. (9+3)/4=3 2. (3+1)/2=2 The same goes whith 25 and 49. So we have 2,4,7.... 1^2+1^2+1^2+1^2=2^2 which is not a surpise, since 4 times any square is an even square 4*(1^2)=2^2 2^2+2^2+2^2+2^2=4^2 or 4*2^2=4^2 4^2+4^2+4^2+1^2=7^2 May be we can find the rest. As to "six"tuples, like 5^2+5^2+5^2+3^2+4^2=10^2, it does not work. Sergey. =============================================== zetacooking Message 2 of 9 Sep 4, 2012 ----------------------------------------------- Hi Sergey, There are two related theorems on this: Every number (which includes all squares) can be written as the sum of four squares. This is Lagrange's four square theorem. Legendre's three square theorem states that every number can be written as the sum of three squares, except for those of the form (4^n)(8k-1). Now 4^n will be even for all n and 8k-1 will be odd for all k so that in order for this form to be a square we need 4^n=e^2 and 8k-1=o^2. Solutions are easy to find for e^2, such as n=(1,2,4...). Checking the form required for the odd square, we have o^2 = (2m-1)^2 = 4m^2-4m+1 So we need 8k-1 = 4m^2-4m+1 8k = 4m^2-4m+2 4k = 2m^2-2m+1 4k = 2(m^2-m)+1 which is impossible because the left is always even and the right is always odd. Since the excluded numbers cannot be square, every square can be written as the sum of three square numbers. This does not imply that every square is a potential space diagonal for a perfect cuboid, since the theorem allows for using 0^2, giving a^2+0^2+0^2 = a^2. Every square can even be written as the sum of one square but this theorem is so trivial, who would want their name attached to it? :) Zero is an invalid number for an n-tuple. For Pythagorean relationships, we need the additional constraint that the squares being summed are above zero. What can say about the squares that can be written as the sum of two or three or m positive squares? How many non-zero squares, at most, are needed, which will sum to every square? David --- In UnsolvedProblems@yahoogroups.com, "sergey_beliy@..." <sergey_beliy@...> wrote: > > Hi.May be it some sort of intresting. > > Among a^2+b^2+c^2+d^2=e^2's i knew only 1^2+1^2+1^2+1^2=2^2. > > Here are some more. > > We are looking for e's. > > We have odd squares 9,25,49...... > > 1. 9 is the 5th odd number: (9+1)/2=5 > 2. in its turn 5 is the 3rd odd number: (5+1)/2=3 > 3. in its turn 3 is the 2nd odd number: (3+1)/2=2 > > or we can do it in 2 steps > > 1. (9+3)/4=3 > 2. (3+1)/2=2 > > The same goes whith 25 and 49. > > So we have 2,4,7.... > > 1^2+1^2+1^2+1^2=2^2 which is not a surpise, since 4 times any square is an even square 4*(1^2)=2^2 > > 2^2+2^2+2^2+2^2=4^2 or 4*2^2=4^2 > > 4^2+4^2+4^2+1^2=7^2 > > May be we can find the rest. > > As to "six"tuples, like 5^2+5^2+5^2+3^2+4^2=10^2, it does not work. > > Sergey. > =============================================== Mark Message 3 of 9 Sep 4, 2012 ----------------------------------------------- David asked, "What can say about the squares that can be written as the sum of two or three or m positive squares? How many non-zero squares, at most, are needed, which will sum to every square?" I just wrote a program, and it appears that for any square greater than 3^2, four non zero squares are needed to sum to the square. (When only three squares are used, the square of numbers of the form 2^n and 5*2^n (and only those numbers) cannot be written as the sum of three (non zero) squares. ) There are some fascinating properties of how many different ways a square can be written as the sum of four non zero squares. 2^2 = 1^2 + 1^2 + 1^2 + 1^2; Obviously 2^2 can only be represented as the sum of four positive squares one way. 3^2 cannot be written as the sum of four positive squares. 4^2 = 2^2 + 2^2 + 2^2 + 2^2. Only one way. 5^2 = 1^2 + 2^2 + 2^2 + 4^2. Only one way. 6^2 = 1^2 + 1^2 + 3^2 + 5^2 = 3^2 + 3^2 + 3^2 + 3^2 . Two ways. And so on. Let n^2 be written as the sum of four positive squares k ways. Then here's data for n up to 120, in the form [n,k] : [1, 0] [2, 1] [3, 0] [4, 1] [5, 1] [6, 2] [7, 2] [8, 1] [9, 2] [10, 5] [11, 3] [12, 2] [13, 5] [14, 8] [15, 9] [16, 1] [17, 7] [18, 10] [19, 9] [20, 5] [21, 16] [22, 12] [23, 13] [24, 2] [25, 19] [26, 18] [27, 22] [28, 8] [29, 20] [30, 32] [31, 23] [32, 1] [33, 35] [34, 25] [35, 42] [36, 10] [37, 32] [38, 31] [39, 51] [40, 5] [41, 38] [42, 55] [43, 42] [44, 12] [45, 80] [46, 43] [47, 50] [48, 2] [49, 63] [50, 62] [51, 83] [52, 18] [53, 63] [54, 75] [55, 91] [56, 8] [57, 103] [58, 65] [59, 77] [60, 32] [61, 83] [62, 74] [63, 144] [64, 1] [65, 127] [66, 116] [67, 99] [68, 25] [69, 151] [70, 133] [71, 111] [72, 10] [73, 117] [74, 102] [75, 217] [76, 31] [77, 163] [78, 163] [79, 137] [80, 5] [81, 204] [82, 121] [83, 150] [84, 55] [85, 207] [86, 133] [87, 238] [88, 12] [89, 172] [90, 253] [91, 228] [92, 43] [93, 271] [94, 157] [95, 256] [96, 2] [97, 204] [98, 198] [99, 332] [100, 62] [101, 221] [102, 263] [103, 230] [104, 18] [105, 488] [106, 197] [107, 247] [108, 75] [109, 257] [110, 283] [111, 384] [112, 8] [113, 275] [114, 325] [115, 371] [116, 65] [117, 464] [118, 240] [119, 374] [120, 32] Notice that when n is a power of two then its square can be expressed as the sum of four non zero squares only one way. Also notice that when n is even then 2n, 4n, 8n, 16n, ect will have the same k value as n. Also notice that if n is a prime and increases, then k increases. Will this always hold? Don't know. (Interestingly it doesn't hold when it is expressed as the sum of three squares!) Mark --- In UnsolvedProblems@yahoogroups.com, "zetacooking" <hartwelld@...