Can both $x^2 + y+2$ and $y^2+4x$ be squares?

Question

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.

Answer 1

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