Put $\mathrm{A}=\mathrm{a}-1$ and the equation becomes $\mathrm{A}(2 \mathrm{~A}+1)=\mathrm{b}(3 \mathrm{~b}+1)$. Let $\mathrm{d}$ be the greatest common divisor of A and $b$. Put $A=d x, b=d y$. Then $x(2 d x+1)=y(3 d y+1)$. Since $x$ and $y$ are coprime, $\mathrm{x}$ must divide $3 \mathrm{dy}+1$. So put $3 \mathrm{dy}+1=\mathrm{nx}$. Then $2 \mathrm{dx}+1=\mathrm{ny}$. Solving for $\mathrm{x}$ and $\mathrm{y}$ in terms of $n$ and $d$ we get $x=(n+3 d) /\left(n^{2}-6 d^{2}\right), y=(n+2 d) /\left(n^{2}-6 d^{2}\right)$.

So we would certainly be home if we could show that there were infinitely many solutions to $\mathrm{n}^{2}-6 \mathrm{~d}^{2}=1$. It is not hard to find the first few: $1^{2}-6.0^{2}=1,5^{2}-6 \cdot 2^{2}=1,49^{2}-6 \cdot 20^{2}=1$. We notice that $49^{2}=2.5^{2}-1$, so we wonder whether $\mathrm{n}=2.49^{2}-1$ might be another solution and indeed we find it gives $d=1960=2 \cdot 49.20$. This suggests we try $\left(2 n^{2}-1\right)^{2}-6(2 n d)^{2}=4 n^{4}-$ $4 n^{2}+1-24 n^{2} d^{2}=4 n^{2}\left(n^{2}-6 d^{2}-1\right)+1=1$. So there are indeed infinitely many solutions to $n^{2}$ $-6 \mathrm{~d}^{2}=1$ and we are done.