Competition problem — Number theory

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Number theory Difficulty 5.7 AIME, harder Prove it Brazil

We call a number pal if it doesn't have a zero digit and the sum of the squares of the digits is a perfect square. For example, $2115522$ is pal (because $2^2 + 1^2 + 1^2 + 5^2 + 5^2 + 2^2 + 2^2 = 8^2$ ) but $304$ and $12$ are not pal.

a. What is the greatest two-digit pal number?

b. Does there exist a $2011$ -digit pal number?

Solution

a. First notice that $86$ is pal. Then it's not hard to check by hand that every number from $87$ to $99$ is not pal.

b. The answer is yes. First consider the $2011$ -digit number $11\ldots1$ . The sum of its digits is $2011$ . The smallest perfect square greater than $2011$ is $45^2 = 2025$ . Since $2025 - 2011 = 14$ and $14 = 2 \cdot (2^2 - 1^2) + (3^2 - 1^2)$ , we can exchange two $1$ s by two $2$ s and one $1$ by one $3$ . So we obtain the pal number $11\ldots123$ .

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Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.