Which are those positive integers $a, b$ for which

$$
a^{2}-4 b \text { and } b^{2}-4 a \text { are both }
$$

perfect squares?
I. Solution. Since the roles of $a$ and $b$ in the problem are completely symmetric, we can assume that $a \leq b$. Then $a^{2}-4 b \leq a^{2}-4 a$, and since $a^{2}-4 b$ is a perfect square and thus non-negative, it follows that $a^{2}-4 a \geq 0$. According to the problem's conditions, $a$ is positive, so $a \geq 4$.

We know that $4 \leq a \leq b$. If $b=4$, then $a=4$, and this is also a solution to the problem. If $b>4$, that is, $b \geq 5$, then $b^{2}-4 a$ can be squeezed between two nearby perfect squares: $b^{2}>b^{2}-4 a \geq b^{2}-4 b>(b-3)^{2}$. Since $b^{2}-4 a$ is also a perfect square, only two possibilities remain: $b^{2}-4 a=(b-1)^{2}$ or $b^{2}-4 a=(b-2)^{2}$. In the former case, $4 a=2 b-1$, where the left side is even and the right side is odd, so this cannot occur. One case remains: $b^{2}-4 a=(b-2)^{2}$, from which $a=b-1$. According to the condition, then $a^{2}-4 b=(b-1)^{2}-4 b=(b-3)^{2}-8$ is a perfect square. If $1 \leq b \leq 5$, then the value of this expression is negative; when $b=7$, it is 8, which is not a perfect square; but when $b=6$, it is indeed a perfect square. Finally, for $b \geq 8$, the difference between $(b-3)$ and the nearest perfect square, $(b-4)^{2}$, is $2 b-7 \geq 9$, so in this case $(b-3)^{2}-8$ cannot be a perfect square. In addition to the solution $a=b=4$, we have thus found one more solution: $b=6, a=5$, in which case $b^{2}-4 a=6^{2}-4 \cdot 5=4^{2}$ and $a^{2}-4 b=5^{2}-4 \cdot 6=1$, indeed perfect squares. Clearly, the symmetric pair of this, the pair $a=6$, $b=5$, is also a solution.

II. Solution. It is known that if $A$ and $B$ are integers, then the solutions of the equation $x^{2}-A x+B=0$ are exactly integers if the discriminant of the equation, $A^{2}-4 B$, is a perfect square. Our problem can thus be rephrased as: for which positive integers $a$ and $b$ will the solutions of both the equation $x^{2}-a x+b=0$ and the equation $y^{2}-b y+a=0$ be integers? Let $x_{1}$ and $x_{2}$ be the roots of the former, and $y_{1}$ and $y_{2}$ be the roots of the latter. According to the relationships between the roots and coefficients,

$$
x_{1}+x_{2}=a=y_{1} y_{2} \quad \text { and } \quad y_{1}+y_{2}=b=x_{1} x_{2}
$$

and since $a>0, b>0$, $x_{1}, x_{2}, y_{1}, y_{2}$ are also positive. The problem is thus to determine those positive integers $x_{1}, x_{2}, y_{1}, y_{2}$ for which $x_{1}+x_{2}=y_{1} y_{2}$ and $y_{1}+y_{2}=x_{1} x_{2}$. This was precisely the 1st problem of the 1987 Kürschák competition (1. KöMaL 1988/2. p. 50.), and according to its solution, essentially two quadruples of numbers satisfy this: one is $x_{1}=x_{2}=y_{1}=y_{2}=2$, in which case $a=b=4$, and the other is $x_{1}=1, x_{2}=5, y_{1}=2, y_{2}=3$ and its symmetric variations $(5, 1, 2, 3; 5, 1, 3, 2; 1, 5, 3, 2; 2, 3, 1, 5; 2, 3, 5, 1; 3, 2, 1, 5; 3, 2, 5, 1$). In this case, $a=6$ and $b=5$ (or $a=5$ and $b=6$).

These are clearly solutions, and there are no others.