# finding all squares

• Jan 16th 2012, 05:53 PM
Veronica1999
finding all squares
Could I get some help on this problem?

Thanks.
• Jan 16th 2012, 07:31 PM
alexmahone
Re: finding all squares
Let $n^2+9n+14=a^2$, where $a$ is an integer.

$n^2+9n+14-a^2=0$

Let us now calculate the discriminant of this quadratic equation in $n$.

$\Delta=81-4(14-a^2)=25+4a^2=5^2+(2a)^2$

In order for $n$ to be an integer, we need $\Delta$ to be a perfect square.

The only Pythagorean triple that has 5 as one of the smaller values is 5, 12, 13.

So, $a=\pm 6$

$n^2+9n+14=36$

$n^2+9n-22=0$

$(n+11)(n-2)=0$

$n=-11,\ 2$