> wrote: > > Hi Sergey, > > There are two related theorems on this: > > Every number (which includes all squares) can be written as the sum of four squares. This is Lagrange's four square theorem. > > Legendre's three square theorem states that every number can be written as the sum of three squares, except for those of the form (4^n)(8k-1). > > Now 4^n will be even for all n and 8k-1 will be odd for all k so that in order for this form to be a square we need 4^n=e^2 and 8k-1=o^2. Solutions are easy to find for e^2, such as n=(1,2,4...). Checking the form required for the odd square, we have > > o^2 = (2m-1)^2 = 4m^2-4m+1 > > So we need > > 8k-1 = 4m^2-4m+1 > > 8k = 4m^2-4m+2 > > 4k = 2m^2-2m+1 > > 4k = 2(m^2-m)+1 > > which is impossible because the left is always even and the right is always odd. Since the excluded numbers cannot be square, every square can be written as the sum of three square numbers. > > This does not imply that every square is a potential space diagonal for a perfect cuboid, since the theorem allows for using 0^2, giving a^2+0^2+0^2 = a^2. Every square can even be written as the sum of one square but this theorem is so trivial, who would want their name attached to it? :) > > Zero is an invalid number for an n-tuple. For Pythagorean relationships, we need the additional constraint that the squares being summed are above zero. > > What can say about the squares that can be written as the sum of two or three or m positive squares? How many non-zero squares, at most, are needed, which will sum to every square? > > David > > > --- In UnsolvedProblems@yahoogroups.com, "sergey_beliy@" <sergey_beliy@> wrote: > > > > Hi.May be it some sort of intresting. > > > > Among a^2+b^2+c^2+d^2=e^2's i knew only 1^2+1^2+1^2+1^2=2^2. > > > > Here are some more. > > > > We are looking for e's. > > > > We have odd squares 9,25,49...... > > > > 1. 9 is the 5th odd number: (9+1)/2=5 > > 2. in its turn 5 is the 3rd odd number: (5+1)/2=3 > > 3. in its turn 3 is the 2nd odd number: (3+1)/2=2 > > > > or we can do it in 2 steps > > > > 1. (9+3)/4=3 > > 2. (3+1)/2=2 > > > > The same goes whith 25 and 49. > > > > So we have 2,4,7.... > > > > 1^2+1^2+1^2+1^2=2^2 which is not a surpise, since 4 times any square is an even square 4*(1^2)=2^2 > > > > 2^2+2^2+2^2+2^2=4^2 or 4*2^2=4^2 > > > > 4^2+4^2+4^2+1^2=7^2 > > > > May be we can find the rest. > > > > As to "six"tuples, like 5^2+5^2+5^2+3^2+4^2=10^2, it does not work. > > > > Sergey. > > > =============================================== zetacooking Message 4 of 9 Sep 5, 2012 ----------------------------------------------- Excellent analysis, Mark. Interesting results and a well presented report of your findings. And fast, too! I think posting your source code here would have educational value. Your sequence does not exist on OEIS. I encourage you to submit it. David --- In UnsolvedProblems@yahoogroups.com, "Mark" <mark.underwood@...> wrote: > > David asked, "What can say about the squares that can be written as the sum of two or three or m positive squares? How many non-zero squares, at most, are needed, which will sum to every square?" > > I just wrote a program, and it appears that for any square greater than 3^2, four non zero squares are needed to sum to the square. > > (When only three squares are used, the square of numbers of the form 2^n and 5*2^n (and only those numbers) cannot be written as the sum of three (non zero) squares. ) > > There are some fascinating properties of how many different ways a square can be written as the sum of four non zero squares. > > 2^2 = 1^2 + 1^2 + 1^2 + 1^2; Obviously 2^2 can only be represented as the sum of four positive squares one way. > > 3^2 cannot be written as the sum of four positive squares. > > 4^2 = 2^2 + 2^2 + 2^2 + 2^2. Only one way. > > 5^2 = 1^2 + 2^2 + 2^2 + 4^2. Only one way. > > 6^2 = 1^2 + 1^2 + 3^2 + 5^2 = 3^2 + 3^2 + 3^2 + 3^2 . Two ways. > > And so on. > > Let n^2 be written as the sum of four positive squares k ways. Then here's data for n up to 120, in the form [n,k] : > > [1, 0] [2, 1] [3, 0] [4, 1] [5, 1] [6, 2] [7, 2] [8, 1] [9, 2] [10, 5] [11, 3] [12, 2] [13, 5] [14, 8] [15, 9] [16, 1] [17, 7] [18, 10] [19, 9] [20, 5] [21, 16] [22, 12] [23, 13] [24, 2] [25, 19] [26, 18] [27, 22] [28, 8] [29, 20] [30, 32] [31, 23] [32, 1] [33, 35] [34, 25] [35, 42] [36, 10] [37, 32] [38, 31] [39, 51] [40, 5] [41, 38] [42, 55] [43, 42] [44, 12] [45, 80] [46, 43] [47, 50] [48, 2] [49, 63] [50, 62] [51, 83] [52, 18] [53, 63] [54, 75] [55, 91] [56, 8] [57, 103] [58, 65] [59, 77] [60, 32] [61, 83] [62, 74] [63, 144] [64, 1] [65, 127] [66, 116] [67, 99] [68, 25] [69, 151] [70, 133] [71, 111] [72, 10] [73, 117] [74, 102] [75, 217] [76, 31] [77, 163] [78, 163] [79, 137] [80, 5] [81, 204] [82, 121] [83, 150] [84, 55] [85, 207] [86, 133] [87, 238] [88, 12] [89, 172] [90, 253] [91, 228] [92, 43] [93, 271] [94, 157] [95, 256] [96, 2] [97, 204] [98, 198] [99, 332] [100, 62] [101, 221] [102, 263] [103, 230] [104, 18] [105, 488] [106, 197] [107, 247] [108, 75] [109, 257] [110, 283] [111, 384] [112, 8] [113, 275] [114, 325] [115, 371] [116, 65] [117, 464] [118, 240] [119, 374] [120, 32] > > Notice that when n is a power of two then its square can be expressed as the sum of four non zero squares only one way. > > Also notice that when n is even then 2n, 4n, 8n, 16n, ect will have the same k value as n. > > Also notice that if n is a prime and increases, then k increases. Will this always hold? Don't know. (Interestingly it doesn't hold when it is expressed as the sum of three squares!) > > Mark > > > > > --- In UnsolvedProblems@yahoogroups.com, "zetacooking" <hartwelld@> wrote: > > > > Hi Sergey, > > > > There are two related theorems on this: > > > > Every number (which includes all squares) can be written as the sum of four squares. This is Lagrange's four square theorem. > > > > Legendre's three square theorem states that every number can be written as the sum of three squares, except for those of the form (4^n)(8k-1). > > > > Now 4^n will be even for all n and 8k-1 will be odd for all k so that in order for this form to be a square we need 4^n=e^2 and 8k-1=o^2. Solutions are easy to find for e^2, such as n=(1,2,4...). Checking the form required for the odd square, we have > > > > o^2 = (2m-1)^2 = 4m^2-4m+1 > > > > So we need > > > > 8k-1 = 4m^2-4m+1 > > > > 8k = 4m^2-4m+2 > > > > 4k = 2m^2-2m+1 > > > > 4k = 2(m^2-m)+1 > > > > which is impossible because the left is always even and the right is always odd. Since the excluded numbers cannot be square, every square can be written as the sum of three square numbers. > > > > This does not imply that every square is a potential space diagonal for a perfect cuboid, since the theorem allows for using 0^2, giving a^2+0^2+0^2 = a^2. Every square can even be written as the sum of one square but this theorem is so trivial, who would want their name attached to it? :) > > > > Zero is an invalid number for an n-tuple. For Pythagorean relationships, we need the additional constraint that the squares being summed are above zero. > > > > What can say about the squares that can be written as the sum of two or three or m positive squares? How many non-zero squares, at most, are needed, which will sum to every square? > > > > David > > > > > > --- In UnsolvedProblems@yahoogroups.com, "sergey_beliy@" <sergey_beliy@> wrote: > > > > > > Hi.May be it some sort of intresting. > > > > > > Among a^2+b^2+c^2+d^2=e^2's i knew only 1^2+1^2+1^2+1^2=2^2. > > > > > > Here are some more. > > > > > > We are looking for e's. > > > > > > We have odd squares 9,25,49...... > > > > > > 1. 