Berov s.L.

Seryozha chose two different natural numbers $a$ and $b$. He wrote down four numbers in his notebook: $a, a+2, b$ and $b+2$. Then he wrote on the board all six pairwise products of the numbers from the notebook. What is the maximum number of perfect squares that can be among the numbers on the board?
Note that no two squares of natural numbers differ by 1: $x^{2}-y^{2}=(x-y)(x+y)$, and the second bracket is greater than one. Therefore, the numbers

$a(a+2)=(a+1)^{2}-1$ and $b(b+2)=(b+1)^{2}-1$ are not squares.

The numbers $a b$ and $a(b+2)$ cannot both be squares; otherwise, their product $a^2 b(b+2)$ would also be a square, and then the number $b(b+2)$ would be a square as well. Similarly, among the numbers $(a+2) b$ and $(a+2)(b+2)$, at most one can be a square. In total, there are no more than two squares on the board.

Two squares can be obtained, for example, with $a=2$ and $b=16$: then $a(b+2)=6^{2}$ and $(a+2) b=8^{2}$.

## Answer

Two squares.