Primenumbers Yahoo Groups Generalized Proof re Square Triangular Numbers =============================================== ramsey2879 Message 1 of 2 May 27, 2005 ----------------------------------------------- Generalized Proof that an infinite number of square triangular numbers exist. I prove below that in general that if n^2 – T(q) = K, there are an infinite number of distinct pairs n` and q` such that (n`)^2 –T(q) = K. First a little history of the problem. It is widely recognized that in general, any odd square minus 1, divided by 8 is a triangular number, i.e. (2n +1)^2 – 1 = 8* T(n). Thus to find square triangular numbers can be reduced to solving the Pell equation 8n^2 + 1 = m^2. Solutions along this line are given at http://www.cut-the-knot.org/do_you_know/triSquare.shtml , http://mathpages.com/home/kmath159.htm , and http://mathforum.org/library/drmath/view/55927.html . It has been shown in these sites that if S_n is the series of numbers that are squared and Q_n is the corresponding series such that (S_n)^2 = T (Q_n) then S_n = 6*S_(n-1) – S_(n-2) and Q_n = 6*Q_(n-1) – Q_(n-2). However, these solutions are quite involved and are not capable of being generalized to other cases where (S_n)^2 – T(Q_n) <> 0. A related web site http://www.cut-the- knot.org/do_you_know/triSquare2.shtml takes advantage of this solution and a relationship between the two series S_n) and Q_n to give two recursive formulas for finding other values of Q_n and S_n when a single value of S_n and a corresponding value of Q_n are known. This too is restricted to the case K = 0. The purpose of this paper is to prove the infinity of solutions more general expression, including K <> 0, i.e. if there is one solution to n^2 – T(q) = K for the integers n and q, then there are an infinite number of such solutions as well as to show how the recursive formulas also work for complex as well as real numbers, and to show other related results. First I developed two algebraic identities that I got by studying the differences between squares and triangular numbers and noting the patterns. They are: 1) (a-b)^2 = T(2a-b) +T(b) – 2*T(a) 2) 2*T(a) – T(b) = 2*T(a_1) – T(b_1) where a_1 = 3a – 2b and b_1 = 3b – 4a –1. By manipulation of the above identities, generalized expressions for the recursive series S_n and Q_n can be found. The resulting initial conditions are, S_0 = a +b +1, S_1 = -a+b, Q_0 = 2*a+b+1, and Q_1 = 2*a -b (where a and b can be any number including a complex number or a radical). Regardless of the values for a and b, the resulting values of S_n and Q_n solve the equation (S_n)^2 = T_(Q_n) – K(a,b) for all n where K is a constant that is equal to 2*T_a – T_b. This is true even for complex numbers "a" and "b". Proof The reader may easily verify identities 1) and 2) by expansion and collection of like terms. Identity 2 gives the recursive relation as follows: a_0 = a b_0 = b a_1 = 3a-2b b_1 = 3b-4a-1 a_2 = 3*(a_1) – 2*(b_1) = 3*(3a-2b) – 2*(3b-4a-1) = 17a – 12b +2 = 6*a_1 – a_0 + 2 b_2 = 3*(b_1) – 4*(a_1) – 1 = 3*(3b-4a-1) –4*(3a-2b) –1 = 17b – 24a – 4 = 6*b_1 – b_0 + 2 Since the same relationships above are calculated regardless of the initial choices of a and b, a_n = 6*a_(n-1) – a_(n-2) +2 and b_n = 6*b_(n-1) – b_(n-2) +2 for all n. The corresponding series of a_n and b_n can be worked both backward and forward infinitely. Although, a single series set exists for