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Every proof I've seen of Euler's Theorem (that $\gcd(a,m) = 1 \implies a^{\phi(m)} \equiv 1 \pmod m$) involves the fact that the units of $\mathbb{Z}/m\mathbb{Z}$ form a group of order $\phi(m)$. While this is a perfectly good proof, I have to wonder if it was the one that Euler used. I know that there are fairly old precursors to group theory, but it still seems incongruous.

Thus, my question is: Are there proofs of Euler's Theorem that do not use group/ring theory? In particular, what proof (if any; I don't know whether Euler actually discovered the theorem) did Euler use himself?

Martin Sleziak
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Calvin McPhail-Snyder
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  • Only tangentially related, but: Is there a way to write congruences in LaTeX so that they look nice? The spacing in the question is suboptimal. – Calvin McPhail-Snyder Jul 09 '11 at 18:37
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    use \bmod for no space, and \pmod for parenthesis, i.e. $1 \pmod m$ – Harry Stern Jul 09 '11 at 18:44
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    Euler gave essentially the group-theoretic proof in Chapter 7 of his *Tractatus de numerorum doctrina*, 1849 (according to Weil). – Bill Dubuque Jul 09 '11 at 18:51
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    http://f2.org/maths/nthproof.html check this link –  Jul 09 '11 at 19:08
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    possible duplicate of [Proof of the Euler Generalisation of Fermat's Little Theorem using modular arithmetic](http://math.stackexchange.com/questions/14084/proof-of-the-euler-generalisation-of-fermats-little-theorem-using-modular-arithm). I remember having written the same answer earlier and now I have found the question –  Jul 09 '11 at 19:08
  • @Siv Wasn't your answer simply copied from the prior answer? It appears to be word-for-word identical. – Bill Dubuque Jul 09 '11 at 21:28
  • @Bill: I have proofs of some nice theorems typeset on my comp. And copy pasted it from there. I think I must have done the same for the earlier question as well. No wonder they are identical. –  Jul 09 '11 at 21:37
  • @Siv I've been in the same boat. Alas, often it's quicker to search one's own archives rather than to attempt to use the impotent MSE search tools (Googling with site: ... often works better). – Bill Dubuque Jul 09 '11 at 21:50
  • I wouldn't call this a dupe, as that one only allows a specific kind of proof and this one disallows a specific kind of proof. – Aryabhata Jul 10 '11 at 01:26

1 Answers1

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Consider the set of all numbers less than $n$ and relatively prime to it. Let $\{a_1,a_2,...,a_{\varphi(n)}\}$ be this set.

Consider a number $c < n$ and relatively prime to it i.e. $c \in \{a_1,a_2,\ldots,a_{\varphi(n)}\}$.

First observe that for any $a_i$, $c a_{i} \equiv a_{j} \pmod{n}$ for some $j$. (True since $c$ and $a_i$ are themselves relatively prime to $n$, their product has to be relatively prime to $n$. This follows immediately from the definition).

If $c a_{i} \equiv c a_{j} \pmod{n}$ then $a_i = a_j$. (True as cancellation can be done since $c$ is relatively prime to $n$).

Hence, if we now consider the set $\{ca_1,ca_2,...,ca_{\varphi(n)}\}$ this is just a permutation of the set $\{a_1,a_2,...,a_{\varphi(n)}\}$.

Thereby, we have $\displaystyle \prod_{k=1}^{\varphi(n)} ca_k \equiv \prod_{k=1}^{\varphi(n)} a_k \pmod{n}$.

Hence, we get $\displaystyle c^{\varphi(n)} \prod_{k=1}^{\varphi(n)} a_k \equiv \prod_{k=1}^{\varphi(n)} a_k \pmod{n}$.

Now, note that $\displaystyle \prod_{k=1}^{\varphi(n)} a_k$ is relatively prime to $n$ and hence you can cancel them on both sides to get

$$c^{\varphi(n)} \equiv 1 \pmod{n}$$ whenever $(c,n) = 1$.

kevinji
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    +1, very nice. By the way, this proof is in the Wikipedia article on Euler's theroem. – joriki Jul 09 '11 at 18:56
  • But that *is* essentially the group theoretic way - which the OP seeks to avoid. – Bill Dubuque Jul 09 '11 at 19:01
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    Yes. I think the correct answer may just be "the theorem is fundamentally group-theoretic, so all the proofs are fundamentally group-theoretic (or are disguised versions of such)", but I was hoping for some confirmation. – Calvin McPhail-Snyder Jul 09 '11 at 19:34
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    @Bill I disagree that this is the group theoretic proof. This argument only works in an abelian group. – David E Speyer Jul 09 '11 at 20:07
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    I do think that the first point is that Euler _did_ write as group-theoretic a proof as was possible at that date, and the second point is that, in any case, this result is a poster-child for group theory in number theory. – paul garrett Jul 09 '11 at 20:09
  • @David Of course I know that. But the question concerns only abelian groups. – Bill Dubuque Jul 09 '11 at 21:29
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    This answer doesn't explicitly use any abstract algebra or require any such knowledge on the part of the reader. As paul garrett says, if you know any group theory then inevitably you will see the result itself in that light. If this is not a satisfactory answer to the question, then perhaps the question should be clarified: what would constitute a less algebraic proof than this? – Pete L. Clark Jul 10 '11 at 04:20
  • FWIW, this is a very standard proof, probably the most common proof in elementary number theory texts. Although now that I think about it, I remember that George Andrews' book gives a combinatorial proof of Fermat's Little Theorem. Maybe that argument could be adapted here as well. I remember thinking when I read that argument "Wow, this is longer and more complicated than any five other proofs of FLT I've ever seen, but at least he's having fun." – Pete L. Clark Jul 10 '11 at 04:24
  • BTW, I (re)designed and taught a number theory course which does use some abstract algebra, but not very much, and in particular for which some students have seen little or no group theory. So I wrote up some "crash course" style notes on basic group theory, including Lagrange's Theorem, which I proved (**how else**?) using cosets. After a long time though I realized that I only ever used the special case of Lagrange's Theorem that the order of an element in a finite **abelian** group divides the order of the group. This of course can be proven by exactly the same argument as given here. – Pete L. Clark Jul 10 '11 at 04:28
  • When I teach the course I again I am planning on throwing away all references to "cosets" but instead just prove the above result -- which I have in my mind taken to calling **Lagrange's Little Theorem** -- in the standard "number theoretic" way. In this way elementary number theory is applied to algebra, which is applied to slightly less elementary number theory. (After a while one begins to disbelieve in a distinction between elementary number theory and elementary algebra. I would love to teach a course that does both at once...) – Pete L. Clark Jul 10 '11 at 04:33
  • I think this answer, coupled with Pete's first comment, is definitely a good answer. While it still looks like the group-theoretic proof, that's only because I know group theory. – Calvin McPhail-Snyder Jul 10 '11 at 18:20
  • @Calvin: You might want to take a look at the proof by Arturo here. http://math.stackexchange.com/questions/14084/proof-of-the-euler-generalisation-of-fermats-little-theorem-using-modular-arithm/14086#14086. But again it depends on how you want to view it. –  Jul 10 '11 at 18:26
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    @PeteL.Clark: when I have taught undergraduate abstract algebra, early in the course I prove two results about every element $g$ in a finite abelian group $G$: (i) $g^{|G|} = 1$ and (ii) the order of $g$ divides $|G|$. First I prove (i) and then I prove (ii) using (i). Later on I prove these same results in any finite group, but I have to switch the order of reasoning, to get (ii) first and then (i), since the method of cosets naturally leads to (ii). – KCd Nov 23 '13 at 01:57