What are some elementary results (number theory) using theorems that went long-unproven? 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.
 A: 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.$\square$
A: 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).
A: 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.
A: 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).
