MATH for FUN - A place to post and work math problems. Yahoo Group Re: [MATH for FUN] Pythagorean Theorem and prime factors. =============================================== MorphemeAddict@wmconnect.com Message 1 of 9 Feb 20, 2009 ----------------------------------------------- In a message dated 2009-02-19 23:17:07 Eastern Standard Time, no_reply@yahoogroups.com writes: > If you take the prime factors of a^2, b^2, and c^2, the sum of > the different prime factors of a^2 and the different prime factors of > b^2 is equal to the sum of the different prime factors of c^2. > Examples: > 3^2+4^2=5^2, 3^1+2^2=5^2, 3+2=5. I don't see where you get 3^1+2^2=5^2, since 3^1+2^2=5^1. > > 6^2+8^2=10^2, 2*3+2^6=2^2*5^2, 2+3+2=5+2 > > 12^2+16^2=20^2, 2^4*3^2+2^8=2^4*5^2, 2+3+2=5+2 > > Just leave out "^n" as the powers of the primes and add the primes > together. Is there a case where this doesn't work? Is this already > known and what's it called? > > Jason > The latter two examples are simply multiples of the first two, so I don't think they're very good examples. I haven't figured out what you're doing, but does it work with, e.g., 5^2+12^2=13^2? stevo =============================================== Edward Moore Message 2 of 9 Feb 20, 2009 ----------------------------------------------- You might want to try some triangles and triples that aren't similar to the 3:4:5 triangle before you get to excited - it doesn't seem to work for 5-12-13. Ed ________________________________ From: bqllpd To: mathforfun@yahoogroups.com Sent: Thursday, February 19, 2009 11:16:49 PM Subject: [MATH for FUN] Pythagorean Theorem and prime factors. This is something I found by accident last night when messing with my HP calc. We all know a^2+b^2=c^2 has positive integer solutions for a, b and c. If you take the prime factors of a^2, b^2, and c^2, the sum of the different prime factors of a^2 and the different prime factors of b^2 is equal to the sum of the different prime factors of c^2. Examples: 3^2+4^2=5^2, 3^1+2^2=5^2, 3+2=5. 6^2+8^2=10^2, 2*3+2^6=2^2* 5^2, 2+3+2=5+2 12^2+16^2=20^ 2, 2^4*3^2+2^8= 2^4*5^2, 2+3+2=5+2 Just leave out "^n" as the powers of the primes and add the primes together. Is there a case where this doesn't work? Is this already known and what's it called? Jason =============================================== cino hilliard Message 3 of 9 Feb 20, 2009 ----------------------------------------------- 5^2 + 12^2 = 13^2 To: mathforfun@yahoogroups.com From: no_reply@yahoogroups.com Date: Fri, 20 Feb 2009 04:16:49 +0000 Subject: [MATH for FUN] Pythagorean Theorem and prime factors. =============================================== slim_the_dude Message 4 of 9 Feb 20, 2009 ----------------------------------------------- If I'm understanding you correctly, would this be a couterexample? If I'm understanding you correctly, would this be a couterexample? 12² + 5² = 13² prime factors of 12 are 2 and 3 prime factors of 5 are 5 preim factors of 13 are 13, but 2+3+5 <> 13 =============================================== slim_the_dude Message 5 of 9 Feb 20, 2009 ----------------------------------------------- Sorry; I somehow did not see all the replys before I posted this. =============================================== Cino Hilliard Message 6 of 9 Feb 20, 2009 ----------------------------------------------- Me too. Nevertheless, the idea is interesting. I took the liberty of writing a program in Pari to examine these numbers. Some output is below. You will notice that I used a=2mu,b=m^2-u^2 and c=m^2+u^2 to generate the list. I also took the liberty of submitting a sequence with reference to the origional post. Notice there is only 1 prime value for c. Also there are other forms besides multiples of 3,4,5 triples such as 360,319,481. pythsum(n) = { local(a,b,cm,u,s); for(m=1,n+n, for(u=1,n, a=2*m*u; b=m^2-u^2; c=m^2+u^2; fa=ifactord(a); fb=ifactord(b); fc=ifactord(c); s=0; s2=0; for(a1=1,length(fa),s+=fa[a1]); for(b1=1,length(fb),s+=fb[b1]); for(c1=1,length(fc),s2+=fc[c1]); if(s==s2&&b>0, /* print(a","b","c"," fa" + "fb" = "fc); */ if(a pythsum(20) 4^2+3^2=5^2,[2] + [3] = [5] 6^2+8^2=10^2,[2, 3] + [2] = [2, 5] 16^2+12^2=20^2,[2] + [2, 3] = [2, 5] 24^2+32^2=40^2,[2, 3] + [2] = [2, 