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.

$3n+1=x^2$
$4n+1=y^2$
$4y^2-3x^2=1$
Put $z=2y $
$z^2-3x^2=1$
One of solution is: $(z_0,x_0)=(2,1) $
Other solutions are given by:
$(z_n+ \sqrt3 y_n)=(2+ \sqrt 3)^{n} $
From above we get ,
$z_{n+1}=7z_n+12y_n $
And
$y_{n+1}=4z_n+7y_n $
Or (in terms of $x_n $);
$x_{n+1}=7x_n+6y_n $
$y_{n+1}=8x_n+7y_n $
Now,
$x_{n+1} \equiv -y_n $ mod 7
$y_{n+1} \equiv x_n $   mod 7
${y_{n+1}}^2-{x_{n+1}}^2 \equiv n $
$\equiv {x_n}^2 -{y_n}^2 \equiv  -n $
$\equiv 0 $
