<s>[INST] If 4n+1 and 3n+1 are both perfect sqares, then 56|n. How can I prove this? 
Prove that if $n$ is a natural number and $(3n+1)$ & $(4n+1)$ are both perfect squares, then $56$ will divide $n$.

Clearly we have to show that $7$ and $8$ both will divide $n$.
I considered first $3n+1=a^2$ and $4n+1=b^2$. $4n+1$ is a odd perfect square.
 - so we have $4n+1\equiv 1\pmod{8}$; from this $2|n$ so $3n+1$ is a odd perfect square.
 - so $3n+1\equiv 1\pmod{8}$ so $8|n$ but I can't show $7|n$. How do I show this?
Thanks for the help.
 [/INST] For $7 | n$, consider adding $4n+1$ and $3n+1$.
Observe that we get $7n + 2 = x^2 + y^2$, where we can let $4n+1 = x^2$, and $3n+1 = y^2$. Now, we see that mod $7$, the only quadratic residues are $0$, $1$, $2$, $4$. The only way to get two squares summing to 2 mod 7 is either if both are congruent to $1$ mod $7$, or if one is $0$ mod $7$ and one is $2$ mod $7$.
If both are congruent to $1$ mod $7$, we have n is $0$ mod $7$, so we need to show that the other case is not possible.
We now try to multiply $4n+1$ by $3n+1$, to get $12n^2 + 7n + 1$. This mod 7 is $5n^2 + 1$. If one of the squares if $0$ mod 7, it means that the product is also 0 mod 7, meaning that $5n^2 + 1 \equiv 0 $ mod $7$. This means that $5n^2 \equiv 6 $ mod $7$, and hence that $n^2 \equiv 5 $ mod $7$, which is an impossibility due to quadratic residues mod 7.
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