<s>[INST] Existence proof of infinite solutions of a diophantine equation 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 $y$. 
 [/INST] 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.
</s>