1. ## prove

if 3n+1 and 4n+1 are perfect squares then prove that n is divisible by 56.

2. Let

$\displaystyle 3n+1\ =\ a^2$

$\displaystyle 4n+1\ =\ b^2$

The second equation implies $\displaystyle b$ is odd. Hence $\displaystyle n$ is even (if $\displaystyle n$ were odd, then $\displaystyle 4n+1\equiv5\,(\bmod\,8)$ which is not possible because perfect squares cannot be congruent to 5 modulo 8). So $\displaystyle a$ is odd.

Subtracting the equations gives $\displaystyle n\,=\,b^2-a^2\,=\,(b-a)(b+a).$ Since $\displaystyle b-a$ and $\displaystyle b+a$ are both even, $\displaystyle n$ is divisible by 4. In fact, one of $\displaystyle b-a$ and $\displaystyle b+a$ must be divisible by 4. For, suppose $\displaystyle b-a$ is not divisible by 4. Then $\displaystyle b-a\equiv2\,(\bmod\,4)$ and $\displaystyle 2a\equiv2\,(\bmod\,4)$ (since $\displaystyle a$ is odd) and so $\displaystyle b+a=(b-a)+2a\equiv0\,(\bmod\,4).$ Hence $\displaystyle n$ is divisible by 8.

It remains to show that $\displaystyle n$ is divisible by 7. There may be a simple way but the only way I can do it is by contradiction. Now, by checking $\displaystyle 1^2,\,2^2,\,\ldots,\,6^2$ in turn, observe that a perfect square cannot be congruent to 5 or 6 modulo 7. If $\displaystyle n\equiv1,3,4,6\,(\bmod\,7),$ then one of $\displaystyle 3n+1$ and $\displaystyle 4n+1$ would be congruent to 5 or 6 mod 7, so we can eliminate these. We are left with $\displaystyle n\equiv\pm2\,(\bmod\,7).$ In this case, one of $\displaystyle 3n+1$ and $\displaystyle 4n+1$ is divisible by 7, and so their product is divisible by 7 and must be a perfect square. Hence

$\displaystyle 12n^2+7n+1-m^2\ =\ 0$

for some integer $\displaystyle m$ divisible by 7. The discriminant of this quadratic in $\displaystyle n$ must be a perfect square; this is $\displaystyle 49-48(1-m^2)=48m^2-1.$ But if $\displaystyle m$ is divisible by 7, then $\displaystyle 48m^2-1\equiv6\,(\bmod\,7)$ which is impossible. Thus $\displaystyle n$ must be divisible by 7.