Solution

The equation can be manipulated into $(a+b)(a-b)=-2(a-b)$. If $b=a$ then we cannot divide both sides, but $a$ is an integer so $b$ is an integer too. Since we have covered the $a=b$ case, we can assume $a\ne b$ to get $a+b=-2$. Since the difference between any two integers is an integer, and $-2$ and $a$ are integers, $b$ must be an integer. edit: sniped by somebody with a better solution than me

Quadratic formula yields $b=\frac{-2\pm\sqrt{4+4(a^2+2a)}}{2}=\frac{-2\pm 2(a+1)}{2}=-1\pm (a+1)$, which is clearly an integer.

We get $b^2+2b-a^2-2a=0$, so $b=\frac{-2\pm\sqrt{4+4a^2+8a}}2$, and since $4+4a^2+8a=4(a+1)^2$, $b=-1\pm(a+1)$, which is always an integer.

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Let $k=a^2+2a.$ Applying the quadratic formula of $b^2+2b-k$ gives $$b = \frac{-2 \pm \sqrt{4+4k}}{2} = -1 \pm 2\sqrt{1+k}.$$Since $1+k = (a+1)^2,$ if $a$ is an integer, $b= -1 \pm 2\sqrt{1+k} = -1 \pm (2a+1)$ is also an integer. $\blacksquare$