Question

The number of integers $n$ for which $n^2-4n+46$ is a perfect square, is

• A. $0$
• B. $2$
• C. $6$
• D. $8$

Answer 1

The correct option is A $0$

Let $n^2-4n+46=k^2,$ where $k\in I$
$\Rightarrow (n-2{)}^2+42=k^2\Rightarrow k^2-(n-2{)}^2=42\Rightarrow (k+n-2)(k-n+2)=a\times b$, where $ab=42$
So, possible pairs of $(a,b)$ are $(1,42),(2,21),(3,14),(6,7),(7,6),(14,3),(21,2),(42,1)$

Let $k+n-2=a$
and $k-n+2=b$
$\Rightarrow n=\frac{a-b}{2}+2$

For $n$ to be an integer, $a-b$ must be even.
So, no such $n$ is possible.