# Can $n+1$ , $2n+1$ , $3n+1$ all be perfect squares , if $n$ is a positive integer?

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 ?

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

## 1 Answer

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 $n=0$.

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