9 is the 5th odd number: (9+1)/2=5 > > > 2. in its turn 5 is the 3rd odd number: (5+1)/2=3 > > > 3. in its turn 3 is the 2nd odd number: (3+1)/2=2 > > > > > > or we can do it in 2 steps > > > > > > 1. (9+3)/4=3 > > > 2. (3+1)/2=2 > > > > > > The same goes whith 25 and 49. > > > > > > So we have 2,4,7.... > > > > > > 1^2+1^2+1^2+1^2=2^2 which is not a surpise, since 4 times any square is an even square 4*(1^2)=2^2 > > > > > > 2^2+2^2+2^2+2^2=4^2 or 4*2^2=4^2 > > > > > > 4^2+4^2+4^2+1^2=7^2 > > > > > > May be we can find the rest. > > > > > > As to "six"tuples, like 5^2+5^2+5^2+3^2+4^2=10^2, it does not work. > > > > > > Sergey. > > > > > > =============================================== Mark Message 5 of 9 Sep 5, 2012 ----------------------------------------------- Hi David, I put the following text into a file called david : for (n=1,120, k=0; for(s1=1,sqrt((n^2)/4), for(s2=s1,sqrt((n^2 - s1^2)/3) , for(s3=s2,sqrt((n^2-s1^2 - s2^2)/2), if(issquare(n^2-s1^2-s2^2-s3^2),k++)))) ; print1([n,k]" ")) Then at the gp pari prompt I put in \r david and out it comes. On my old computer it took about 7 seconds to cover the numbers up to 120. I'm surprised the sequence is not in OEIS. I may put it in, but I would prefer if you would since it was your essential idea after all! Personally I find the sequence for the primes the most interesting and may investigate it further at some later time. Mark --- In UnsolvedProblems@yahoogroups.com, "zetacooking" <hartwelld@...> wrote: > > Excellent analysis, Mark. Interesting results and a well presented report of your findings. And fast, too! > > I think posting your source code here would have educational value. > > Your sequence does not exist on OEIS. I encourage you to submit it. > > David > > --- In UnsolvedProblems@yahoogroups.com, "Mark" <mark.underwood@> wrote: > > > > David asked, "What can say about the squares that can be written as the sum of two or three or m positive squares? How many non-zero squares, at most, are needed, which will sum to every square?" > > > > I just wrote a program, and it appears that for any square greater than 3^2, four non zero squares are needed to sum to the square. > > > > (When only three squares are used, the square of numbers of the form 2^n and 5*2^n (and only those numbers) cannot be written as the sum of three (non zero) squares. ) > > > > There are some fascinating properties of how many different ways a square can be written as the sum of four non zero squares. > > > > 2^2 = 1^2 + 1^2 + 1^2 + 1^2; Obviously 2^2 can only be represented as the sum of four positive squares one way. > > > > 3^2 cannot be written as the sum of four positive squares. > > > > 4^2 = 2^2 + 2^2 + 2^2 + 2^2. Only one way. > > > > 5^2 = 1^2 + 2^2 + 2^2 + 4^2. Only one way. > > > > 6^2 = 1^2 + 1^2 + 3^2 + 5^2 = 3^2 + 3^2 + 3^2 + 3^2 . Two ways. > > > > And so on. > > > > Let n^2 be written as the sum of four positive squares k ways. Then here's data for n up to 120, in the form [n,k] : > > > > [1, 0] [2, 1] [3, 0] [4, 1] [5, 1] [6, 2] [7, 2] [8, 1] [9, 2] [10, 5] [11, 3] [12, 2] [13, 5] [14, 8] [15, 9] [16, 1] [17, 7] [18, 10] [19, 9] [20, 5] [21, 16] [22, 12] [23, 13] [24, 2] [25, 19] [26, 18] [27, 22] [28, 8] [29, 20] [30, 32] [31, 23] [32, 1] [33, 35] [34, 25] [35, 42] [36, 10] [37, 32] [38, 31] [39, 51] [40, 5] [41, 38] [42, 55] [43, 42] [44, 12] [45, 80] [46, 43] [47, 50] [48, 2] [49, 63] [50, 62] [51, 83] [52, 18] [53, 63] [54, 75] [55, 91] [56, 8] [57, 103] [58, 65] [59, 77] [60, 32] [61, 83] [62, 74] [63, 144] [64, 1] [65, 127] [66, 116] [67, 99] [68, 25] [69, 151] [70, 133] [71, 111] [72, 10] [73, 117] [74, 102] [75, 217] [76, 31] [77, 163] [78, 163] [79, 137] [80, 5] [81, 204] [82, 121] [83, 150] [84, 55] [85, 207] [86, 133] [87, 238] [88, 12] [89, 172] [90, 253] [91, 228] [92, 43] [93, 271] [94, 157] [95, 256] [96, 2] [97, 204] [98, 198] [99, 332] [100, 62] [101, 221] [102, 263] [103, 230] [104, 18] [105, 488] [106, 197] [107, 247] [108, 75] [109, 257] [110, 283] [111, 384] [112, 8] [113, 275] [114, 325] [115, 371] [116, 65] [117, 464] [118, 240] [119, 374] [120, 32] > > > > Notice that when n is a power of two then its square can be expressed as the sum of four non zero squares only one way. > > > > Also notice that when n is even then 2n, 4n, 8n, 16n, ect will have the same k value as n. > > > > Also notice that if n is a prime and increases, then k increases. Will this always hold? Don't know. (Interestingly it doesn't hold when it is expressed as the sum of three squares!) > > > > Mark > > > > > > > > > > --- In UnsolvedProblems@yahoogroups.com, "zetacooking" <hartwelld@> wrote: > > > > > > Hi Sergey, > > > > > > There are two related theorems on this: > > > > > > Every number (which includes all squares) can be written as the sum of four squares. This is Lagrange's four square theorem. > > > > > > Legendre's three square theorem states that every number can be written as the sum of three squares, except for those of the form (4^n)(8k-1). > > > > > > Now 4^n will be even for all n and 8k-1 will be odd for all k so that in order for this form to be a square we need 4^n=e^2 and 8k-1=o^2. Solutions are easy to find for e^2, such as n=(1,2,4...). Checking the form required for the odd square, we have > > > > > > o^2 = (2m-1)^2 = 4m^2-4m+1 > > > > > > So we need > > > > > > 8k-1 = 4m^2-4m+1 > > > > > > 8k = 4m^2-4m+2 > > > > > > 4k = 2m^2-2m+1 > > > > > > 4k = 2(m^2-m)+1 > > > > > > which is impossible because the left is always even and the right is always odd. Since the excluded numbers cannot be square, every square can be written as the sum of three square numbers. > > > > > > This does not imply that every square is a potential space diagonal for a perfect cuboid, since the theorem allows for using 0^2, giving a^2+0^2+0^2 = a^2. Every square can even be written as the sum of one square but this theorem is so trivial, who would want their name attached to it? :) > > > > > > Zero is an invalid number for an n-tuple. For Pythagorean relationships, we need the additional constraint that the squares being summed are above zero. > > > > > > What can say about the squares that can be written as the sum of two or three or m positive squares? How many non-zero squares, at most, are needed, which will sum to every square? > > > > > > David > > > > > > > > > --- In UnsolvedProblems@yahoogroups.com, "sergey_beliy@" <sergey_beliy@> wrote: > > > > > > > > Hi.May be it some sort of intresting. > > > > > > > > Among a^2+b^2+c^2+d^2=e^2's i knew only 1^2+1^2+1^2+1^2=2^2. > > > > > > > > Here are some more. > > > > > > > > We are looking for e's. > > > > > > > > We have odd squares 9,25,49...... > > > > > > > > 1. 