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

Question

This is my solution not sure if it is correct though.

Solution: For the expression to be an integer, $(n-4)\sqrt{n-19}$ should be a perfect square. For $(n-4)\sqrt{n-19}$ to be a perfect square, $(n-4)^2(n-19)$ should be a perfect square.

Now consider some perfect square $k^2$ where $k\in\Bbb{Z}$. If I multiply it by some integer $r$ the perfect square will no longer be a perfect square unless $r=1,r=0,k=0$.

We see that above, $(n-4)^2$ is a perfect square and $n-19$ is some integer. So for $(n-4)^2(n-19)$ to be a perfect square, $(n-4)^2$ must be zero or $(n-19)$ must be zero or one.

As a result there are three possible $n$'s such that the expression is an integer, $n=4,n=19,n=20$.

Edit: This solution is wrong!

The product of two perfect squares is also a perfect square , so your statement above in the third paragraph is wrong, $r$ can be any perfect square. — Sarvesh Ravichandran Iyer, Jul 03 '20 at 04:34
The answer may be correct but the solution is definitely wrong. For example the phrase starting with "If I multiply it by some integer r the perfect square will no longer be a perfect square unless..." is wrong. — markvs, Jul 03 '20 at 04:36
Do you consider $n=4$ to be a solution? It does need to go through the complex numbers to calculate the subterm $\sqrt{n-19}$ but the eventual value of the entire expression is 0. — Daniel Schepler, Jul 03 '20 at 05:31

dimitri301 · Accepted Answer · 2020-07-03T05:24:49.543

2

If you start with $$n=s^2+19$$ and substitute that in and just follow the algebra you get 20 and 244. $$\sqrt{(n-4)\sqrt{n-19}}=\sqrt{(s^2+15)s}=q$$ Then$$(s^2+15)s=q^2$$ $$s^3+15 s=q^2$$ Then this means that $$s|q$$ so $$q=s t$$ Then you get $$s^2+15=s t^2$$ by which you infer that $$s|15$$ You then try all the factors of 15 and get 20 or 244.

edited Jul 03 '20 at 05:24

answered Jul 03 '20 at 04:59

dimitri301

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5

How about four? It is a solution right? – Khant Rain Jul 03 '20 at 05:34
I think $st^2$ should be $s^2t^2$. So $s^2|15$. – Khant Rain Jul 03 '20 at 05:39
No, the $s$ gets cancelled both the sides – Anand Jul 03 '20 at 05:40
So the solutions are $n=19,n=20,n=244$? – Khant Rain Jul 03 '20 at 05:44
Yeah, 19 works. – dimitri301 Jul 03 '20 at 05:49

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

1 Answers1