12
$\begingroup$

Let $p$ be a fixed odd prime. Let $x,y\in \mathbb{Z}_+$ such that $\sqrt{x}+\sqrt{y}<\sqrt{2p}$. Prove that $$\sqrt{x}+\sqrt{y}\le \sqrt{\frac{p-1}{2}}+\sqrt{\frac{p+1}{2}}.$$

Any ideas at all? This seems extremely difficult to do using elementary methods.

Note: It is from the 2015 Moldova TST (IMO selection test). The original problem was: Let $p$ be a fixed odd prime and $x,y\in \mathbb{Z}_+$. Find the minimum positive value of $\sqrt{2p}-\sqrt{x}-\sqrt{y}$.

EDIT! : This is apparently an old IMO Shortlist problem which Moldova recycled for their TST. I note that my reformulation of it was correct; see here for a solution: See here.

$\endgroup$
  • $\begingroup$ The first thing I'd try is squaring everything. If $\sqrt x + \sqrt y < \sqrt{2p}$, is it necessarily also true $x + y < 2p$? This is just an idea, I haven't thought it all the way through. $\endgroup$ – John-Luke Mar 31 '15 at 21:07
  • $\begingroup$ You may want to write your answer in the comments so that this question is marked as answered. Relevant meta: meta.math.stackexchange.com/questions/1559/… $\endgroup$ – Element118 Nov 14 '15 at 8:50
  • $\begingroup$ Here should be proved a more general claim. $\endgroup$ – Alex Ravsky May 29 '17 at 1:12
1
$\begingroup$

Let's suppose that the problem is not true and there exists $x,y$ such that $$\sqrt{\frac{p-1}{2}}+\sqrt{\frac{p+1}{2}}<\sqrt{x}+\sqrt{y}<\sqrt{2p}\qquad (1)$$ First, note that if $x+y\le p$ then $$\sqrt{x}+\sqrt{y}\le\sqrt{x}+\sqrt{p-x}\le\sqrt{\frac{p-1}{2}}+\sqrt{\frac{p+1}{2}}$$ Since $\sqrt{x}+\sqrt{p-x}$ is concave function and maximal at $x=p/2$. Squaring both sides of $(1)$ gives $$p+\sqrt{p^2-1}<x+y+2\sqrt{xy}<2p$$ Now, because we showed $x+y>p$, let $\epsilon=p-\sqrt{p^2-1}$ then it is enough to prove that there is no integer $n, m$ with $m<p$ and $$m-\epsilon<2\sqrt{n}<m\qquad (2)$$ Now, suppose that $m$ is even and there exists $k$ such that $2k=m$. Then we get $k<p/2$, also $n<k^2$, which is $n\le k^2-1$ and $$2k-\epsilon <2 \sqrt{n}\le 2\sqrt{k^2-1}$$ $$2(k-\sqrt{k^2-1})<p-\sqrt{p^2-1}$$ However, since $f(x)=\sqrt{x}-\sqrt{x-1}$ is strictly decreasing function for positive $x$, therefore $$\begin{align}2(k-\sqrt{k^2-1})& > 2\left(\frac{p}{2}-\sqrt{\frac{p^2}{4}-1}\right)\\& =p-\sqrt{p^2-4}\\& > p-\sqrt{p^2-1}\end{align}$$ And this is contradiction.

Now we suppose that $m$ is odd and there exists $k$ such that $m=2k+1$. Also we get $4n < 4k^2+4k+1$, so $n \le k^2+k$. Also, from $2k+1-\epsilon<2\sqrt{n}\le2\sqrt{k^2+k}$,$$2k+1-2\sqrt{k^2+k}<p-\sqrt{p^2-1}$$ However, this cannot be true, because $$\begin{align}2k+1-2\sqrt{k^2+k}& = 2k+1-\sqrt{(2k+1)^2-1}\\& =m-\sqrt{m^2-1}\\& > p-\sqrt{p^2-1}\end{align}$$

Therefore, this is contradiction and there exists no integers $m,n$ with $(2)$, and it follows that there exists no integers $x,y$ satisfying $(1)$. Proved!

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.