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.

Let $3n+1=K^2$ and $4n+1=P^2$, where $P$ and $K$ are integers. Therefore, we know that $4(K^2)-3(P^2)=1$. Rearranging terms gives us $K^2= \frac {1+3P^2}4$. By testing values and guesswork we know that $P=13$ and $K=15$. Therefore $3n+1=169$, and then solving for $n$ gives us $n=56$.
