If $a$( or $b$) equals $0$, then $b$($a$) is, evident, perfect square. Else we can think, that $|a|\geq |b|$. If $a=1$ then $b=0$ - it was earlier. If $a=-1$ then $b=2>1=|-1|$ - no. If $|a|\geq 2$, then $(|a|-4)^2<a^2-4|a|\leq a^2+4b\leq a^2+4|a|<(|a|+2)^2$ since $a^2+4b$ and a both odd or even $a^2+4b$ equals $a^2$ or $(|a|-2)^2$. But in the 1st variant $b=0$, hence we have 2nd variant and $b=1-|a|$ $b^2+4a=a^2+1-2|a|+4a$. 1) $a=k>0$ $b^2+4a=(1+k)^2$ and $a^2+4b=(k-2)^2$. 2) $a<0$ $b^2+4a=a^2-6|a|+1=(|a|-3)^2-8$ $a=-6$ $b=-5$ - it's true($5^2-4*6=1^2, 6^2-4*5=4^2$). Also we get answers: $a=0, b=k^2, k\geq 0;$ $a=k^2, b=0, k>0;$ $a=k, b=1-k, k>0;$ $a=1-k, b=k, k>0; k\in\mathbb{Z};$ $a=-5, b=-6; a=-6, b=-5$

What about leftout solution (-4, -4) ?

Suppose $ |a|\ge |b|$. -$ b=0$ then $ a=k^{2}$ -$ |b|\ge 1$. Case 1: $ a=b$ then $ a^{2}+4a=k^{2},k\in\mathbb{Z},k\ge 0$ so $ (a+2)^{2}-k^{2}=4\Leftrightarrow (a+2-k)(a+2+k)=4\Rightarrow a+2-k=2=a+2-k\Leftrightarrow k=0$ then $ a=b=-4$ Case 2 : $ a=-b$ then $ a^{2}-4a=k^{2},k\in\mathbb{Z},k\ge 0$ similar Case 1, not exist $ k$ so that the condition . Case 3 : $ |a|\not= |b|$ Same to pavel kozlov'solution

To Pavel Koslov solution If lal >= 4 then the inequality ( lal - 4 )^2 =< a^2 - 4*lal is true but it can be untrue in the case of lal >= 2. We need consider cases lal=2 and lal=3 separately

$$(3,-2)$$is solution

Though this solution is not mine, I put it. Observe that the two expression is symmetric. If $(x,y)$ is a solution then $(y,x)$ is also a solution. If $b=0$, then $a$ has to be a perfect square. Therefore, $(a,b)=(k^2,0), (0,k^2)$ where $k$ is any arbitrary integer. Now consider that $b\neq 0$ and assume WLOG that $|b|\leq |a|$. Consider the following quadratic equation \begin{eqnarray} x^2+ax-b \end{eqnarray}Notice that this quadratic equation has two non-zero real roots $x_1,x_2.$ $\left(x=\frac{-a\pm \sqrt{a^2+4a}}{2}\right)$ is an integer. By Vieta's formula we have $x_1+x_2=-a, x_1x_2=-b$. The euqality $$\frac{1}{|x_1|}+\frac{1}{|x_2|}\geq |\frac{1}{x_1}+\frac{1}{x_2}|\geq \frac{|a|}{|b|}\geq 1$$Hence we can say that there has at least one root, say $x_1$ such that $x_1\leq 2$. Then there are $4$ possibilities. 1) $x_1=2$. Substituting this in to equation (1) we have $b=2a+4$. Then $b^2+4a=(2a+4)^2+4a=(2a+5)^2-9$. The only solution for the equation $y^2=x^2-9$ is $(3,0)$ where $|y|\leq |x|$. Therefore we have $2a+5=\pm 3$. Now substituting $a=-1$ we have, $b^24=0\Longrightarrow b=\pm 2$ ($-2$ is rejected because we have considered $|b|\leq |a|$). So we get $(a,b)=(-1,2)$. Similarly $(a,b)=(-4,-4)$. 2) $x_1=-2$. Substituting this we have $b=4-2a$. So we have $b^2+4a=(2a-3)^2+7$. The equation $x^2+7=y^2$ has one of the solutions $(3,4)$. Therefore $2a-3=\pm 3$. From this we have $(a,b)=(3,-2)$. 3) $x_1=-1$. Substituting this we have $b=1-a$. Then $a^2+4b=(a-2)^2, b^2+4a=(a+1)^2$. Therefore we get $(a,b)=(k,1-k)$ where $k$ is an integer. 4) $x_1=1$. Substituting this we have $b=a+1$. Hence $b^2+4a=(a+3)^2-8$. The equation $y^2=x^2-8$ has one of solutions $(3,1)$. So $2a-3=\pm 3$. From this we have $(a,b)=(-6,-5)$. Hence all the solutions are $(a,b)=(-4,-4), (-1,2), (3,-2), (-6,-5), (k,1-k), (k^2,0)$ and by symmetry $(a,b)=(2,-1), (-2,3), (-5,-6), (1-k,k), (0,k^2)$ where $k$ is an arbitrary integer.

If $|a|=|b|$, we have $a^2+4b=b^2+4b,b^2+4a=a^2+4a$ $\because b^2+4b+4=(b+2)^2$ is a perfect square $\therefore (b+2)^2=4,b=0,-4$ We can prove $a=0,-4$ in this way. With $|a|=|b|,(a,b)=(0,0),(-4,-4)$ If $|a|\neq |b|$, let $|a|>|b|$, we have $|a|^2-4(|a|-1)\leq a^2+4b\leq |a|^2+4(|a|+1)<(|a|+2)^2$ $\therefore (|a|-2)^2\leq a^2+4b<(|a|+2)^2$, with $a^2+4b\equiv a\pmod{2},a^2+4b=(|a|-2)^2\text{ or }a^2$ 1. $a^2+4b=a^2$, then $b=0, 4a$ is a perfect square $\Rightarrow a$ is a perfect square $\therefore (a,b)=(k^2,0),k\in\mathbb{N}^{*}$ 2. $a^2+4b=(|a|-2)^2=|a|^2-4|a|+4\Rightarrow b=1-|a|\leq 0$ $\boxed{1} a=-|a|,b^2+4a=(1-|a|)^2-4|a|=|a|^2-6|a|+1=(|a|-3)^2-8$ $\Rightarrow |a|=6\Rightarrow (a,b)=(-6,-5)$ $\boxed{2} a=|a|,b^2+4a=(1-|a|^2)+4|a|=(1+|a|)^2$ is a perfect square, so that $(a,b)=(t,1-t),t\in\mathbb{N}^{*}$ $\therefore (a,b)=(0,0),(-4,-4),(k^2,0),(0,k^2),(-6,-5),(-5,-6),(t,1-t),(1-t,t),k,t\in\mathbb{N}^{*}$