Primenumbers Yahoo Groups Generalized Proof re Square Triangular Numbers =============================================== 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. >