Perfect Sqaure

We have $n^{2}+45=x^{2}$ where $x^{2}$ is the perfect square. This is rearranged to $45=x^{2}-n^{2}$ , which can be factorised to $45=(x+n)(x-n)$ . Now consider all possible factors of 45. These are: $45$ x $1$ , $15$ x $3$ and $9$ x $5$ . Since $x+n>x-n$ , set $x+n$ equal to the larger factor. This gives $x+n=45$ and $x-n=1$ , so $x=23, n=22$ . Taking the next factor pair, $x+n=15$ and $x-n=3$ , so $x=9, n=6$ . Finally, $x+n=9$ and $x-n=5$ , $x=7, n=2$ . The possible values of $n$ are therefore $2, 6$ and $22$ .