9 is the 5th odd number: (9+1)/2=5 > > > > 2. in its turn 5 is the 3rd odd number: (5+1)/2=3 > > > > 3. in its turn 3 is the 2nd odd number: (3+1)/2=2 > > > > > > > > or we can do it in 2 steps > > > > > > > > 1. (9+3)/4=3 > > > > 2. (3+1)/2=2 > > > > > > > > The same goes whith 25 and 49. > > > > > > > > So we have 2,4,7.... > > > > > > > > 1^2+1^2+1^2+1^2=2^2 which is not a surpise, since 4 times any square is an even square 4*(1^2)=2^2 > > > > > > > > 2^2+2^2+2^2+2^2=4^2 or 4*2^2=4^2 > > > > > > > > 4^2+4^2+4^2+1^2=7^2 > > > > > > > > May be we can find the rest. > > > > > > > > As to "six"tuples, like 5^2+5^2+5^2+3^2+4^2=10^2, it does not work. > > > > > > > > Sergey. > > > > > > > > > > =============================================== Mark Message 6 of 9 Sep 5, 2012 ----------------------------------------------- OK this is neat. From the data I derived a formula for the number of ways the square of an odd prime n can be written as the sum of four non zero squares: f = floor((n^2 + 4n + 24)/48) So given that n is an odd prime, k is the actual count of the number of ways that four non zero squares can sum to n^2, and f is from the formula. [n,k,f] : [3, 0, 0] [5, 1, 1] [7, 2, 2] [11, 3, 3] [13, 5, 5] [17, 7, 7] [19, 9, 9] [23, 13, 13] [29, 20, 20] [31, 23, 23] [37, 32, 32] [41, 38, 38] [43, 42, 42] [47, 50, 50] [53, 63, 63] [59, 77, 77] [61, 83, 83] [67, 99, 99] [71, 111, 111] [73, 117, 117] [79, 137, 137] [83, 150, 150] [89, 172, 172] [97, 204, 204] [101, 221, 221] [103, 230, 230] [107, 247, 247] [109, 257, 257] [113, 275, 275] [127, 347, 347] [131, 368, 368] [137, 402, 402] [139, 414, 414] [149, 475, 475] [151, 488, 488] [157, 527, 527] [163, 567, 567] [167, 595, 595] [173, 638, 638] [179, 682, 682] [181, 698, 698] [191, 776, 776] [193, 792, 792] [197, 825, 825] [199, 842, 842] Is this known? I don't know... And I'm certainly not up for a proof hehe! Mark --- In UnsolvedProblems@yahoogroups.com, "Mark" <mark.underwood@...> wrote: > > > > > Hi David, > > I put the following text into a file called david : > > for (n=1,120, k=0; for(s1=1,sqrt((n^2)/4), for(s2=s1,sqrt((n^2 - s1^2)/3) , for(s3=s2,sqrt((n^2-s1^2 - s2^2)/2), if(issquare(n^2-s1^2-s2^2-s3^2),k++)))) ; print1([n,k]" ")) > > Then at the gp pari prompt I put in > > \r david > > and out it comes. On my old computer it took about 7 seconds to cover the numbers up to 120. > > I'm surprised the sequence is not in OEIS. > > I may put it in, but I would prefer if you would since it was your essential idea after all! > > Personally I find the sequence for the primes the most interesting and may investigate it further at some later time. > > Mark > > > --- In UnsolvedProblems@yahoogroups.com, "zetacooking" <hartwelld@> wrote: > > > > Excellent analysis, Mark. Interesting results and a well presented report of your findings. And fast, too! > > > > I think posting your source code here would have educational value. > > > > Your sequence does not exist on OEIS. I encourage you to submit it. > > > > David > > > > --- In UnsolvedProblems@yahoogroups.com, "Mark" <mark.underwood@> wrote: > > > > > > David asked, "What can say about the squares that can be written as the sum of two or three or m positive squares? How many non-zero squares, at most, are needed, which will sum to every square?" > > > > > > I just wrote a program, and it appears that for any square greater than 3^2, four non zero squares are needed to sum to the square. > > > > > > (When only three squares are used, the square of numbers of the form 2^n and 5*2^n (and only those numbers) cannot be written as the sum of three (non zero) squares. ) > > > > > > There are some fascinating properties of how many different ways a square can be written as the sum of four non zero squares. > > > > > > 2^2 = 1^2 + 1^2 + 1^2 + 1^2; Obviously 2^2 can only be represented as the sum of four positive squares one way. > > > > > > 3^2 cannot be written as the sum of four positive squares. > > > > > > 4^2 = 2^2 + 2^2 + 2^2 + 2^2. Only one way. > > > > > > 5^2 = 1^2 + 2^2 + 2^2 + 4^2. Only one way. > > > > > > 6^2 = 1^2 + 1^2 + 3^2 + 5^2 = 3^2 + 3^2 + 3^2 + 3^2 . Two ways. > > > > > > And so on. > > > > > > Let n^2 be written as the sum of four positive squares k ways. Then here's data for n up to 120, in the form [n,k] : > > > > > > [1, 0] [2, 1] [3, 0] [4, 1] [5, 1] [6, 2] [7, 2] [8, 1] [9, 2] [10, 5] [11, 3] [12, 2] [13, 5] [14, 8] [15, 9] [16, 1] [17, 7] [18, 10] [19, 9] [20, 5] [21, 16] [22, 12] [23, 13] [24, 2] [25, 19] [26, 18] [27, 22] [28, 8] [29, 20] [30, 32] [31, 23] [32, 1] [33, 35] [34, 25] [35, 42] [36, 10] [37, 32] [38, 31] [39, 51] [40, 5] [41, 38] [42, 55] [43, 42] [44, 12] [45, 80] [46, 43] [47, 50] [48, 2] [49, 63] [50, 62] [51, 83] [52, 18] [53, 63] [54, 75] [55, 91] [56, 8] [57, 103] [58, 65] [59, 77] [60, 32] [61, 83] [62, 74] [63, 144] [64, 1] [65, 127] [66, 116] [67, 99] [68, 25] [69, 151] [70, 133] [71, 111] [72, 10] [73, 117] [74, 102] [75, 217] [76, 31] [77, 163] [78, 163] [79, 137] [80, 5] [81, 204] [82, 121] [83, 150] [84, 55] [85, 207] [86, 133] [87, 238] [88, 12] [89, 172] [90, 253] [91, 228] [92, 43] [93, 271] [94, 157] [95, 256] [96, 2] [97, 204] [98, 198] [99, 332] [100, 62] [101, 221] [102, 263] [103, 230] [104, 18] [105, 488] [106, 197] [107, 247] [108, 75] [109, 257] [110, 283] [111, 384] [112, 8] [113, 275] [114, 325] [115, 371] [116, 65] [117, 464] [118, 240] [119, 374] [120, 32] > > > > > > Notice that when n is a power of two then its square can be expressed as the sum of four non zero squares only one way. > > > > > > Also notice that when n is even then 2n, 4n, 8n, 16n, ect will have the same k value as n. > > > > > > Also notice that if n is a prime and increases, then k increases. Will this always hold? Don't know. (Interestingly it doesn't hold when it is expressed as the sum of three squares!) > > > > > > Mark > > > > > > > > > > > > > > > --- In UnsolvedProblems@yahoogroups.com, "zetacooking" <hartwelld@> wrote: > > > > > > > > Hi Sergey, > > > > > > > > There are two related theorems on this: > > > > > > > > Every number (which includes all squares) can