TITLE: Using Overpowered Theorems to Solve Easy Problems
QUESTION [41 upvotes]: I thought it would be interesting to start a thread about using overpowered theorems to solve easy problems. Two examples come to mind. Post your favorite example of problem and solution!
1). $\sqrt[3]{2}$ is irrational.
Proof: Suppose $\sqrt[3]{2} = \frac{a}{b}$ for $a,b \in \mathbb{N}$. Then, $a^3=2b^3 = b^3+b^3$, contradicting Fermat's Last Theorem.
2). There are infinitely many primes.
Proof: By the Prime Number Theorem, the asymptotic density of primes is $\frac{x}{\ln(x)}$, so by L'Hopitals Rule, $\lim_{x \to \infty} \frac{x}{\ln(x)} = \lim_{x \to \infty} \frac{1}{1/x} = \lim_{x \to \infty} x = \infty$, so there must be infinitely many primes.

REPLY [8 votes]: Here, we will prove by exhaustion that the solution for $x$ in $ax-b=0$ is $x=\frac ba$ if $a$ and $b$ are integers.
First, apply the rational roots theorem to see that all rational roots must be of the form $x=\pm\frac pq$, where $p$ and $q$ are the factors of $b$ and $a$ respectively.  Since this must be done in general, we can't directly factor $a$ and $b$, so we just cheat by noting that all factors must be whole numbers between $1$ and $a$ or $b$ inclusively.  Thus,
$$x\stackrel?=\pm\begin{cases}\frac11&\frac21&\frac31&\dots&\frac b1\\\frac12&\frac22&\frac32&\dots&\frac b2\\\frac13&\frac23&\frac33&\dots&\frac b3\\\vdots&\vdots&\vdots&\ddots&\vdots\\\frac1a&\frac2a&\frac3a&\dots&\frac ba\end{cases}$$
And since there is a finite amount of cases ($a\times b$ at worst), a proof by exhaustion is possible.  Going through, we find $x=\frac ba$ to be one such solution, and by the fundamental theorem of algebra, there is at most one distinct root, so all other solutions we may have found in our trial and error are equal to this one.