Prove that $\sqrt {2^n-1}$ is irrational for every integer $ n>1$

I tried assuming it was equal to $\frac p q $.

I get $2^nq^2-q^2 = p^2 $

But I don't see where to go from there.


For $n\geq2$, $2^n-1\equiv 3\pmod 4$, unlike a perfect square.

  • $\begingroup$ To note for the questioner, the square root of non-square number is irrational. $\endgroup$ – Hanul Jeon Aug 4 '16 at 16:37
  • $\begingroup$ So you are using that $\sqrt a$ is either an integer or an irrational number $\endgroup$ – punctured dusk Aug 4 '16 at 16:37
  • $\begingroup$ @HanulJeon I was actually just waiting for the questioner to ask if he doesn't get the hint :) $\endgroup$ – Landon Carter Aug 4 '16 at 16:39
  • $\begingroup$ Wow, you guys are fast! $\endgroup$ – Francois Grandchamp Aug 4 '16 at 16:44
  • $\begingroup$ I didn't see that $2^n-1$ was congruent to 3 mod 4. I will look into why the square root of such numbers are always irrational $\endgroup$ – Francois Grandchamp Aug 4 '16 at 16:47

Landon Carter's answer is direct and has the elegance of simplicity. Here is a partial answer for the case $n=2k$, an even number. I too use the fact that if the square root is rational it has to be an integer.

So we assume $2^{2k}-1$ is a square and will get a contradiction. That is, $(2^k-1)(2^k+1)$ is a square. The these bracketed quantities are both odd and differ by 2, hence have no common factors.

So this forces both the numbers i.e., $2^{2k}\pm1$ to be perfect squares. AT the beginning the squares $1$ and $4$ differ by $3$, afterwards squares have to differ by more than 3.

Here we have two squares differing by 2. Contradiction.


To continue your idea, rather than to restart and do one of the other correct answers.

$2^nq^2 - q^2 = p^2$.

Assume $p = 1$

The $2^nq^2 - q^2 = 1$. Then $2^n - 1 = 1/q^2$ is an integer. So $q=1$. So $2^n - 1 = 1$. So $2^n = 2$ so $n = 1 \not > 1$.

Assume $p \ne 1$.

Let $k$ be a prime factor of $p$. Then either $k|q^2$ which isn't possible as we assumed (or should have assumed) $p/q$ is in lowest terms, or $k|2^n - 1$.

This is true for all prime factors so $p^2|2^n -1$.

So $q^2\frac{2^n - 1}{p^2} = 1$ and $\frac{2^n - 1}{p^2} \in \mathbb Z$ so $\frac{2^n - 1}{p^2} = 1/q^2 \in \mathbb Z$ so $q = \pm 1$ and

$2^n - 1 = p^2$.

$p$ is clearly odd so let $p = 2p' + 1$

$2^n = 4p'^2 + 4p' + 2$

$4 \not \mid 4p'^2 + 4p +2 = 2^n$ so $n < 2$. So $n \le 1$. So $n \not > 1$.


It is clearly equivalent to prove that the diophantine equation $$2^n=x^2+1$$ has no integer solution for $n\gt 1$ Notice that $x$ must be odd so $x=2m+1$ hence $$2^n=(4m^2+4m+1)+1=4m^2+4m+2\iff2^{n-1}=2m^2+2m+1$$ This is absurde.

  • $\begingroup$ It was obviously a lapsus taking $x=2n+1$ with the same symbol $n$ that the exponent. This maked, I hope you change your downvote because if not you are doing the worst in maths (¿what is it?) $\endgroup$ – Piquito Aug 4 '16 at 17:40
  • $\begingroup$ Not sure why it should by clear or obvious that $2^n = x^2 +1$ having no rational solutions is equivalent to showing it has no integer solutions. It is, but I don't don't see why it is "clear". Unless you assume $x^2 = p/q$ implies $p/q$ is an integer. Which does need to be proven at some point or another. Probably around the same time as the student would be asked this problem. $\endgroup$ – fleablood Aug 4 '16 at 20:25
  • $\begingroup$ Do you have in hand an example of $2^n−1$ rational non integer? I want to confess frankly that I wanted to write yesterday a comment like this: “Do not be a coward and justify your downvote”. But I censured myself and decided not to write it, I gave it as past history, without any bitterness, believe me, please. $\endgroup$ – Piquito Aug 5 '16 at 23:12
  • $\begingroup$ Of course, I don't. But it wouldn't be obvious to a student at this level that there can't be. " If $\sqrt{m}$ is rational then $m$ is a perfect square" is something a student needs to be but hasn't nescessarily yet proven. Why are you talking about downvoting? I didn't downvote you. $\endgroup$ – fleablood Aug 5 '16 at 23:20
  • $\begingroup$ Anyway, what I have told you about my censured comment is true and this is not necessarily directed to you. Regards. $\endgroup$ – Piquito Aug 6 '16 at 2:02

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