Find all integers n such that $\sqrt{n-4 \sqrt{n-19}}$ is also an integer

Question

The following puzzle is from the December 2012 issue of the Eureka journal (published by The Cambridge University Mathematical Society).

Find all integers n such that $\sqrt{n-4 \sqrt{n-19}}$ is also an integer.

To be accepted, your answer must show that you have found all integers satisfying the given condition.

Answer 1

If $\sqrt{n-4\sqrt{n-19}}$ is an integer, then $\sqrt{n-19}$ must also be an integer, because square roots of integers are either integer or irrational. Let us call this integer $m=\sqrt{n-19}$. We rewrite the expression in terms on $m$:
$$\sqrt{m^2+19-4m}$$ which must be an integer. Therefore,
$$m^2+19-4m$$ must be a perfect square or, equivalently,
$$(m-2)^2+15$$
must be a perfect square. As $(m-2)^2$ is also a perfect square, we must find two perfect squares whose difference is 15. As the differences between consecutive perfect squares increase monotonically, it is easy to see that there are only two pairs of perfect squares that differ by 15: $(1, 16)$ and $(49, 64)$. The smaller square is $(m-2)^2$, so we have four options:
- $m-2=-1 \implies m=1 \implies n=20$.
- $m-2=1 \implies m=3 \implies n=28$.
- $m-2=-7 \implies m=-5 \implies n=44$. This solution is not valid.
- $m-2=7 \implies m=9 \implies n=100$.

So there are three solutions: $n=20$, $n=28$, $n=100$. The value $n=44$ is not valid because it would require the negative square root of $n-19$.