XI APMO

Without loss of generality, assume that $|b|\le |a|$. If $b=0$, then $a$ must be a perfect square. So $(a=k^2,b=0)$ for each $k\in \mathbb{Z}$ is a solution.

Now we consider the case $b\ne 0$. Because $a^2+4b$ is a perfect square, the quadratic equation

$$x^2+ax-b=0\text{\hspace{1em}(*)}$$

has two non-zero integral roots $x_1,x_2$.

Then $x_1+x_2=-a$ and $x_1{x}_2=-b$, and from this it follows that $$\frac{1}{|x_1|}+\frac{1}{|x_2|}\ge \left|\frac{1}{x_1}+\frac{1}{x_2}\right|=\frac{|a|}{|b|}\ge 1$$

Hence there is at least one root, say $x_1$, such that $|x_1|\le 2$.

There are the following possibilities.

(1) $x_1=2$. Substituting $x_1=2$ into () we get $b=2a+4$. So we have $b^2+4a=(2a+4{)}^2+4a=4a^2+20a+16=(2a+5{)}^2-9$. It is easy to see that the solution in non-negative integers of the equation $x^2-9=y^2$ is $(3,0)$. Hence $2a+5=\pm 3$. From this we obtain $a=-4,b=-4$ and $a=-1,b=2$. The latter should be rejected because of the assumption $|a|\ge |b|$.

(2) $x_1=-2$. Substituting $x_1=-2$ into () we get $b=4-2a$. Hence $b^2+4a=4a^2-12a+16=(2a-3{)}^2+7$. It is easy to show that the solution in non-negative integers of the equation $x^2+7=y^2$ is $(3,4)$. Hence $2a-3=\pm 3$. From this we obtain $a=3,b=-2$.

(3) $x_1=1$. Substituting $x_1=1$ into (*) we get $b=a+1$. Hence $b^2+4a=a^2+6a+1=(a+3{)}^2-8$. It is easy to show that the solution in non-negative integers of the equation $x^2-8=y^2$ is $(3,1)$. Hence $a+3=\pm 3$. From this we obtain $a=-6,b=-5$.

(4) $x_1=-1$. Substituting $x_1=-1$ into $(\ast )$ we get $b=1-a$. Then $a^2+4b=(a-2{)}^2$, $b^2+4a=(a+1{)}^2$. Consequently, $a=k,b=1-k$ ($k\in \mathbb{Z}$) is a solution.

Testing these solutions and by symmetry we obtain the following solutions $$(-4,-4),(-5,-6),(-6,-5),(0,k^2),(k^2,0),(k,1-k)$$ where $k$ is an arbitrary integer. (Observe that the solution $(3,-2)$ obtained in the second possibility is included in the last solution as a special case.)

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Alternative solution.

Without loss of generality assume that $|b|\le |a|$. Then $a^2+4b\le a^2+4|a|<a^2+4|a|+4=(|a|+2{)}^2$. Given that $a^2+4b$ is a perfect square and since $a^2+4b$ and $a^2$ have the same parity then $a^2+4b\ne (|a|+1{)}^2$, so $$a^2+4b\le a^2\text{\hspace{1em}(1)}$$

Case 1. $a^2+4b=a^2$. Then $b=0$ and $a$ must be a perfect square. So $a=k^2,b=0$ ($k\in \mathbb{Z}$) is a solution.

Case 2. $a^2+4b=(|a|-2{)}^2$. Then $b=1-|a|$, therefore $b^2+4a=a^2-2|a|+4a+1$ must be a perfect square.

If $a>0$ then $b^2+4a=(a+1{)}^2$ is a perfect square for each $a\in \mathbb{Z}$. Consequently $a=k$ and $b=1-k$ ($k\in {\mathbb{Z}}^{+}$) is a solution.

If $a<0$ then $b^2+4a=m^2-6m+1$ must be a perfect square, where $m=-a>0$. For $m\ge 8$ $$(m-3{)}^2>m^2-6m+1>(m-4{)}^2$$ therefore $m<8$. If $m=1,2,3,4,5$ then $m^2-6m+1<0$. If $m=6$, $m^2-6m+1=1$ is a perfect square thus $a=-6$ and $b=-5$ is a solution. If $m=7$, $m^2-6m+1=8$ is not a perfect square.

Case 3. $a^2+4b\le (|a|-4{)}^2$. Since $|b|\le |a|$ then $b\ge -|a|$, thus $a^2-4|a|\le a^2+4b\le (|a|-4{)}^2$. It follows that $|a|\le 4$. We have following possibilities:

(a) $|a|=4$. Then $16+4b=0$ or $b=-4$. Thus $b^2+4a=16\pm 16$ must be a perfect square. So $a=-4$ and $b=-4$.

(b) $|a|=3$. In this case $a^2+4b=9+4b\le 1$, then $9+4b=0$ or $9+4b=1$. The equation $9+4b=0$ does not have integer solutions. The solution of the second equation is $b=-2$. Then $b^2+4a=4\pm 12$ must be a perfect square, thus $a=3$.

(c) $|a|=2$. $a^2+4b=4+4b\le 4$. Since $4+4b$ is even and must be a perfect square then $4+4b=4$ or $4+4b=0$. Therefore $b=0$ or $b=-1$. If $b=0$, $b^2+4a=\pm 8$ is not a perfect square. If $b=-1$ then $b^2+4a=1\pm 8$ is a perfect square if $a=2$. Thus $a=2$ and $b=-1$ is a solution.

(d) $|a|=1$. Then $a^2+4b=4b+1\le 9$. Since $4b+1$ must be an odd perfect square then $4b+1=1$ or $4b+1=9$. So $b=0$ or $b=2$. If $b=0$, $b^2+4a=\pm 4$, then $a=1$. If $b=2$ then $a=-1$, but this is not possible because $|b|\le |a|$. Thus $a=1$ and $b=0$ is a solution in this case.

(e) $|a|=0$. Since $|b|\le |a|$ then $b=0$.

Testing these solutions and by symmetry we obtain the following solutions: $$(k^2,0),(0,k^2),(k,1-k),(-6,-5),(-5,-6),(-4,-4)$$ where $k$ is an arbitrary integer. Note that if $(k,1-k)$ is a solution with $k>0$, then taking $t=1-k$, $k=1-t$, so $(1-t,t)$ is solution. Thus by symmetry $(k,1-k)$ is a solution for any integer.