Could I get some help on this problem?
Thanks.
Let $\displaystyle n^2+9n+14=a^2$, where $\displaystyle a$ is an integer.
$\displaystyle n^2+9n+14-a^2=0$
Let us now calculate the discriminant of this quadratic equation in $\displaystyle n$.
$\displaystyle \Delta=81-4(14-a^2)=25+4a^2=5^2+(2a)^2$
In order for $\displaystyle n$ to be an integer, we need $\displaystyle \Delta$ to be a perfect square.
The only Pythagorean triple that has 5 as one of the smaller values is 5, 12, 13.
So, $\displaystyle a=\pm 6$
$\displaystyle n^2+9n+14=36$
$\displaystyle n^2+9n-22=0$
$\displaystyle (n+11)(n-2)=0$
$\displaystyle n=-11,\ 2$