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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 ?

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  • $\begingroup$ related question $\endgroup$ – lulu Jun 8 '18 at 13:30
  • $\begingroup$ @lulu Is there a general way to find the solutions of such systems (in this case $n+1=a^2$ , $2n+1=b^2$ , $3n+1=c^2$) ? $\endgroup$ – Peter Jun 8 '18 at 13:32
  • $\begingroup$ Only method I know is to change this into a Pell problem. $\endgroup$ – lulu Jun 8 '18 at 13:50
  • $\begingroup$ Consideration of quadratic residues modulo $3$ and $4$ gives that $n$ must be a multiple of $12$. $\endgroup$ – Connor Harris Jun 8 '18 at 13:50
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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 $n=0$.

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    $\begingroup$ It can also be shown without using any maths as advanced as elliptic curves, as in the sources linked to from this thread. $\endgroup$ – Rosie F Jun 8 '18 at 16:42

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