Track / Stage 6 / 16 of 400 #1496 of 2444

Problem 1496

National Olympiad, first round

Number theory Difficulty 6.0 Prove it Russian Mathematical Olympiad · Russia

Sergey chose two different positive integers $a$ and $b$ . Then he calculated all six pairwise products of the four numbers $a$ , $a+2$ , $b$ , and $b+2$ . Find the maximal number of perfect squares among the six calculated numbers. (S. Berlov)

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

Answer. Two.

Note that no two squares of natural numbers differ by $1$ , since $x^2 - y^2 = (x - y)(x + y)$ , where the second factor 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. Moreover, the numbers $ab$ and $a(b+2)$ cannot both be squares, otherwise their product $a^2 \cdot b(b+2)$ would also be a square, and then $b(b+2)$ would also be a square. Similarly, among the numbers $(a+2)b$ and $(a+2)(b+2)$ , at most one can be a square. Thus, there are at most two squares among the six numbers.

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

Remark. There are other examples, for instance, $(a, b) = (6, 96)$ .

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.