# Task 9.1

For which natural numbers $n$ is the expression $n^{2}-4 n+11$ a square of a natural number?

## Number of points 7
Answer:

for $n=5$

## Solution

Let $n^{2}-4 n+11=t^{2}$.

Note that $n^{2}-4 n+4=(n-2)^{2}$ is also the square of some integer $r=n-2$, which is less than $t$.

We get that $t^{2}-r^{2}=(t+r)(t-r)=7$.

The numbers $(t+r)$ and $(t-r)$ are natural and the first is greater than the second.

Thus, $(t+r)=7$, and $(t-r)=1$.

Solving this system, we get $t=4, r=3$, which gives $n=5$.