Can both $x^2 + y+2$ and $y^2+4x$ be squares? Prove that there exist no positive integers $x$ and $y$ such that both $x^2+y+2$ and $y^2+4x$ are perfect squares.
I thought I could perhaps solve this by square bounding but I couldn't get anywhere with it.
Thanks in advance for any help.
 A: Thanks to Erick Wong for setting me on the right track.
Assume for sake of contradiction that $x^2+y+2$ and $y^2+4x$ are both perfect squares.
Then as $y$ is a positive integer, $x^2+y+2 \geq (x+1)^2=x^2 +2x+1$
$$\implies y+2\geq 2x+1$$
$$y+1 \geq 2x \; \; \; \; \; (1)$$
Following a similar argument, as $x$ is a positive integer, $y^2+4x \geq (y+1)^2=y^2+2y+1$
$$\implies 4x \geq 2y+1$$
But, $4x$ is even and $2y+1$ is odd so equality can never hold and hence
$$y^2+4x \geq (y+2)^2=y^2+4y+4$$
$$\implies x \geq y+1 \; \; \; \; (2)$$
Combining (1) and (2) we have that $x \geq y+1 \geq 2x$
$$\implies x \geq 2x$$
Which is a contradiction as $x \geq 1$.
Hence there are no positive integers $x$ and $y$ satisfying the requirements. QED
A: Maybe it is better to solve such a system the system of Diophantine equations:
$$\left\{\begin{aligned}&x^2+yz+2z^2=q^2\\&y^2+4xz=r^2\end{aligned}\right.$$
Then the solution can be written.
$$x=p^3+4kp^2-4pk^2-k^3$$
$$y=p^3+142kp^2+49pk^2+4k^3$$
$$z=4p(5k^2+53kp+138p^2)$$
$$q=781p^3+350kp^2+44pk^2+k^3$$
$$r=47p^3+106kp^2+39pk^2+4k^3$$
$p,k$ - integers asked us.
