## Task A-1.8. (20 points)

Determine all natural numbers $n$ for which $n^{2}-440$ is a perfect square.
## Solution.

Let $n^{2}-440$ be the square of a number $k, k \in \mathbb{N}_{0}$.

From $n^{2}-440=k^{2}$, it follows that $(n-k)(n+k)=440$.

Since the right side of the equation is an even number, the product $(n-k)(n+k)$ must also be even, so at least one of the numbers $n-k, n+k$ must be even. This is only possible if $n$ and $k$ have the same parity, and then both numbers $n-k, n+k$ are even.

(5 points)

The number $440=2 \cdot 2 \cdot 2 \cdot 5 \cdot 11$ can be expressed as the product of two even numbers in four ways: $440=2 \cdot 220=4 \cdot 110=10 \cdot 44=20 \cdot 22$.

(5 points)

Since $n-k \leq n+k$, we get four systems:
a) $n-k=2$
c) $n-k=10$
$n+k=220$
$n+k=44$
b) $n-k=4$
d) $n-k=20$
$n+k=110$
$n+k=22$

From these systems, we obtain all the desired numbers $n$: $111, 57, 27, 21$.

(8 points)

Note: For each individual correct solution (e.g., $n=21$), give 2 points.

## MUNICIPAL/SCHOOL COMPETITION IN MATHEMATICS

## 2nd grade - high school - A category

## January 29, 2009.