5] 36^2+27^2=45^2,[2, 3] + [3] = [3, 5] 64^2+48^2=80^2,[2] + [2, 3] = [2, 5] 54^2+72^2=90^2,[2, 3] + [2, 3] = [2, 3, 5] 96^2+128^2=160^2,[2, 3] + [2] = [2, 5] 144^2+108^2=180^2,[2, 3] + [2, 3] = [2, 3, 5] 256^2+192^2=320^2,[2] + [2, 3] = [2, 5] 216^2+288^2=360^2,[2, 3] + [2, 3] = [2, 3, 5] 324^2+243^2=405^2,[2, 3] + [3] = [3, 5] 360^2+319^2=481^2,[2, 3, 5] + [11, 29] = [13, 37] 384^2+512^2=640^2,[2, 3] + [2] = [2, 5] 576^2+432^2=720^2,[2, 3] + [2, 3] = [2, 3, 5] 486^2+648^2=810^2,[2, 3] + [2, 3] = [2, 3, 5] 392^2+735^2=833^2,[2, 7] + [3, 5, 7] = [7, 17] 638^2+720^2=962^2,[2, 11, 29] + [2, 3, 5] = [2, 13, 37] 60^2+899^2=901^2,[2, 3, 5] + [29, 31] = [17, 53] 1024^2+768^2=1280^2,[2] + [2, 3] = [2, 5] 864^2+1152^2=1440^2,[2, 3] + [2, 3] = [2, 3, 5] 1296^2+972^2=1620^2,[2, 3] + [2, 3] = [2, 3, 5] 1440^2+1276^2=1924^2,[2, 3, 5] + [2, 11, 29] = [2, 13, 37] Enjoy, Cino =============================================== zelkomir Message 7 of 9 Feb 20, 2009 ----------------------------------------------- Lets make some Pythagoreans tercets: (m^2+n^2)^2 - (m^2-n^2)^2 = (2mn)^2 n=1 m=2 (m^2+n^2) = 5 (m^2-n^2) = 3 {3 + 2 = 5} //Lets mark this tercet as tercet "A" 2mn = 4 ----------------------------------------------- n=1 m=4 (m^2+n^2) = 17 (m^2-n^2) = 15 {2 + 3 + 5 != 17} 2mn = 8 ------------------------------------------------- n=2 m=3 (m^2+n^2) = 13 {2 + 3 + 5 !=13} (m^2-n^2) = 5 2mn = 12 ------------------------------------------------- n=2 m=5 (m^2+n^2) = 29 (m^2-n^2) = 21 {2 + 5 + 3 + 7 != 29} 2mn = 20 ------------------------------------------------- As we can see your suppose is true in one from this four cases. Why? Your suppose is true if: the sum of the different prime factors of (m^2+n^2) is equal to the sum of the different prime factors of (2mn + (m^2-n^2)). You'r right, as my knowledge now, only for the tercets which are multipliers of tercet A. 6,8,10 9,12,15 {3 + (3 + 2) = 3 + 5} 12,16,20 {(2+3) + 2 = 2 + 5} etc. etc. =============================================== PhilM4@aol.com Message 8 of 9 Feb 20, 2009 ----------------------------------------------- Quick summary of the posts to which I am responding: The first post posited that in integral Pythagorean triples, the sum of the prime factors of the legs equaled the sum of the prime factors of the hypotenuse. Notably, in a 3-4-5 triangle, 3 + 2 = 5, and in a 6-8-10 triangle, (2 + 3) + 2 = (2 + 5). Subsequent posters have noted that there are exceptions, but another posted seemed to suggest that the rule does work for all 3-4-5 triangles. This is the point to which I object. 3-4-5 triangles do NOT always work. If the lengths are multiplied by, say, 7, we arrive at a 21-28-35 triangle. The sums of the primes are then (3 + 7) + (2 + 7) =19 for the legs and (5 + 7) = 12 for the hypotenuse. Nor is this result at all a surprise. The extension of 3-4-5 to 6-8-10 involves multiplying by 2 or, in essence, adding a factor two to the 3 and 5 sides. This balances out. The same is true when 3 and 4 are the multipliers (i.e., 9-12-15 and 12-16-20). For that matter, the distinct prime factors add up for multipliers like 8, 16, 32, 2^n and 9, 27, 81, and 3^n (among others). =============================================== cino hilliard Message 9 of 9 Feb 21, 2009 ----------------------------------------------- Indeed, let the triple a,b,c have prime divisors without multiplicity, a1,a2..al,b1,b2..bm,c1,c2..cn. Now in Jason's query, a^2+b^2=c^2 breaks down to a1+a2...+al + b1+b2...+bm = c1+c2+...+cn. (1) If we multiply a,b,c by any combination of factor in either a or b, we have 2 possibilities. a1+a2..+al + b1+b2..bm + a1+a2+..ai = c1+c2+...+ a1+a2+..ai a1+a2..+al + b1+b2..bm + b1+b2+..bj = c1+c2+...+ b1+b2+..bj for i <=l and j<=m. So essentially we are adding the same thing to both sides which means the new factors will add up to the same thing. Now if we multiply the triple by a number that contains a prime factor p not in the set {a1,a2..,al,b1,b2..,bm}, then we will be adding 2p to the left side and p to the right. This is only true for p = 0 which is not a prime factor. Have fun, Cino =============================================== Cached by Georg Fischer at Nov 14 2019 12:46 with clean_yahoo.pl V1.4