<s>[INST] If $\sqrt{k-9}$ and $\sqrt{k+36}$ are both integers, the number of possible values of $k$ is? If $\sqrt{k-9}$ and $\sqrt{k+36}$ are both integers, the number of possible values of $k$ is: A) $1$; B) $2$; C) $3$; D) more than $3$.
I started to try different values of $k$ in the first root ($13, 18, 25, 34$) and see if $\sqrt{k+36}$ is an integer. So far I found $k=13$ is a solution. Are there any other solutions? Also, I don't think I am doing it in the fastest way.
 [/INST] Let's first solve $a^2-b^2=45$ with $a:=\sqrt{k+36},\,b:=\sqrt{k-9}$ both integers $\ge0$. Since $(a-b)(a+b)=3^2\times 5$ must be a factorisation into two odd positive factors with $a-b\le a+b$, either $$a-b=1,\,a+b=45\implies a=23,\,b=22\implies k=b^2+9=493,$$ or $$a-b=3,\,a+b=15\implies b=6\implies k=45,$$ or $$a-b=5,\,a+b=9\implies b=2\implies k=13.$$So the answer is C).
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