Square triangular number
In mathematics, a square triangular number (or triangular square number) is a number which is both a triangular number and a perfect square. There are an infinite number of square triangular numbers; the first few are 0, 1, 36, 1225, 41616, 1413721, 48024900, 1631432881, 55420693056, 1882672131025 (sequence A001110 in OEIS).
Write Nk for the kth square triangular number, and write sk and tk for the sides of the corresponding square and triangle, so that
We define the triangular root of a triangular number to be . From this definition and the quadratic formula, we get Therefore, is triangular if and only if is square, and naturally is square and triangular if and only if is square, i. e., there are numbers and such that . This is an instance of the Pell equation, with n=8. All Pell equations have the trivial solution (1,0), for any n; this solution is called the zeroth, and indexed as . If we let denote the k'th non-trivial solution to any Pell equation for a particular n, it can be shown by the method of descent that and . Hence there are an infinity of solutions to any Pell equation for which there is one non-trivial one, which holds whenever n is not a square. The first non-trivial solution when n=8 is easy to find: it is (3,1). A solution to the Pell equation for n=8 yields a square triangular number and its square and triangular roots as follows: and Hence, the first square triangular number, derived from (3,1), is 1 (how exciting!), and the next, derived from (17,6) (=6×(3,1)-(1,0)), is 36.
Other equivalent formulas (obtained by expanding this formula) that may be convenient include
The corresponding explicit formulas for sk and tk are :13
With a bit of algebra this becomes
and then letting x = 2t + 1 and y = 2s, we get the Diophantine equation
and therefore all solutions are given by
There are many identities about the Pell numbers, and these translate into identities about the square triangular numbers.
A. V. Sylwester gave a short proof that there are an infinity of square triangular numbers, to wit:
If the triangular number n(n+1)/2 is square, then so is the larger triangular number
We know this result has to be a square, because it is a product of three squares: 2^2 (by the exponent), (n(n+1))/2 (the n'th triangular number, by proof assumption), and the (2n+1)^2 (by the exponent). The product of any numbers that are squares is naturally going to result in another square, which can best be proven by geometrically visualizing the multiplication as the multiplying of a NxN box by an MxM box, which is done by placing one MxM box inside each cell of the NxN box, naturally producing another square result.
The triangular roots are alternately simultaneously one less than a square and twice a square, if k is even, and simultaneously a square and one less than twice a square, if k is odd. Thus, , and In each case, the two square roots involved multiply to give and 
and In other words, the difference between two consecutive square triangular numbers is the square root of another square triangular number.
The generating function for the square triangular numbers is:
As becomes larger, the ratio approaches and the ratio of successive square triangular numbers approaches . The table below shows values of between 0 and 7.
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According to the records, it was presented to the St. Petersburg Academy on May 4, 1778.
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