Could I get some help on this problem?

Thanks.

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- Jan 16th 2012, 05:53 PMVeronica1999finding all squares
Could I get some help on this problem?

Thanks. - Jan 16th 2012, 07:31 PMalexmahoneRe: finding all squares
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, $\displaystyle a=\pm 6$

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

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

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

$\displaystyle n=-11,\ 2$