K = 0, 1, 2 and 5, this is not so for the case of K = 6, 20, etc. For instance for the case of K = 6 there are two sets of distinct series for instance, with a_0 = 2, b_0 = 0 and the other starting with a_0 = b_0 = 3. Certain values of K, e.g. 4, 7, 8, 13, 16 etc., have no solution to K = 2*T(a) – T(b) in integers. For identity 1, we let the value of T(b) – 2*T(a) = -K. By identity 2, T(b_n) – 2 * T(a_n) = -K for all n. Since identity 1 works for all a and b we can make the substitution b_n` = -(b_n+1) without changing the value of K since T(b) = T(-b-1). Thus (a_n + b_n +1)^2 = T(2*a_n + b_n + 1) - K for all n S_n = a_n + b_n +1 S_0 = a+b+1 S_1 = -a + b S_2 = -7a + 5b –1 S_n = 6*S_(n-1) – S_(n-2) for all n > 1 Q_n = 2a_n + b_n +1 Q_0 = 2a + b +1 Q_1 = 2a – b Q_2 = 10a-7b +1 Q_n = 6*Q_(n-1) – Q_(n-2) +2 for all n > 1 Again, this works for all a and b, both real or complex. If we know S_0 and S_1, the solution of the resulting simultaneous equations gives: a = (S_0 – S_1 –1)/2 and b = (S_0 + S_1-1)/2. For instance the first two square triangular numbers, not counting 0, are 1 and 36. Thus S_0 = +/- 1 and S_1 = +/- 6, which gives two basic solutions for Q_n and K. That is K can be 0 or 9 in this case since 1 = T_1-0 and 36 = T_8-0; and since 1 = T_4 – 9 and 36 = T_9 – 9. The other two solutions simply give S_n = - S_n and T_n = T_(-n-1) etc. =============================================== ramsey2879 Message 2 of 2 Oct 10, 2011 ----------------------------------------------- In the prior post on this topic, I showed that S(n)=6S(n-1) - S(n-2) and that Q(n) = 6Q(n-1)-Q(n-2) + 2. It can further be shown that for S(1) = Sqrt(a), S(2) = Sqrt(b) that S(3) = 6S(2) - S(1) = Sqrt(34b - a + 2*[(S(1)^2) -6S(1)*S(2) + (S(2)^2)]). The term within the brackets, S(1)^2 -6S(1)*S(2) + S(2)^2, is a constant, K, for all a,b, as adjacent terms in the series S(n) = 6S(n-1)-S(n-2) --- In Triangular_and_Fibonacci_Numbers@yahoogroups.com, "ramsey2879" wrote: > > Generalized Proof that an infinite number of square triangular > numbers exist. > > I prove below that in general that if n^2 – T(q) = K, there are an > infinite number of distinct pairs n` and q` such that (n`)^2 –T(q) = > K. First a little history of the problem. > It is widely recognized that in general, any odd square minus 1, > divided by 8 is a triangular number, i.e. (2n +1)^2 – 1 = 8* T(n). > Thus to find square triangular numbers can be reduced to solving the > Pell equation 8n^2 + 1 = m^2. Solutions along this line are given > at http://www.cut-the-knot.org/do_you_know/triSquare.shtml , > http://mathpages.com/home/kmath159.htm , and > http://mathforum.org/library/drmath/view/55927.html . It has been > shown in these sites that if S_n is the series of numbers that are > squared and Q_n is the corresponding series such that (S_n)^2 = T > (Q_n) then S_n = 6*S_(n-1) – S_(n-2) and Q_n = 6*Q_(n-1) – Q_(n-2) +2. > However, these solutions are quite involved and are not capable of > being generalized to other cases where (S_n)^2 – T(Q_n) <> 0. A > related web site http://www.cut-the- > knot.org/do_you_know/triSquare2.shtml takes advantage of this > solution and a relationship between the two series S_n) and Q_n to > give two recursive formulas for finding other values of Q_n and S_n > when a single value of S_n and a corresponding value of Q_n are > known. This too is restricted to the case K = 0. The purpose