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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.

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

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  • $\begingroup$ 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
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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.

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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 $n \not > 1$.

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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.

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  • $\begingroup$ 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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