Question

The sum of integral values of $n$ for which $n^2+17n+75$ is a perfect square is

• A. $-17$
• B. $-15$
• C. $-14$
• D. $0$

Answer 1

The correct option is A $-17$

$\text{Let}{n}^2+17n+75=m^2,m\in Z$
$\Rightarrow n^2+17n+(75-m^2)=0$
$\Rightarrow n=\frac{-17\pm \sqrt{{17}^2-4(75-m^2)}}{2}=\frac{-17\pm \sqrt{4m^2-11}}{2}$
Hence $4m^2-11$ should be an odd perfect square if $n$ has to be an integer.
So, $4m^2-11=p^2$ where $p$ is an odd integer.
$\Rightarrow (2m-p)(2m+p)=11$

$\begin{array}{cccc}2m-p& 2m+p& m& p\\ 1& 11& 3& 5\\ 11& 1& 3& -5\\ -1& -11& -3& -5\\ -11& -1& -3& 5\end{array}$

Possible value of $m$ is $3,-3.$

$\therefore n=\frac{-17\pm \sqrt{36-11}}{2}=-11,-6$