Prove that if $2n+1$ and $3n+1$ are both perfect squares then $40|n$. 
Prove that if $2n+1$ and $3n+1$ are both perfect squares then $40|n$.

First, I took
$$2n+1 \equiv x^2 \equiv 0, 1 \pmod 4$$
which showed that $n$ was even.
Now, 
$$3n + 1 \equiv y^2 \equiv 0, 1, 4 \pmod 8$$ But since $n$ is even, we get that $8|n$.
So, now any square $\equiv 0, 1, 4, 5, 6, 9 \pmod{10}$. So, I tested $2n+1$ and $3n+1$ for all numbers from 0 to 9. For only two, 0 and 5, were both of them ending with the residues above mentioned. So, I finally proved that $5|n$.
So, $40|n$.
Is my proof correct?
Also, my proof is too roundabout and lengthy. I had to write many programs to take different modulos. Can anyone suggest a more elegant proof especially for the second part when I have to show that $5|n$? 
Thanks.

Your proof is good.
I suppose you could be more direct by pointing out:
For any natural $m$, $m^2 \equiv 0,1,4 \mod 8$ and $m^2 \equiv 0,1,-1 \mod 5$.
So $2n+1$ and $3n+1$ being perfect squares implies:
a) $2n \equiv -1, 0 ,3 \mod 8$.
As $2n$ is even $2n \equiv -1, 3 \mod 8$ are impossible so $2n \equiv 0 \mod 8$ so $n \equiv 0, 4 \mod 8$.
b) $3n \equiv -1, 0, 3 \mod 8$.
As $n \equiv 0, 4 \mod 8$, $3n \equiv 0, 4 \mod 8$ (respectively).  As $3n \equiv 4 \mod 8$ isn't possible $n \equiv 4 \mod 8$ isn't possible so $n \equiv 0 \mod 8.$
c) $2n \equiv -1, 0, -2 \mod 5 \equiv 4, 0, -1 \mod 5$.
So $n \equiv 2, 0, -1 \mod 5$
d) $3n \equiv -1, 0, -2 \mod 5 \equiv -6, 0, 3 \mod 5$
So $n \equiv -2, 0, 1 \mod 5$
The only compatibility between c) and d) is $n \equiv 0 \mod 5$
So we have $8|n$ and $5|n$ so $40|n$.
