Two squares and n

Note that $x^2 \equiv 0,1$ or $4 \pmod 5$ for all integer $x.$ In other words, squares leave the remainder $0, 1$ or $4$ when divided by $5.$

If $n \equiv 1$ or $3 \pmod 5$ , then $2n+1\equiv 3$ or $2 \pmod 5$ and $2n+1$ cannot be a perfect square as $3,2$ are not quadratic residues modulo $5$ . Similarly, if $n \equiv 2$ or $4 \pmod 5$ then $3n+1 \equiv 2$ or $3 \pmod 5$ and $3n+1$ cannot be a perfect square.

Thus, if both $2n+1$ and $3n+1$ are perfect squares, then $n \equiv 0 \pmod 5$ .

Also $2n+1\equiv 1 \pmod 8$ as odd square integers leave the remainder $1$ when divided by $8.$ It follows that $2n+1\equiv 1\pmod 8 \implies n\equiv 4\pmod 8$ or $n\equiv 0 \pmod 8$ . But, if $n\equiv 4\pmod 8 \implies 3n+1\equiv 5\pmod 8$ and thus cannot be a square as $5$ is not a quadratic residue modulo $8$ .

Thus, if both $2n+1$ and $3n+1$ are perfect squares, then $n \equiv 0 \pmod 8$ .

From the above results we conclude that $5\mid n$ and $8\mid n$ which implies $40\mid n$ . Checking $n=40$ gives $2n+1=81=9^2$ and $3n+1=121=11^2$ which are indeed perfect squares.

Therefore the minimum value of $n$ for which both $2n+1$ and $3n+1$ are perfect squares is $\boxed {n=40}$ .

For both 2n+1 and 3n+1 to be perfect squares, 2n+1 must be of the form $(8p+1)^2$ or $(8p-1)^2$ . Then n=32 $p^2$ +8p or n=32 $p^2$ -8p. With p=1, n =40 or 24. The option n=24 doesn't work. Therefore minimum of n is 40.