be written as the sum of four squares. This is Lagrange's four square theorem. > > > > > > > > Legendre's three square theorem states that every number can be written as the sum of three squares, except for those of the form (4^n)(8k-1). > > > > > > > > Now 4^n will be even for all n and 8k-1 will be odd for all k so that in order for this form to be a square we need 4^n=e^2 and 8k-1=o^2. Solutions are easy to find for e^2, such as n=(1,2,4...). Checking the form required for the odd square, we have > > > > > > > > o^2 = (2m-1)^2 = 4m^2-4m+1 > > > > > > > > So we need > > > > > > > > 8k-1 = 4m^2-4m+1 > > > > > > > > 8k = 4m^2-4m+2 > > > > > > > > 4k = 2m^2-2m+1 > > > > > > > > 4k = 2(m^2-m)+1 > > > > > > > > which is impossible because the left is always even and the right is always odd. Since the excluded numbers cannot be square, every square can be written as the sum of three square numbers. > > > > > > > > This does not imply that every square is a potential space diagonal for a perfect cuboid, since the theorem allows for using 0^2, giving a^2+0^2+0^2 = a^2. Every square can even be written as the sum of one square but this theorem is so trivial, who would want their name attached to it? :) > > > > > > > > Zero is an invalid number for an n-tuple. For Pythagorean relationships, we need the additional constraint that the squares being summed are above zero. > > > > > > > > What can say about the squares that can be written as the sum of two or three or m positive squares? How many non-zero squares, at most, are needed, which will sum to every square? > > > > > > > > David > > > > > > > > > > > > --- In UnsolvedProblems@yahoogroups.com, "sergey_beliy@" <sergey_beliy@> wrote: > > > > > > > > > > Hi.May be it some sort of intresting. > > > > > > > > > > Among a^2+b^2+c^2+d^2=e^2's i knew only 1^2+1^2+1^2+1^2=2^2. > > > > > > > > > > Here are some more. > > > > > > > > > > We are looking for e's. > > > > > > > > > > We have odd squares 9,25,49...... > > > > > > > > > > 1. 9 is the 5th odd number: (9+1)/2=5 > > > > > 2. in its turn 5 is the 3rd odd number: (5+1)/2=3 > > > > > 3. in its turn 3 is the 2nd odd number: (3+1)/2=2 > > > > > > > > > > or we can do it in 2 steps > > > > > > > > > > 1. (9+3)/4=3 > > > > > 2. (3+1)/2=2 > > > > > > > > > > The same goes whith 25 and 49. > > > > > > > > > > So we have 2,4,7.... > > > > > > > > > > 1^2+1^2+1^2+1^2=2^2 which is not a surpise, since 4 times any square is an even square 4*(1^2)=2^2 > > > > > > > > > > 2^2+2^2+2^2+2^2=4^2 or 4*2^2=4^2 > > > > > > > > > > 4^2+4^2+4^2+1^2=7^2 > > > > > > > > > > May be we can find the rest. > > > > > > > > > > As to "six"tuples, like 5^2+5^2+5^2+3^2+4^2=10^2, it does not work. > > > > > > > > > > Sergey. > > > > > > > > > > > > > > > =============================================== zetacooking Message 7 of 9 Sep 5, 2012 ----------------------------------------------- Au contraire, Mark, it was your idea to count the number of ways. It belongs to you and, as we see already, opens up a field of exploration which at present you are better able to explore than I. This may lead to an entire collection of sequences for OEIS. Here's some identities that may help prove why your function might be valid: n^2 + 4n + 24 can be written as an alternating series of 4 squares = (n+8)^2 - (n+6)^2 + n^2 - 2^2 which is the sum of the difference of 2 pairs of squares 2 ways: = ((n+8)^2 - (n+6)^2) + (n^2 - 2^2) = 4(n+7) + (n+2)(n-2) = ((n+8)^2 - 2^2) + (n^2 - (n+6)^2) = (n+10)(n+6) - 12(n+3) Also note: 48 = 4^2 + 4^2 + 4^2 = 6^2 + 2^2 + 2^2 + 2^2 There may be a possibility for extending your results into something relevant for the GC problem. Although it has little to do with squares, it shares the characteristic of the size of the argument corresponding to a generally increasing number of solutions. A popular visualization of this tendency is the Goldbach comet. I wonder if your sequences would also display bands if graphed similarly. David --- In UnsolvedProblems@yahoogroups.com, "Mark" <mark.underwood@...> wrote: > > > OK this is neat. From the data I derived a formula for the number of ways the square of an odd prime n can be written as the sum of four non zero squares: > > f = floor((n^2 + 4n + 24)/48) > > So given that n is an odd prime, k is the actual count of the number of ways that four non zero squares can sum to n^2, and f is from the formula. [n,k,f] : > > > [3, 0, 0] [5, 1, 1] [7, 2, 2] [11, 3, 3] [13, 5, 5] [17, 7, 7] [19, 9, 9] [23, 13, 13] [29, 20, 20] [31, 23, 23] [37, 32, 32] [41, 38, 38] [43, 42, 42] [47, 50, 50] [53, 63, 63] [59, 77, 77] [61, 83, 83] [67, 99, 99] [71, 111, 111] [73, 117, 117] [79, 137, 137] [83, 150, 150] [89, 172, 172] [97, 204, 204] [101, 221, 221] [103, 230, 230] [107, 247, 247] [109, 257, 257] [113, 275, 275] [127, 347, 347] [131, 368, 368] [137, 402, 402] [139, 414, 414] [149, 475, 475] [151, 488, 488] [157, 527, 527] [163, 567, 567] [167, 595, 595] [173, 638, 638] [179, 682, 682] [181, 698, 698] [191, 776, 776] [193, 792, 792] [197, 825, 825] [199, 842, 842] > > > Is this known? I don't know... > > And I'm certainly not up for a proof hehe! > > > Mark > > > --- In UnsolvedProblems@yahoogroups.com, "Mark" <mark.underwood@> wrote: > > > > > > > > > > Hi David, > > > > I put the following text into a file called david : > > > > for (n=1,120, k=0; for(s1=1,sqrt((n^2)/4), for(s2=s1,sqrt((n^2 - s1^2)/3) , for(s3=s2,sqrt((n^2-s1^2 - s2^2)/2), if(issquare(n^2-s1^2-s2^2-s3^2),k++)))) ; print1([n,k]" ")) > > > > Then at the gp pari prompt I put in > > > > \r david > > > > and out it comes. On my old computer it took about 7 seconds to cover the numbers up to 120. > > > > I'm surprised the sequence is not in OEIS. > > > > I may put it in, but I would prefer if you would since it was your essential idea after all! > > > > Personally I find the sequence for the primes the most interesting and may investigate it further at some later time. > > > > Mark > > > > > > --- In UnsolvedProblems@yahoogroups.com, "zetacooking" <hartwelld@> wrote: > > > > > > Excellent analysis, Mark. Interesting results and a well presented report of your findings. And fast, too! > > > > > > I think posting your source code here would have educational value. > > > > > > Your sequence does not exist on OEIS. I encourage you to submit it. > > > > > > David > > > > > > --- In UnsolvedProblems@yahoogroups.com, "Mark" <mark.underwood@> wrote: > > > > > > > > David asked, "What can say about the squares that can be