Pell Fermat equation If $3n + 1$ and $4n + 1$ are perfect squares then $56$ divides $n$.
I somehow proved $n$ is even so proceeding further proved that $8$ divides $n$ but don't know how to manage for $7$.
 A: Let $3k+1=x^2,4k+1=y^2$ As you proved $8|k$.
Notice that $z^2-3y^2=1\ \ (*)$ when $z=2a$. All solutions of this Pell equation are obtained by substitution of $(a_n,b_n)$ in the equation: $(2+\sqrt{3})^n=a_n+\sqrt{3}b_n$. Because $z=2a$, w're looking for solutions with even $a_n$.
Let's prove that $a_n$ is even if and only if $n$ is odd (you can use induction) but here we will use the binomial formula :
$$(2+\sqrt{3})^n=\sum_{k=0}^n 2^{n-k}(\sqrt 3)^{k}=\sum_{k=0,k\, even}^n 2^{n-k}(\sqrt 3)^{k}+(\sum_{k=0,k\, odd}^n 2^{n-k}(\sqrt 3)^{\frac{k-1}{2}})\sqrt 3 $$. 
We observe that all terms in $a_n$ are even except when $k$ is even and $n-k=0$ we will have the term $3^{\frac{k}{2}}$ which makes $a_n$ odd. So finally $a_n$ is even if and only is $n$ is odd, hence all solutions of our equation $(*)$ are given by
$(2+\sqrt{3})^{2n+1}=z_n+\sqrt{3}y_n$.
We can easily get the recursions:
$z_{n+1}=2x_{n+1}=14x_n+12y_n,  y_{n+1}=8x_n+7y_n.$ 
This means that if $3k+1, 4k+1$ are squares then there exist $n$ such that $3k+1=x_n^2, 4k+1=y_n^2$ or $k=y_n^2-x_n^2$, the question now is equivalent to prove for all $n$ we have $7|y_n^2-x_n^2$.
By induction on $n$ , for $n=0$  note that $ y_0^2-x_0^2\equiv 1^2-1^2 \equiv 0\pmod{7}$
Suppose that $y_n^2-x_n^2 \equiv 0\pmod{7}$ , So $x_{n+1}=7x_n+6y_n \equiv -y_n\pmod{7}$ and $y_{n+1}=8x_n+7y_n\equiv x_n\pmod{7}$,we get $y_{n+1}^2-x_{n+1}^2 \equiv y_n^2-x_n^2 \equiv 0\pmod{7}$.
