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As the title states, I'm supposed to prove for distinct primes $p_1,p_2 >2$ the primes dividing $2^{p_1}-1$ and $2^{p_2}-1$ are distinct.

The Wikipedia page on Theorems about Mersenne Numbers states that #6: if $m$ and $n$ are natural numbers then $m$ and $n$ are coprime if and only if $2^m-1$ and $2^n-1$ are coprime, the only one without proof. I believe I have to prove the forward $(\Longrightarrow)$ direction.

edit: this was my own work for a homework question, and I was mainly wondering if the proof was okay. there is no proof directly provided on Wikipedia, and I didn’t find the links to the proof the clearest either. this was the clearest proof to me at the time and wanted someone else’s feedback because I couldn’t believe my proof was right.

My attempt at a proof (I use $a,b$ instead of $p_1,p_2$ or $m,n$):

Suppose $(a,b)=1$ and $(2^a - 1, 2^b -1) = d$.

Since $(a,b)=1$ there exist $x,y \in \mathbb{Z}$ s.t. $ax+by=1$.

Since $a,b > 2$, $x$ and $y$ cannot both be greater than zero, otherwise $ax+by > 2+2 >1$. Neither can $x$ or $y$ be equal to zero, otherwise $x$ or $y$ would $\not\in\mathbb{Z}$, or $ax+by=0\not=1$. Neither can $x$ and $y$ be less than zero, otherwise then $ax+by <0<1$.

Thus one of $x$ or $y$ must be positive, while the other negative.

Suppose $ax>0$ then $by<0$, then $d|2^{ax}-1$ and $d|2^{-by}-1$ since

$$2^{ax}-1 = (2^a -1)((2^a)^{x-1}+(2^a)^{x-2}+\cdots+1)$$

$$2^{-by}-1 = (2^b -1)((2^b)^{-y-1}+(2^b)^{-y-2}+\cdots+1)$$

and $d|2^a -1$ and $d|2^b -1$ by definition. Then $2^{ax} - 1 = d\alpha$ and $2^{-by} -1 = d\beta$ and then $2^{ax}-2^{-by} = 2^{-by}(2^{ax+by}-1)=2^{-by}(2^1-1)=2^{-by} = d(\alpha -\beta)$ thus $d|2^{-by}$ and $d$ must be a power of $2$. But $d|2^a-1$ where $2^a-1$ is odd, and the only odd power of $2$ is $2^0=1$, hence $d=1$. If $(a,b)=1$ then $(2^a -1 ,2^b-1)=d=1$.

Is my proof rigorous or am I missing something?

no lemon no melon
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    Looks good to me. You have a typo in the line "Suppose $ax>0$ then $bx<0$...." which should say "... then $by<0$...". – DanielWainfleet Feb 28 '21 at 06:34
  • @DanielWainfleet Is there anything special as to avoiding the only even prime $p=2$? I was looking at it and would including $2$ only change $>$ to $\geq$ in $a,b>2$ and $ax+by>2+2>1$? Thus the question I'm asked to solve might be over specific or a typo, and we could include $2$ as well? Sweet catch btw thx – no lemon no melon Feb 28 '21 at 06:40
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    On re-reading it, I think you can say that the case $1=\min(a,b)$ is trivial as it implies $\min(2^a-1,2^b-1)=1$. If $1<\min(a,b)$ then WLOG $by<0<ax$ and the rest of the proof works as is, including the case $a=2\lor b=2$. – DanielWainfleet Mar 01 '21 at 19:32
  • @DanielWainfleet thanks – no lemon no melon Mar 01 '21 at 19:37
  • As you probably know, it is unknown whether there are infinitely many Mersenne $primes$, but quite a few Mersenne numbers are known to be prime, including the largest explicitly known primes. Prior to the computer era, (& prior to many advances in modern Number Theory) the largest explicitly known prime was $2^{127}-1.$ – DanielWainfleet Mar 01 '21 at 19:49
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    No I was not aware, neat: a Mersenne prime $(2^{127} -1)$, made by another Mersenne prime $(127=128-1 = 2^7 -1)$ made by another Mersenne prime $(7=8-1 = 2^3 -1)$, and calculating $2^{127}$ is doable by hand but cumbersome at best, that's impressive that they knew $2^{127}-1$ was prime without computers! – no lemon no melon Mar 01 '21 at 20:13
  • And $ 7$ is made by the Mersenne prime $3=2^2-1!.$ I never noticed any of that. So the next one at the top should be $2^{2^{127}-1}-1;??$ :). – DanielWainfleet Mar 01 '21 at 21:34
  • that would blow out the largest known Mersenne prime by far! $2^{127} > 10^{38}$ how could the ancients know $2^{127}-1$ was prime? – no lemon no melon Mar 01 '21 at 22:23
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    https://mathshistory.st-andrews.ac.uk/Biographies/Lucas/ – DanielWainfleet Mar 03 '21 at 04:35
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    https://math.stackexchange.com/questions/4045202/mersenne-primes-before-computers – no lemon no melon Mar 03 '21 at 04:42
  • @BillDubuque I meant to close the question yesterday, but hastily pressed no, I voted yesterday to close yesterday, I was vote #2 – no lemon no melon May 12 '21 at 20:07

2 Answers2

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Your proof is fine, though you might also need to mention the case $ax < 0$ and $by > 0$ (though immediate).

By Wikipedia standards, most theorems are accompanied by a proof, or if not possible, a reference to the proof would be provided. In this case, the statement of this theorem is accompanied with a "[20]", which links to an archived version of Will Edgington's Mersenne Page. The various Lemmas in the section Mersenne Number Proofs supplies a proof to the claim.

Discussions on MSE that are also useful include:

player3236
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Looks rigorous enough. But there's a more elementary proof with a slightly more general definition of Mersenne numbers( where exponents are natural numbers):

$$2(2^n-1)+1=2^{n+1}-1$$ since $$2^1-1=1$$ is a sequence member, iteration produces the general recurrence :$$2^{n+m}-1=2^m(2^n-1)+(2^m-1)$$ This means for two Mersennes to share a factor, the Mersenne with exponent their difference in exponent must share it.

Euclidean gcd is based on taking the difference of two values, and saying the gcd must be a factor. But, applying that in the exponent repeatedly, gives us the fact that two coprime exponent Mersenne numbers, can only share a factor if: $$2^1-1=1$$ shares it. Therefore they can only share the factor 1.

Roddy MacPhee
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