I was wondering if there exists a proof from abstract algebra of this theorem. It seems like there should be. In fact, I've tried to come up with one and it always seems I'm coming close, but I can't do it.
Could you provide one?
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1It think you mean to ask about $2^{n} - 1$, not $2^{n-1}$. $2^{n-1}$ is prime if and only if $n = 2$. – Ben Grossmann Nov 28 '16 at 15:26
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Do you mean the Mersenne Primes, i.e. $2^n-1$? – AlgorithmsX Nov 28 '16 at 15:27
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Well, what's wrong with "if $n=pq$ then $2^{n-1}$ is divisible by $2^p-1$ and by $2^q-1$"? – barak manos Nov 28 '16 at 15:30
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1The main issue here is: What do you consider as sufficiently abstract to fit your desire? – MooS Nov 28 '16 at 15:35
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If $n$ is not prime, then $2^n - 1$ is not prime.
Suppose that $n = ab$ for integers $a,b\geq2$. Then note that we can factor $$ x^n - 1 = (x^a)^b - 1 = (x^a - 1)([\text{stuff}]) $$ which means that $$ 2^n - 1 = (2^a)^b - 1 = (2^a - 1)([\text{stuff}]) $$ where stuff is, notably, an integer and bigger than $1$. It follows that $2^n-1$ is not prime.
Ben Grossmann
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We can eliminate the crufty "stuff" by using modular arithmetic, e.g. see my answer. – Bill Dubuque Nov 28 '16 at 19:18
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1Ha! Of course it's not your "stuff" but, rather, the division algorithm's, viz. the crufty quotients - which we don't need when we are only interested in the remainder. One could call modular arithmetic the art of eliminating crufty quotients. – Bill Dubuque Nov 28 '16 at 19:43
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Hint $\,\ a^{\large k}\!-\!1\mid a^{\large kn}\!-\!1\ $ by $\ {\rm mod}\,\ \color{#c00}{a^{\large k}\!-\!1}\!:\,\ a^{\large kn}\equiv \overbrace{(\color{#c00}{a^{\large k}})^{\large n}\equiv \color{#c00}1^{\large n}}^{\Large (\color{#c00}{a^k\,\ \equiv\,\ 1})^{\Large n}}\equiv 1\,$ by Congruence Rules.
This is a special case of the Factor Theorem $\ b\!-\!1\mid f(b)-f(1)\,$ for $\,f(x)\,$ an integer coef polynomial, since, like above $\ {\rm mod}\,\ b\!-\!1\!:\,\ \color{#c00}{b\equiv 1}\,\Rightarrow\, f(\color{#c00}b)\equiv f(\color{#c00}1).\, $ In the OP $\ f = x^{\large n},\ \ b = a^{\large k}.\ $ The prior implication is a special case of one of the linked Congruence Rules, namely the Polynomial Congruence Rule. whereas the OP is simply the special case when the polynomial is a power $\,f(x) = x^n\,$ (Power Congruence Rule) - clear from the overbraced powering in the proof.
Thus the proof follows very naturally from the viewpoint of modular arithmetic or congruences. They have the advantage that one doesn't need to compute any quotients (only remainders, or congruent numbers), which greatly simplifies many proofs.
Remark $\ $ One can prove further that $\, (a^{\large k}\!-\!1,a^{\large j}\!-\!1) = a^{\large (k,j)}\!-\!1\,$ where $\,(x,y):=\gcd(x,y).\,$ This is a special case of a strong divisibility sequence.
Bill Dubuque
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