written as the sum of two or three or m positive squares? How many non-zero squares, at most, are needed, which will sum to every square?" > > > > > > > > I just wrote a program, and it appears that for any square greater than 3^2, four non zero squares are needed to sum to the square. > > > > > > > > (When only three squares are used, the square of numbers of the form 2^n and 5*2^n (and only those numbers) cannot be written as the sum of three (non zero) squares. ) > > > > > > > > There are some fascinating properties of how many different ways a square can be written as the sum of four non zero squares. > > > > > > > > 2^2 = 1^2 + 1^2 + 1^2 + 1^2; Obviously 2^2 can only be represented as the sum of four positive squares one way. > > > > > > > > 3^2 cannot be written as the sum of four positive squares. > > > > > > > > 4^2 = 2^2 + 2^2 + 2^2 + 2^2. Only one way. > > > > > > > > 5^2 = 1^2 + 2^2 + 2^2 + 4^2. Only one way. > > > > > > > > 6^2 = 1^2 + 1^2 + 3^2 + 5^2 = 3^2 + 3^2 + 3^2 + 3^2 . Two ways. > > > > > > > > And so on. > > > > > > > > Let n^2 be written as the sum of four positive squares k ways. Then here's data for n up to 120, in the form [n,k] : > > > > > > > > [1, 0] [2, 1] [3, 0] [4, 1] [5, 1] [6, 2] [7, 2] [8, 1] [9, 2] [10, 5] [11, 3] [12, 2] [13, 5] [14, 8] [15, 9] [16, 1] [17, 7] [18, 10] [19, 9] [20, 5] [21, 16] [22, 12] [23, 13] [24, 2] [25, 19] [26, 18] [27, 22] [28, 8] [29, 20] [30, 32] [31, 23] [32, 1] [33, 35] [34, 25] [35, 42] [36, 10] [37, 32] [38, 31] [39, 51] [40, 5] [41, 38] [42, 55] [43, 42] [44, 12] [45, 80] [46, 43] [47, 50] [48, 2] [49, 63] [50, 62] [51, 83] [52, 18] [53, 63] [54, 75] [55, 91] [56, 8] [57, 103] [58, 65] [59, 77] [60, 32] [61, 83] [62, 74] [63, 144] [64, 1] [65, 127] [66, 116] [67, 99] [68, 25] [69, 151] [70, 133] [71, 111] [72, 10] [73, 117] [74, 102] [75, 217] [76, 31] [77, 163] [78, 163] [79, 137] [80, 5] [81, 204] [82, 121] [83, 150] [84, 55] [85, 207] [86, 133] [87, 238] [88, 12] [89, 172] [90, 253] [91, 228] [92, 43] [93, 271] [94, 157] [95, 256] [96, 2] [97, 204] [98, 198] [99, 332] [100, 62] [101, 221] [102, 263] [103, 230] [104, 18] [105, 488] [106, 197] [107, 247] [108, 75] [109, 257] [110, 283] [111, 384] [112, 8] [113, 275] [114, 325] [115, 371] [116, 65] [117, 464] [118, 240] [119, 374] [120, 32] > > > > > > > > Notice that when n is a power of two then its square can be expressed as the sum of four non zero squares only one way. > > > > > > > > Also notice that when n is even then 2n, 4n, 8n, 16n, ect will have the same k value as n. > > > > > > > > Also notice that if n is a prime and increases, then k increases. Will this always hold? Don't know. (Interestingly it doesn't hold when it is expressed as the sum of three squares!) > > > > > > > > Mark > > > > > > > > > > > > > > > > > > > > --- In UnsolvedProblems@yahoogroups.com, "zetacooking" <hartwelld@> wrote: > > > > > > > > > > Hi Sergey, > > > > > > > > > > There are two related theorems on this: > > > > > > > > > > Every number (which includes all squares) can be written as the sum of four squares. This is Lagrange's four square theorem. > > > > > > > > > > Legendre's three square theorem states that every number can be written as the sum of three squares, except for those of the form (4^n)(8k-1). > > > > > > > > > > Now 4^n will be even for all n and 8k-1 will be odd for all k so that in order for this form to be a square we need 4^n=e^2 and 8k-1=o^2. Solutions are easy to find for e^2, such as n=(1,2,4...). Checking the form required for the odd square, we have > > > > > > > > > > o^2 = (2m-1)^2 = 4m^2-4m+1 > > > > > > > > > > So we need > > > > > > > > > > 8k-1 = 4m^2-4m+1 > > > > > > > > > > 8k = 4m^2-4m+2 > > > > > > > > > > 4k = 2m^2-2m+1 > > > > > > > > > > 4k = 2(m^2-m)+1 > > > > > > > > > > which is impossible because the left is always even and the right is always odd. Since the excluded numbers cannot be square, every square can be written as the sum of three square numbers. > > > > > > > > > > This does not imply that every square is a potential space diagonal for a perfect cuboid, since the theorem allows for using 0^2, giving a^2+0^2+0^2 = a^2. Every square can even be written as the sum of one square but this theorem is so trivial, who would want their name attached to it? :) > > > > > > > > > > Zero is an invalid number for an n-tuple. For Pythagorean relationships, we need the additional constraint that the squares being summed are above zero. > > > > > > > > > > What can say about the squares that can be written as the sum of two or three or m positive squares? How many non-zero squares, at most, are needed, which will sum to every square? > > > > > > > > > > David > > > > > > > > > > > > > > > --- In UnsolvedProblems@yahoogroups.com, "sergey_beliy@" <sergey_beliy@> wrote: > > > > > > > > > > > > Hi.May be it some sort of intresting. > > > > > > > > > > > > Among a^2+b^2+c^2+d^2=e^2's i knew only 1^2+1^2+1^2+1^2=2^2. > > > > > > > > > > > > Here are some more. > > > > > > > > > > > > We are looking for e's. > > > > > > > > > > > > We have odd squares 9,25,49...... > > > > > > > > > > > > 1. 9 is the 5th odd number: (9+1)/2=5 > > > > > > 2. in its turn 5 is the 3rd odd number: (5+1)/2=3 > > > > > > 3. in its turn 3 is the 2nd odd number: (3+1)/2=2 > > > > > > > > > > > > or we can do it in 2 steps > > > > > > > > > > > > 1. (9+3)/4=3 > > > > > > 2. (3+1)/2=2 > > > > > > > > > > > > The same goes whith 25 and 49. > > > > > > > > > > > > So we have 2,4,7.... > > > > > > > > > > > > 1^2+1^2+1^2+1^2=2^2 which is not a surpise, since 4 times any square is an even square 4*(1^2)=2^2 > > > > > > > > > > > > 2^2+2^2+2^2+2^2=4^2 or 4*2^2=4^2 > > > > > > > > > > > > 4^2+4^2+4^2+1^2=7^2 > > > > > > > > > > > > May be we can find the rest. > > > > > > > > > > > > As to "six"tuples, like 5^2+5^2+5^2+3^2+4^2=10^2, it does not work. > > > > > > > > > > > > Sergey. > > > > > > > > > > > > > > > > > > > > > =============================================== marku606 Message 8 of 9 Sep 5, 2012 ----------------------------------------------- Hi David, I've since checked all the primes from 3 up to 997 and the formula agrees with the algorithm count in all cases. Why it works remains to be seen. But it is known that half of the primes - the primes of the form 4n+1 - can be written as the sum of two squares in exactly one way. So I'm sure that such will have a bearing on discovering why the formula works. Since you twisted my arm, I just submitted the sequence (for the primes only) to OEIS. :) (my second sequence!) Mark --- In UnsolvedProblems@yahoogroups.com, "zetacooking" <hartwelld@...