# Triangular square number

A triangular square number is a number which is both a triangular number and a perfect square. There is an infinity of triangular squares, given by the formula

[itex] N_k = {1 \over 32} \left( \left( 1 + \sqrt{2} \right)^{2k} - \left( 1 - \sqrt{2} \right)^{2k} \right)^2 . [itex]

The problem of finding triangular square numbers reduces to Pell's equation in the following way. Every triangular number is of the form n(n − 1)/2. Therefore we seek integers n, m such that

[itex]n(n+1)/2 = m^2.[itex]

With a bit of algebra this becomes

[itex](2n+1)^2=8m^2+1,[itex]

and then letting k = 2n + 1 and h = 2m, we get the Diophantine equation

[itex]k^2=2h^2+1[itex]

which is an instance of Pell's equation.

The kth triangular square Nk is equal to the sth perfect square and the tth triangular number, such that

[itex] s(N) = \sqrt{N}, [itex]
[itex] t(N) = \lfloor \sqrt{2 N} \rfloor. [itex]

t is given by the formula

[itex] t(N_k) = {1 \over 4} \left[ \left( \left( 1 + \sqrt{2} \right)^k + \left( 1 - \sqrt{2} \right)^k \right)^2 - \left( 1 + (-1)^k \right)^2 \right] [itex].

As k becomes larger, the ratio t/s approaches the square root of two:

[itex] \begin{matrix} N=1 & s=1 & t=1 & t/s=1 \\ N=36 & s=6 & t=8 & t/s = 1.3333333 \\ N=1225 & s=35 & t=49 & t/s = 1.4 \\ N=41616 & s=204 & t=288 & t/s = 1.4117647 \\ N=1,413,721 & s=1189 & t=1681 & t/s = 1.4137931 \\ N=48,024,900 & s=6930 & t=9800 & t/s = 1.4141414 \\ N=1,631,432,881 & s=40391 & t=57121 & t/s = 1.4142011 \end{matrix} [itex]

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