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Firstly, I do not mind if there are examples from fields other than number theory! This was just the first field, and where I think the richest examples, may come from.

Now to elaborate on the title, if only a little: Are there any elementary number theory proofs that have since been made using theorems (such as Fermat's Last Theorem or Goldbach's Weak Theorem $^{1}$) that went unproven for a long period of time? Or, in general, elementary proofs using theorems that required an enormous amount of machinery (algebraic geometry, modularity theorem, etc.)?

I wouldn't mind seeing elementary proofs of a result that had been already proven. But new proofs would also be very intriguing. Part of the reason for thinking in this way is this: I wonder where people in the previous era of mathematics would have looked next had they been able to use these theorems. Thanks for the help.


$^1$ this is a theorem now, right?

Edit: Added the big-list tag and bounty in hopes of more answers.

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  • $\begingroup$ Counts something so silly like Goldbach $\implies$ every even number $\ge 6$ is a sum of two odd numbers? $\endgroup$ – Martín-Blas Pérez Pinilla Jan 13 '15 at 10:16
  • $\begingroup$ Do you need that the proof of the big theorem didn't use the elementary result that you use it to prove? For example (and not a good answer since there was no great delay proving the first result): Cauchy's Theorem implies Cauchy's Theorem For A Triangle, but if you've just spend the day proving the former from the latter, this result is extremely uninteresting! Similarly, FLT proves FLT for the case n=3, proved over 200 years previous to FLT, but again ain't interesting. No example has yet come to mind where the little theorem isn't exactly just a special case of the big one, though. $\endgroup$ – Steve Jessop Jan 13 '15 at 12:42
  • $\begingroup$ @SteveJessop Oh I see. That's interesting but I was hoping for results that were not special cases of the big theorem. I certainly had not thought of that though. $\endgroup$ – Eoin Jan 13 '15 at 21:07
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    $\begingroup$ See mathoverflow.net/questions/42512/… $\endgroup$ – John Gowers Jan 23 '15 at 13:23
  • $\begingroup$ The phrase "elementary number theory proofs" often means something other than proofs of number theory results that are easy to follow. Historically some important results in number theory were proven using complex analysis, and methods of proof that did not entail such excursions outside the properties of integer arithmetic became touted as "elementary" (even if they might be fairly involved, c.f. elementary proofs of the Prime Number Thm.. $\endgroup$ – hardmath Jan 24 '15 at 17:41
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I'm not sure I understand your question correctly. Are you asking for elementary results proved using much more complicated theorems? If so, here's my favourite example.

We will prove that the $n^{th}$ root of $2$ is irrational, for $n>2$

Suppose $\sqrt[n]{2}=\frac{p}{q}$, then $2 = \frac{p^n}{q^n}$, which implies $p^n=q^n+q^n$

This contradicts Fermat's Last Theorem, so $\sqrt[n]{2}$ must be irrational.

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  • $\begingroup$ That's a perfect example of what I was thinking! $\endgroup$ – Eoin Jan 13 '15 at 5:29
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    $\begingroup$ And, of course, as the joke goes, FLT isn't strong enough to prove the irrationality of $\sqrt{2}$ itself :) $\endgroup$ – Thomas Jan 13 '15 at 9:37
  • $\begingroup$ @MathN00b That's correct. However, that's all we needed to prove. $\endgroup$ – hasnohat Jan 18 '15 at 3:12
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In my old paper I used Four Colors Theorem to prove a two dimensional case of the following

Proposition. Every open subset of the space $\Bbb R^m$ can be partitioned into $n$ homeomorphic parts if $n\ge 2^{m+1}-1$ or $m=2$ and $n\ge 4$.

Proof. For every positive integer $k$ let $A_k$ be the set of all points $x\in\Bbb R^m$ such that all coordinates of the point $2^{k-1}x$ are integer. Let $\mathcal W_k$ be the family of sets in the form $\{x+[0;2^{1-k})^m\}$ where $x\in A_k$. The point $x$ we shall call the vertex of the set $x+[0;2^{1-k})^m$. We call the subsets $W_1$ and $W_2$ of $\Bbb R^m$ separated if $\overline{W_1}\cap W_2=\varnothing=\overline{W_2}\cap W_1$. We shall need the following

Lemma. Let $k$ be a positive integer and $x_1$, $x_2$ be the vertices of the sets $W_1,W_2\in\mathcal W_k$ respectively. If $x_2\not\in\overline{W_1}$ and $x_1\not\in\overline{W_2}$ then the sets $W_1$ and $W_2$ are separated.

Proof. Suppose the contrary. Without loss of generality we may suppose that $k=1$, $W_1\not=W_2$ and there is a point $x\in\overline{W_1}\cap W_2$. Let $i$ be an arbitrary index, $1\le i\le m$. Since $x\in\overline{W_1}$ then $x_1^i\le x^i\le x_1^i+1$ where $x^i$ and $x_1^i$ are the $i$-th coordinates of the points $x$ and $x_1$ respectively. Since $x_2^i=\lfloor x^i\rfloor$ then $x_1^i\le x_2^i\le x_1^i+1$. Because this condition holds for every $i$ we have that $x_2\in\overline{W_1}$.$\square$

Lemma implies that for every $k$ and each $W\in\mathcal W_k$ the family $\{W'\in\mathcal W_k:W'$ and $W$ are not separated $\}$ has the size equal to the doubled number of the vertices of the $m$-dimensional cube minus $1$ which is $2^{m+1}-1\le n$.

