Divisibility of $2^x-1 $

Question

This is one of the first questions on round 1 of the Korean Math Olympiad

Find all natural numbers x, y such that $xy=2^x-1$

I "know" that the only solution is $(x,y)=(1,1)$, but I cannot find a proof of this

So $0 \notin \mathbb N$? Because $(0,y)$ with $y \in \mathbb N$ would be also possible — ulead86, Jan 13 '14 at 09:59
@KarthikT i have tried to answer this question by different way,please see it — dato datuashvili, Jan 13 '14 at 10:47

score 2 · Accepted Answer · answered Jan 13 '14 at 10:01

2

Lemma. If $a\not\equiv1,a^n\equiv1$ mod $m$ then $(n,\varphi(m))>1$.

Let $p\mid x$ be the smallest prime divisor. Now $p>2$, and $2^x\equiv1$ mod $p$ implies $(x,p-1)>1$ ...

answered Jan 13 '14 at 10:01

anon

i have tried using arithmetic comparison – dato datuashvili Jan 13 '14 at 10:46
i have deleted my answer,good lucks – dato datuashvili Jan 13 '14 at 11:14
thank you for downvoting,but i have deleted already :D – dato datuashvili Jan 13 '14 at 11:16
This is brilliant! Thanks a lot for this, I knew I had to look for a contradiction, but had no clue as to where to start! – KarthikT Jan 13 '14 at 18:37

1 Answers1