of this > paper is to prove the infinity of solutions more general expression, > including K <> 0, i.e. if there is one solution to n^2 – T(q) = K for > the integers n and q, then there are an infinite number of such > solutions as well as to show how the recursive formulas also work for > complex as well as real numbers, and to show other related results. > > First I developed two algebraic identities that I got by studying the > differences between squares and triangular numbers and noting the > patterns. They are: > > 1) (a-b)^2 = T(2a-b) +T(b) – 2*T(a) > 2) 2*T(a) – T(b) = 2*T(a_1) – T(b_1) where a_1 = 3a – 2b and b_1 > = 3b – 4a –1. > Where T(b) = 2*T(a), (a-b)^2 are the square triangular numbers. > By manipulation of the above identities, generalized expressions for > the recursive series S_n and Q_n can be found. The resulting initial > conditions are, S_0 = a +b +1, S_1 = -a+b, Q_0 = 2*a+b+1, and Q_1 > = 2*a -b (where a and b can be any number including a complex number > or a radical). Regardless of the values for a and b, the resulting > values of S_n and Q_n solve the equation (S_n)^2 = T_(Q_n) – K(a,b) > for all n where K is a constant that is equal to 2*T_a – T_b. This > is true even for complex numbers "a" and "b". > > Proof > The reader may easily verify identities 1) and 2) by expansion and > collection of like terms. Identity 2 gives the recursive relation as > follows: > a_0 = a > b_0 = b > > a_1 = 3a-2b > b_1 = 3b-4a-1 > > a_2 = 3*(a_1) – 2*(b_1) > = 3*(3a-2b) – 2*(3b-4a-1) > = 17a – 12b +2 > = 6*a_1 – a_0 + 2 > > b_2 = 3*(b_1) – 4*(a_1) – 1 > = 3*(3b-4a-1) –4*(3a-2b) –1 > = 17b – 24a – 4 > = 6*b_1 – b_0 + 2 > > Since the same relationships above are calculated regardless of the > initial choices of a and b, a_n = 6*a_(n-1) – a_(n-2) +2 and b_n > = 6*b_(n-1) – b_(n-2) +2 for all n. The corresponding series of a_n > and b_n can be worked both backward and forward infinitely. > Although, a single series set exists for K = 0, 1, 2 and 5, this is > not so for the case of K = 6, 20, etc. For instance for the case of K > = 6 there are two sets of distinct series for instance, with a_0 = 2, > b_0 = 0 and the other starting with a_0 = b_0 = 3. Certain values of > K, e.g. 4, 7, 8, 13, 16 etc., have no solution to K = 2*T(a) – T(b) > in integers. > > For identity 1, we let the value of T(b) – 2*T(a) = -K. By identity > 2, T(b_n) – 2 * T(a_n) = -K for all n. Since identity 1 works for > all a and b we can make the substitution b_n` = -(b_n+1) > without changing the value of K since T(b) = T(-b-1). Thus > > (a_n + b_n +1)^2 = T(2*a_n + b_n + 1) - K for all n > > S_n = a_n + b_n +1 > S_0 = a+b+1 > S_1 = -a + b > S_2 = -7a + 5b –1 > S_n = 6*S_(n-1) – S_(n-2) for all n > 1 > > Q_n = 2a_n + b_n +1 > Q_0 = 2a + b +1 > Q_1 = 2a – b > Q_2 = 10a-7b +1 > Q_n = 6*Q_(n-1) – Q_(n-2) +2 for all n > 1 > > Again, this works for all a and b, both real or complex. If we know > S_0 and S_1, the solution of the resulting simultaneous equations > gives: > a = (S_0 – S_1 –1)/2 and b = (S_0 + S_1-1)/2. For instance the first > two square triangular numbers, not counting 0, are 1 and 36. Thus > S_0 = +/- 1 and S_1 = +/- 6, which gives two basic solutions for Q_n > and K. That is K can be 0 or 9 in this case since 1 = T_1-0 and 36 = > T_8-0; and since 1 = T_4 – 9 and 36 = T_9 – 9. The other two > solutions simply give S_n = - S_n and T_n = T_(-n-1) etc. >