Now define by induction the families $\mathcal V_k\subset\mathcal W_k$. Put $\mathcal V_1=\{W\in\mathcal W_1:W\subset X\}$ and $\mathcal V_k=\{W\in\mathcal W_k:W\subset X\setminus\bigcup_{i=1}^{k-1}\bigcup\mathcal V_i\}$ for $k\ge 2$. Put $\mathcal V=\bigcup\mathcal V_k$. By the construction $X=\bigcup\mathcal V$. If the family $\mathcal V$ is finite then we move the coordinate center by the irrational distance along one of the coordinate axes and repeat the construction obtaining the infinite family $\mathcal V$.

Now we assign to each member $V\in\mathcal V$ one of $n$ colors in such that the following condition holds

(*) every monochromatic sets $V,V'\in\mathcal V$ are separated.

At first we suppose that $n\ge 2^{m+1}-1$. Enumerate every family $\mathcal V_k$ in an arbitrary order. Now we shall color the family $\mathcal V$ in the following way using no more than $n$ colors. Firstly we color all members of the family $\mathcal V_1$ in the enumerated order, then all members of the family $\mathcal V_2$ in the enumerated order and so on. We claim that at each step of the process for the currently colored member $V\in\mathcal V$ there is no more than $2^{m+1}-2<n$ already colored members $W\in\mathcal V$ such that $V$ and $W$ are not separated. Indeed, let $V\in\mathcal V_k$ for some $k$. If $W$ is already colored and $W$ and $V$ are not separated then there is a set $W'\in\mathcal W_k$ such that $W'\subset W$ and the sets $W'$ and $V$ are not separated. Since the family $\mathcal W_k$ is disjoint it implies the claim. So we may color the member $V$ into a color different from the colors of already colored not separated from $V$ members of $\mathcal V$. The constructed coloring satisfies the condition (*).

Let now $m = 2$ and $n\ge 4$. Determine the graph $G$ as follows. As the set of vertices $V(G)$ of the graph $G$ we take the set of {\it the centers} of the closures of the family $\mathcal V$ elements. Let $V_1, V_2\in\mathcal V $ and $c_1, c_2\in V(G)$ be the centers of the squares $\overline{V_1}$ and $\overline{V_2}$ respectively. The vertices $c_1 $ and $c_2$ are connected by an an edge iff the sets $V_1$ and $V_2$ are not separated. As the edge we shall consider a segment $[c_1,c_2]$ of the plane. It can be checked that $[c_1;c_2]\subset V_1\cup V_2$ provided the vertices $c_1$ and $c_2$ are connected. Now we show that the graph $G$ is planar. Suppose that the edges $[c_1;c_2]$ and $[c_3;c_4]$ of the graph $G$ are intersected. Since $[c_1,c_2]\subset V_1\cup V_2$ and $[c_3;c_4]\subset V_3\cup V_4$ and the family $\mathcal V$ is disjoint then $\{c_1,c_2\}\cap \{c_3,c_4\}\not=\varnothing$. Without loss of generality we may suppose that $c_1=c_3$ and $c_2\not=c_4$. Then $c_1$ is the unique common point of the segments $[c_1;c_2]$ and $[c_3;c_4]$. Indeed, otherwise one of these segments would contain the other, which is impossible since the family $\mathcal V$ is disjoint. Therefore the graph $G$ is planar. Since $n\ge 4$ then the vertices of the graph $G$ can be colored in no more than $n$ colours such that arbitrary two monochromatic vertices are not connected by an edge. From here we obtain the coloring of the family $\mathcal V$ in no more than $n$ colors satisfying the condition (*).

For an index $i$ let $\mathcal V^i$ denotes the members of $\mathcal V$ colored into the color $i$. Since the family $\mathcal V$ is infinite then during the coloring we can easily ensure that precisely $n$ colors are used and every family $\mathcal V^i$ for $1\le 1\le n$ is infinite.

Let $x\in X$. Since $X$ is open in $\Bbb R^m$ then there is a number $k$ such that the natural cube neighborhood of $x$ with the side $2^{-k}$ is contained in $X$. Then the set $\bigcup_{j=1}^{k+1}\bigcup\mathcal V_j$ is a neighborhood of $x$ and since each family $\mathcal V_j$ is locally finite then the family $\mathcal V$ is locally finite too. This implies that the family $\mathcal V^i$ is locally finite for each $i$. Since every two members of the family $\mathcal V^i$ are separated then every member of $\mathcal V^i$ is open in the union $\bigcup_{i}\bigcup\mathcal V^i$. Hence for each $i$ the union $\bigcup\mathcal V^i$ is homeomorphic to the space $[0;1)^m\times\Bbb N$ which yields the partition of $X$ onto $n$ homeomorphic parts.$\square$

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There's also a well known overpowered proof of the infinitude of primes:

If $P$ is the set of primes, the Euler product formula tells us $$\prod\limits_{p\in P}\frac{1}{1-\frac{1}{p^2}}=\zeta(2)=\frac{\pi^2}{6}$$

However, $\pi$ is transendental, so $|P|$ cannot be finite (otherwise the product would be rational).

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    $\begingroup$ To be fair, it was known $\pi^2$ was irrational well before it was known that $\pi$ is transcendental. $\endgroup$ – user208259 Jan 22 '15 at 22:37
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If Euclid knew that the series of reciprocals of prime numbers diverges then we would not have his proof that there are infinitely many primes. :-) (see more here).

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