Competition problem — Number theory

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Number theory Difficulty 5.7 AIME, harder Prove it Austria

Anton chooses as starting number an integer $n \ge 0$ which is not a square. Berta adds to this number its successor $n + 1$ . If this sum is a perfect square, she has won. Otherwise, Anton adds to this sum, the subsequent number $n + 2$ . If this sum is a perfect square, he has won. Otherwise, it is again Berta's turn and she adds the subsequent number $n + 3$ , and so on.
Prove that Anton wins with infinitely many starting numbers.

Solution

We will prove that Anton wins for the infinity of starting numbers $3x^2-1$ with $x \ge 1$ .
Since $3x^2 - 1 \equiv 2 \pmod 3$ , it cannot be a perfect square. After Berta adds the subsequent integer $3x^2$ , the sum $6x^2 - 1$ is also $\equiv 2 \pmod 3$ and consequently not a perfect square. Now Anton adds the subsequent number $3x^2 + 1$ and obtains the perfect square $9x^2$ . Therefore, Anton has won and we have found an infinity of possible starting numbers.

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