> wrote: > > Au contraire, Mark, it was your idea to count the number of ways. It belongs to you and, as we see already, opens up a field of exploration which at present you are better able to explore than I. This may lead to an entire collection of sequences for OEIS. > > Here's some identities that may help prove why your function might be valid: > > n^2 + 4n + 24 can be written as an alternating series of 4 squares > > = (n+8)^2 - (n+6)^2 + n^2 - 2^2 > > which is the sum of the difference of 2 pairs of squares 2 ways: > > = ((n+8)^2 - (n+6)^2) + (n^2 - 2^2) = 4(n+7) + (n+2)(n-2) > > = ((n+8)^2 - 2^2) + (n^2 - (n+6)^2) = (n+10)(n+6) - 12(n+3) > > Also note: > > 48 = 4^2 + 4^2 + 4^2 = 6^2 + 2^2 + 2^2 + 2^2 > > There may be a possibility for extending your results into something relevant for the GC problem. Although it has little to do with squares, it shares the characteristic of the size of the argument corresponding to a generally increasing number of solutions. A popular visualization of this tendency is the Goldbach comet. I wonder if your sequences would also display bands if graphed similarly. > > David > > > --- In UnsolvedProblems@yahoogroups.com, "Mark" <mark.underwood@> wrote: > > > > > > OK this is neat. From the data I derived a formula for the number of ways the square of an odd prime n can be written as the sum of four non zero squares: > > > > f = floor((n^2 + 4n + 24)/48) > > > > So given that n is an odd prime, k is the actual count of the number of ways that four non zero squares can sum to n^2, and f is from the formula. [n,k,f] : > > > > > > [3, 0, 0] [5, 1, 1] [7, 2, 2] [11, 3, 3] [13, 5, 5] [17, 7, 7] [19, 9, 9] [23, 13, 13] [29, 20, 20] [31, 23, 23] [37, 32, 32] [41, 38, 38] [43, 42, 42] [47, 50, 50] [53, 63, 63] [59, 77, 77] [61, 83, 83] [67, 99, 99] [71, 111, 111] [73, 117, 117] [79, 137, 137] [83, 150, 150] [89, 172, 172] [97, 204, 204] [101, 221, 221] [103, 230, 230] [107, 247, 247] [109, 257, 257] [113, 275, 275] [127, 347, 347] [131, 368, 368] [137, 402, 402] [139, 414, 414] [149, 475, 475] [151, 488, 488] [157, 527, 527] [163, 567, 567] [167, 595, 595] [173, 638, 638] [179, 682, 682] [181, 698, 698] [191, 776, 776] [193, 792, 792] [197, 825, 825] [199, 842, 842] > > > > > > Is this known? I don't know... > > > > And I'm certainly not up for a proof hehe! > > > > > > Mark > > > > > > --- In UnsolvedProblems@yahoogroups.com, "Mark" <mark.underwood@> wrote: > > > > > > > > > > > > > > > Hi David, > > > > > > I put the following text into a file called david : > > > > > > for (n=1,120, k=0; for(s1=1,sqrt((n^2)/4), for(s2=s1,sqrt((n^2 - s1^2)/3) , for(s3=s2,sqrt((n^2-s1^2 - s2^2)/2), if(issquare(n^2-s1^2-s2^2-s3^2),k++)))) ; print1([n,k]" ")) > > > > > > Then at the gp pari prompt I put in > > > > > > \r david > > > > > > and out it comes. On my old computer it took about 7 seconds to cover the numbers up to 120. > > > > > > I'm surprised the sequence is not in OEIS. > > > > > > I may put it in, but I would prefer if you would since it was your essential idea after all! > > > > > > Personally I find the sequence for the primes the most interesting and may investigate it further at some later time. > > > > > > Mark > > > > > > > > > --- In UnsolvedProblems@yahoogroups.com, "zetacooking" <hartwelld@> wrote: > > > > > > > > Excellent analysis, Mark. Interesting results and a well presented report of your findings. And fast, too! > > > > > > > > I think posting your source code here would have educational value. > > > > > > > > Your sequence does not exist on OEIS. I encourage you to submit it. > > > > > > > > David > > > > > > > > --- In UnsolvedProblems@yahoogroups.com, "Mark" <mark.underwood@> wrote: > > > > > > > > > > David asked, "What can say about the squares that can be written as the sum of two or three or m positive squares? How many non-zero squares, at most, are needed, which will sum to every square?" > > > > > > > > > > I just wrote a program, and it appears that for any square greater than 3^2, four non zero squares are needed to sum to the square. > > > > > > > > > > (When only three squares are used, the square of numbers of the form 2^n and 5*2^n (and only those numbers) cannot be written as the sum of three (non zero) squares. ) > > > > > > > > > > There are some fascinating properties of how many different ways a square can be written as the sum of four non zero squares. > > > > > > > > > > 2^2 = 1^2 + 1^2 + 1^2 + 1^2; Obviously 2^2 can only be represented as the sum of four positive squares one way. > > > > > > > > > > 3^2 cannot be written as the sum of four positive squares. > > > > > > > > > > 4^2 = 2^2 + 2^2 + 2^2 + 2^2. Only one way. > > > > > > > > > > 5^2 = 1^2 + 2^2 + 2^2 + 4^2. Only one way. > > > > > > > > > > 6^2 = 1^2 + 1^2 + 3^2 + 5^2 = 3^2 + 3^2 + 3^2 + 3^2 . Two ways. > > > > > > > > > > And so on. > > > > > > > > > > Let n^2 be written as the sum of four positive squares k ways. Then here's data for n up to 120, in the form [n,k] : > > > > > > > > > > [1, 0] [2, 1] [3, 0] [4, 1] [5, 1] [6, 2] [7, 2] [8, 1] [9, 2] [10, 5] [11, 3] [12, 2] [13, 5] [14, 8] [15, 9] [16, 1] [17, 7] [18, 10] [19, 9] [20, 5] [21, 16] [22, 12] [23, 13] [24, 2] [25, 19] [26, 18] [27, 22] [28, 8] [29, 20] [30, 32] [31, 23] [32, 1] [33, 35] [34, 25] [35, 42] [36, 10] [37, 32] [38, 31] [39, 51] [40, 5] [41, 38] [42, 55] [43, 42] [44, 12] [45, 80] [46, 43] [47, 50] [48, 2] [49, 63] [50, 62] [51, 83] [52, 18] [53, 63] [54, 75] [55, 91] [56, 8] [57, 103] [58, 65] [59, 77] [60, 32] [61, 83] [62, 74] [63, 144] [64, 1] [65, 127] [66, 116] [67, 99] [68, 25] [69, 151] [70, 133] [71, 111] [72, 10] [73, 117] [74, 102] [75, 217] [76, 31] [77, 163] [78, 163] [79, 137] [80, 5] [81, 204] [82, 121] [83, 150] [84, 55] [85, 207] [86, 133] [87, 238] [88, 12] [89, 172] [90, 253] [91, 228] [92, 43] [93, 271] [94, 157] [95, 256] [96, 2] [97, 204] [98, 198] [99, 332] [100, 62] [101, 221] [102, 263] [103, 230] [104, 18] [105, 488] [106, 197] [107, 247] [108, 75] [109, 257] [110, 283] [111, 384] [112, 8] [113, 275] [114, 325] [115, 371] [116, 65] [117, 464] [118, 240] [119, 374] [120, 32] > > > > > > > > > > Notice that when n is a power of two then its square can be expressed as the sum of four non zero squares only one way. > > > > > > > > > > Also notice that when n is even then 2n, 4n, 8n, 16n, ect will have the same k value as