The triangular numbers are 1, 3, 6, 10, 15, 21, 28, 36, 45, ... and so on, given by the formula n(n+1)/2. Notice that the numbers 1 and 36 on this list are perfect squares as well as triangular. A standard problem in elementary number theory is to determine ALL the numbers that are both square and triangular. Thus we want all the solutions of m^2 = n(n+1)/2. Solving this for n using the quadratic formula gives __________ -1 + / 1 + 8m^2 n = ---------------- 2 Obviously this equation will give an integer value of n if and only if the quantity inside the square root is a square, so there must be an integer q such that q^2 - 8 m^2 = 1 Equations of this form are called Pell's equation, and there are infinitely many pairs of integers (q,m) that satisfy this equation. The usual method of determining these solutions is via the continued fraction for sqrt(8) 1 sqrt(8) = 2 + ------------------------ 1 1 + ---------------- 1 4 + ------------ 1 1 + ------- etc. The period of this continued fraction is 2, so every 2nd convergent is q/m where (q,m) satisfy the Pell equation. It follows that q and m satisfy the recurrences q[k] = 6 q[k-1] - q[k-2] m[k] = 6 m[k-1] - m[k-2] where q[k],m[k] denotes the kth solution. By the theory of difference equations we know that q[k] and m[k] must both be of the form A r^k + B s^k where A and B are constants and r and s are the roots of the characteristic equation x^2 - 6 x + 1 = 0 Thus we have r = 3 + 2 sqrt(2) and s = 3 - 2 sqrt(2). For q(k) we see from the initial values that A = B = 1/2, so (3+2sqrt(2))^k + (3-2sqrt(2))^k q[k] = ----------------------------------- 2 and of course the kth value of n is q[k] - 1 (3+2sqrt(2))^k + (3-2sqrt(2))^k - 2 n[k] = ---------- = ------------------------------------ 2 4 where n[k] is the kth solution of our original square-triangular equation n(n+1)/2 = m^2. The first few values of n[k] for k=1,2,.. are 1, 8, 49, 288, 1681, ... Letting u and U denote 3+2sqrt(2) and 3-2sqrt(2) respectively, we can substitute the expression for n[k] into the expression for m^2 to give the formula for the kth square triangular number 1 1 / u^k + U^k - 2 \ / u^k + U^k + 2 \ m^2 = - n (n+1) = - ( -------------- )( -------------- ) 2 2 \ 4 / \ 4 / _ _ 1 | 2 | = ---- | (u^k + U^k) - 4 | 32 |_ _| Expanding the square, and noting that u^2 = 17+12sqrt(2), U^2 = 17-12sqrt(2), and 2uU = 2, we have _ _ n(n+1) 1 | k k | m^2 = ------ = ---- | [17+12sqrt(2)] + [17-12sqrt(2)] - 2 | 2 32 |_ _| Incidentally, the numbers n[k] alternate odd and even, because they safisfy the recurrence n[k] = 6n[k-1] - n[k-2] + 2. Furthermore, from the relation (n)(n+1) = 2m^2 and the fact that n and n+1 are coprime (i.e., have no common factor), it follows that either n or n+1 must be an odd square and the other must be twice a square, accordingly as n is odd or even. So, we've shown that n[k] is a square for every odd index k=2j-1, and of course the quantity 4n is a square if and only if n is a square, so we have the related result that the number 4 n[2j-1] = (3+2sqrt(2))^(2j-1) + (3-2sqrt(2))^(2j-1) - 2 is a square for every positive integer j.

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