Can $a^2+b+2$ and $b^2+4a$ both be perfect squares?

Question

Are there any positive integers $a$ and $b$ so that $a^2+b+2$ and $b^2+4a$ are both perfect squares?

+$1$ Good question. As has already been pointed out (but to prevent future postings), the restriction that the integers be positive cannot be dropped as $a=0$ and $b=2$ provide such an example. — Clayton, Nov 27 '13 at 18:38

score 3 · Accepted Answer · answered Nov 27 '13 at 18:40

3

Suppose that $a\le b$. Since $b^2$ is a perfect square, and $b^2+4a\lt (b+2)^2$, $b^2+4a$ cannot be a perfect square unless $b^2+4a=(b+1)^2$, which is impossible.

Now suppose that $a\gt b$. Then $a^2+b+2$ is too small to be a perfect square, since the smallest perfect square greater than $a^2$ is $a^2+2a+1$.

answered Nov 27 '13 at 18:40

André Nicolas

507,029

That's a very slick argument. – marshall Nov 27 '13 at 19:14

score 2 · Answer 2 · answered Nov 27 '13 at 18:40

No.

$b^2+4a$ is a perfect square $>b^2$, and $b^2+4a \not =(b+1)^2$ (fails by checking parity), so $b^2+4a \geq (b+2)^2$ so $a \geq b+1$

$a^2+b+2$ is a perfect square $>a^2$, so $a^2+b+2 \geq (a+1)^2$ so $b \geq 2a-1$.

Combining, $b \geq 2a-1 \geq 2(b+1)-1=2b+1>b$, a contradiction.

Can $a^2+b+2$ and $b^2+4a$ both be perfect squares?

2 Answers2