$\sqrt{4n}$ is only an integer if $\sqrt{n}$ is an integer

idk this should be in high school olumpiads hso, this is way to hard for msm

i mean i could be wrong, but i thought it was pretty trivial

I mostly agree w/ @3above (who is also @above) but I wanted to make sure my sol was rigorous, so mine is a bit more lengthy.

oh yeah, it works for 1 and 2 and the difference gets exponiationally bigger hence, proved

wait... if n is >=1 There are two cases: Case I: n>1 sqrt(5n)-sqrt(2n) always greater than 1 Therefore there has to be at least an integer between sqrt(2n) and sqrt(5n) Case II: n=1 2 is between sqrt(2) & sqrt(5) Does the above sol work? (Sorry, I don't know latex)

The statement is equivalent to showing there's a perfect square $x^2$ such that $2n<x^2<5n$. We now let $y^2$ be the largest perfect square such that $y^2\leq 2n$. We will give a proof by contradiction. Assume $(y+1)^2\geq 5n$. We get $(y+1)^2=y^2+2y+1=2n+2\sqrt{2n}+1\geq 5n$, which we can manipulate to $9n^2 - 14n + 1\leq 0$. From here, we get that $\frac{22}{81}\leq n\leq \frac{103}{81}$, so we only need to check $n=1$. We see that $\sqrt{2}<4<\sqrt{5}$ works. Hence, it works for all integers $n\geq 1$.

akpi2 wrote: wait... if $n \ge 1$ There are two cases: Case I: $n>1$ $\sqrt{5n}-\sqrt{2n} \ge 1$ Therefore there has to be at least an integer between $\sqrt{2n}$ and $\sqrt{5n}$ Case II: $n=1$ $2$ is between $\sqrt2$ & $\sqrt5$ Does the above sol work? (Sorry, I don't know latex) a) added LaTeX b) Yeah, pretty much; I just feel like the step where you stay $\sqrt{5n} - \sqrt{2n} \ge 1 \implies \exists k \mid k \in \left[\sqrt{2n},\sqrt{5n}\right] \cap \mathbb{Z}$ needs more rigorous justification. (That's what I did in mine.)

Thank you DouDragon!