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

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.

• I may be wrong but in my opinion this is a math problem rather than math puzzle Jul 31, 2022 at 22:29
• @VarunW. On the contrary, it couldn't be clearer that this is recreational maths as from a serious maths perspective this is not in any way an interesting question. Aug 1, 2022 at 0:11
• That shows that it's not "serious maths", but it's consistent with the question being (1) a recreational-maths puzzle suitable for PSE or (2) a basically routine exercise not so suitable for PSE. If you do the Obvious Thing then after 10-20 lines of working the answer comes out pretty straightforwardly, but then the same is true for e.g. plenty of puzzles from the likes of Dudeney. I reckon this one is pretty close to the borderline. (The nearest thing we have to an official policy is at puzzling.meta.stackexchange.com/questions/6030/…) Aug 1, 2022 at 0:21
• @GarethMcCaughan I agree with you that it is not difficult. But using the reasoning you link a sudoku would have to be classified as a non puzzle. I know and value your restraint but would feel more at easy with this community at large if attitudes would lean more to live and let live and less to seek out and destroy. Aug 1, 2022 at 0:46
• For me, this question was anything but routine. I had never seen anything like it before. I enjoyed solving this puzzle so I decided to share it. I have learned that the skill set of the people in PSE varies quite a lot. I hoped that some people would find it interesting and challenging. But I know you can’t please everyone. Aug 1, 2022 at 1:08

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 $$n-19$$.