n. > > > > > > > > > > Also notice that if n is a prime and increases, then k increases. Will this always hold? Don't know. (Interestingly it doesn't hold when it is expressed as the sum of three squares!) > > > > > > > > > > Mark > > > > > > > > > > > > > > > > > > > > > > > > > --- In UnsolvedProblems@yahoogroups.com, "zetacooking" <hartwelld@> wrote: > > > > > > > > > > > > Hi Sergey, > > > > > > > > > > > > There are two related theorems on this: > > > > > > > > > > > > Every number (which includes all squares) can be written as the sum of four squares. This is Lagrange's four square theorem. > > > > > > > > > > > > Legendre's three square theorem states that every number can be written as the sum of three squares, except for those of the form (4^n)(8k-1). > > > > > > > > > > > > Now 4^n will be even for all n and 8k-1 will be odd for all k so that in order for this form to be a square we need 4^n=e^2 and 8k-1=o^2. Solutions are easy to find for e^2, such as n=(1,2,4...). Checking the form required for the odd square, we have > > > > > > > > > > > > o^2 = (2m-1)^2 = 4m^2-4m+1 > > > > > > > > > > > > So we need > > > > > > > > > > > > 8k-1 = 4m^2-4m+1 > > > > > > > > > > > > 8k = 4m^2-4m+2 > > > > > > > > > > > > 4k = 2m^2-2m+1 > > > > > > > > > > > > 4k = 2(m^2-m)+1 > > > > > > > > > > > > which is impossible because the left is always even and the right is always odd. Since the excluded numbers cannot be square, every square can be written as the sum of three square numbers. > > > > > > > > > > > > This does not imply that every square is a potential space diagonal for a perfect cuboid, since the theorem allows for using 0^2, giving a^2+0^2+0^2 = a^2. Every square can even be written as the sum of one square but this theorem is so trivial, who would want their name attached to it? :) > > > > > > > > > > > > Zero is an invalid number for an n-tuple. For Pythagorean relationships, we need the additional constraint that the squares being summed are above zero. > > > > > > > > > > > > What can say about the squares that can be written as the sum of two or three or m positive squares? How many non-zero squares, at most, are needed, which will sum to every square? > > > > > > > > > > > > David > > > > > > > > > > > > > > > > > > --- In UnsolvedProblems@yahoogroups.com, "sergey_beliy@" <sergey_beliy@> wrote: > > > > > > > > > > > > > > Hi.May be it some sort of intresting. > > > > > > > > > > > > > > Among a^2+b^2+c^2+d^2=e^2's i knew only 1^2+1^2+1^2+1^2=2^2. > > > > > > > > > > > > > > Here are some more. > > > > > > > > > > > > > > We are looking for e's. > > > > > > > > > > > > > > We have odd squares 9,25,49...... > > > > > > > > > > > > > > 1. 9 is the 5th odd number: (9+1)/2=5 > > > > > > > 2. in its turn 5 is the 3rd odd number: (5+1)/2=3 > > > > > > > 3. in its turn 3 is the 2nd odd number: (3+1)/2=2 > > > > > > > > > > > > > > or we can do it in 2 steps > > > > > > > > > > > > > > 1. (9+3)/4=3 > > > > > > > 2. (3+1)/2=2 > > > > > > > > > > > > > > The same goes whith 25 and 49. > > > > > > > > > > > > > > So we have 2,4,7.... > > > > > > > > > > > > > > 1^2+1^2+1^2+1^2=2^2 which is not a surpise, since 4 times any square is an even square 4*(1^2)=2^2 > > > > > > > > > > > > > > 2^2+2^2+2^2+2^2=4^2 or 4*2^2=4^2 > > > > > > > > > > > > > > 4^2+4^2+4^2+1^2=7^2 > > > > > > > > > > > > > > May be we can find the rest. > > > > > > > > > > > > > > As to "six"tuples, like 5^2+5^2+5^2+3^2+4^2=10^2, it does not work. > > > > > > > > > > > > > > Sergey. > > > > > > > > > > > > > > > > > > > > > > > > > > > > =============================================== sergey_beliy@ymail.com Message 9 of 9 Sep 7, 2012 ----------------------------------------------- --- In UnsolvedProblems@yahoogroups.com, "zetacooking" <hartwelld@...> wrote: > > Hi Sergey, > > There are two related theorems on this: > > Every number (which includes all squares) can be written as the sum of four squares. This is Lagrange's four square theorem. > > Legendre's three square theorem states that every number can be written as the sum of three squares, except for those of the form (4^n)(8k-1). > > Now 4^n will be even for all n and 8k-1 will be odd for all k so that in order for this form to be a square we need 4^n=e^2 and 8k-1=o^2. Solutions are easy to find for e^2, such as n=(1,2,4...). Checking the form required for the odd square, we have > > o^2 = (2m-1)^2 = 4m^2-4m+1 > > So we need > > 8k-1 = 4m^2-4m+1 > > 8k = 4m^2-4m+2 > > 4k = 2m^2-2m+1 > > 4k = 2(m^2-m)+1 > > which is impossible because the left is always even and the right is always odd. Since the excluded numbers cannot be square, every square can be written as the sum of three square numbers. > > This does not imply that every square is a potential space diagonal for a perfect cuboid, since the theorem allows for using 0^2, giving a^2+0^2+0^2 = a^2. Every square can even be written as the sum of one square but this theorem is so trivial, who would want their name attached to it? :) > > Zero is an invalid number for an n-tuple. For Pythagorean relationships, we need the additional constraint that the squares being summed are above zero. > > What can say about the squares that can be written as the sum of two or three or m positive squares? How many non-zero squares, at most, are needed, which will sum to every square? > > David > > > --- In UnsolvedProblems@yahoogroups.com, "sergey_beliy@" <sergey_beliy@> wrote: > > > > Hi.May be it some sort of intresting. > > > > Among a^2+b^2+c^2+d^2=e^2's i knew only 1^2+1^2+1^2+1^2=2^2. > > > > Here are some more. > > > > We are looking for e's. > > > > We have odd squares 9,25,49...... > > > > 1. 9 is the 5th odd number: (9+1)/2=5 > > 2. in its turn 5 is the 3rd odd number: (5+1)/2=3 > > 3. in its turn 3 is the 2nd odd number: (3+1)/2=2 > > > > or we can do it in 2 steps > > > > 1. (9+3)/4=3 > > 2. (3+1)/2=2 > > > > The same goes whith 25 and 49. > > > > So we have 2,4,7.... > > > > 1^2+1^2+1^2+1^2=2^2 which is not a surpise, since 4 times any square is an even square 4*(1^2)=2^2 > > > > 2^2+2^2+2^2+2^2=4^2 or 4*2^2=4^2 > > > > 4^2+4^2+4^2+1^2=7^2 > > > > May be we can find the rest. > > > > As to "six"tuples, like 5^2+5^2+5^2+3^2+4^2=10^2, it does not work. > > > > Sergey. > > > Hi David . Thanks. I did not know it, but anyway i think part "may be we can find the rest" still holds (could be used). Sergey. =============================================== Cached by Georg Fischer at Nov 14 2019 12:47 with clean_yahoo.pl V1.4