Prove by construction that $x(x+1)=2 y^2$ has an infinite number of solutions where $x$ and $y$ are both positive whole numbers. Do not use an iterative procedure like that for solutions of Pell's equation in your proof.
Background: I came up with a proof of this years ago in which $x$ was a multiple of $32$ and a quadratic or quartic polynomial of an integer parameter. Given that, $x(x+1)/2$ was guaranteed to be a perfect square. I've since encountered another, probably better proof the above but efforts to find my old one have been in vain.
Proof using Pell:
1) Complete the square of the $x$ Side of the equation to get $x^2+x+1/4=2y^2+1/4$.
2) This is equal to $(x+1/2)^2=2y^2+1/4$.
3) Multiply both sides by $4$ to get: $(2x+1)^2=8y^2+1$.
4) Let $m=2x+1$ and $n=2y$. Then we have $m^2-2n^2=1$, Pell's Equation.
5) Given solution ($m,n$), construct a symmetric $2x2$ matrix with ($m,n \, \sqrt{2}$) as the first row.
6) The square of a symmetric matrix is symmetric. The determinant of the square of a matrix is the square of that matrix's determinant. We thus end up with two new integers in the first row ($m^2+2n^2, 2 \, m \, n$) which are again solutions of Pell's equation.
7) This can be iterated to generate a sequence of solutions.
8) $x$ and $y$ can be represented in terms of $m$ and $n$, so we thus have an infinite number of solutions for $x$ and $y$.
If you have one solution to $x^2-dy^2 = 1$, by using
$\begin{array}\\ (x^2-dy^2)(u^2-dv^2) &=x^2u^2-d(x^2v^2+y^2u^2)+d^2y^2v^2\\ &=x^2u^2+d^2y^2v^2\pm 2dxuyv-d(x^2v^2+y^2u^2\pm 2xuyv)\\ &=(xu\pm dyv)^2-d(xv\pm yu)^2\\ \end{array} $
you can construct an arbitrary number of them.
When $d=8$, since $3^2-8\cdot1^2 = 1$, from this first solution you can get $u^2-8v^2 =(3u\pm 8v)^2-8(3v\pm u)^2 $.
Choosing "+" gives Will Jagy's iteration.
Wow, is this not original.