Perfect problem

We have integers $u,v$ such that $2x+1=u^2$ and $3x+1=v^2$ . But then $u^2$ is odd, so $u$ is odd. Thus $u^2 \equiv 1 \pmod{8}$ , and hence $x$ is even, so that $v^2$ is odd, and hence $v$ is odd. Thus $v^2 \equiv 1 \pmod{8}$ , and so $x = v^2 - u^2$ is a multiple of $8$ .

The following table of congruences modulo $5$ $\begin{array}{c|c|c|c} x & 2x+1 & 3x+1 & x^2 \\ \hline 0 & 1 & 1 & 0 \\ 1 & 3 & 4 & 1 \\ 2 & 0 & 2 & 4 \\ 3 & 2 & 0 & 4 \\ 4 & 4 & 3 & 1 \end{array}$ shows that $2x+1$ and $3x+1$ can only both be squares (and hence both congruent to $0,1$ or $4$ modulo $5$ ) if $5$ divides $x$ . Thus we deduce that $40$ must divide $x$ .

Since $x=40$ gives $2x+1 = 9^2$ and $3x+1 = 11^2$ , we see that $\boxed{40}$ is the largest possible value of $y$ .

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it is probably worth noting that there are an infinite number of such integers $x$ . There exists integers $x,u,v$ such that $2x+1 = u^2$ and $3x+1=v$ if and only if there exists a pair of integers $u,v$ such that $3u^2 - 2v^2 = 1$ . If $(u,v)$ is a solution of this second equation, then so is $(5u+4v,6u+5v)$ . Since $(1,1)$ is a solution, we obtain an infinite number of solutions.