Can the numbers $n+1$ , $2n+1$ and $3n+1$ be simultaneously perfect squares for any positive integer $n$ ?
I tried to find that out and arrived at the equation system $$c^2-3a^2=-2$$ $$b^2-2a^2=-1$$ by setting $$n+1=a^2$$ $$2n+1=b^2$$ $$3n+1=c^2$$ and I conjecture that the only solution in positive integers is $c=a=b=1$ , corresponding to $n=0$. But how can I prove this conjecture ? Or did I miss a solution ?
If this happens, then $1, n+1, 2n+1, 3n+1$ is a four-term arithmetic progression of perfect squares.
But it can be shown (painfully, using elliptic curves, as demonstrated at this link - see Theorem 3.4) that all four-term arithmetic progressions of rational perfect squares must be constant, so we have $n=0$.