# Topic 4 Notes
###### tags: `525` `math` `history`

## Fermat

### [Fermat's right triangle theorem (1670)](https://en.wikipedia.org/wiki/Fermat%27s_right_triangle_theorem#:~:text=Fermat's%20right%20triangle%20theorem%20is,complete%20proof%20given%20by%20Fermat.&text=A%20right%20triangle%20and%20a,sides%20commensurate%20with%20each%20other.)

This was stated, but not proved by Fibonacci in response to another problem posed as a challenge. (See the reference above.) Fermat never published a proof of this, but wrote one down in the very same copy of Diophantus's *Arithmetica* where he mentions having a proof of no $n\ge 3$ such that $x^n+y^n=z^n$ has a solution. Apparently, the margin was big enough to hold the present argument.

**Theorem** A right triangle with rational sides cannot have square rational area.

**Proof** First notice that *rational* can be replaced by *integer*. (Easy exercise.) (It is important that Fermat moved the ideas of Diophantus from rationals to integers and hence truly started the field of number theory, wherein primality, parity, etc., become relevant.)

We know exactly what integers can be the legs of a right triangle, namely those of the form $((m^2-n^2)r,(2mn)r,(m^2+n^2)r)$. The area would be $mn(m^2-n^2)r^2=s^2$. Let $d=\gcd(m,n)$, then we could factor out $dr$ from each side and the area would still be a square. So we may assume that $m$ and $n$ are relatively prime.

So now we are in the situation where
$$
A=mn(m-n)(m+n)=a^2
$$
But $m,n,m+n,m-n$ are relatively prime. (Easy exercise.) This means that $m$, $n$, $m-n$ and $m+n$ are perfect squares. Let $m=\bar m^2$, $n=\bar n^2$, $m+n=r^2$ and $m-n=s^2$.

For $A$ to be an integer at all, exactly one of $m$ or $n$ is even and both $m+n$ and $m-n$ are odd. WLOG assume $n$ is even. Thus $r$ and $s$ are odd so $r+s$ and $r-s$ are even and one is divisible by 4 since $(r-s)(r+s)=r^2-s^2=2n=2\bar n^2$ and $\bar n$ is even. Let $u=(r-s)/2$ and $v=(r+s)/2$. We have
$$
u^2+v^2=(r^2-2rs+s^2)/4+(r^2+2rs+s^2)/4=(r^2+s^2)/2=m=\bar m^2.
$$
So $(u,v,m)$ is a Pythagorean triple with area $\frac{1}{2}uv=\frac{1}{8}(r-s)(r+s)=\frac{1}{8}(r^2-s^2)=\frac{1}{4}\bar n^2$.

So we started with a Pythagorean triple $(a,b,c)$ so that $\frac{1}{2}ab$ is a square and got a new Pythagorean triple $(\bar a,\bar b,\bar c)$ where $\bar c<c$ and $\frac{1}{2}\bar a\bar b$ is a square. This leads to an infinite descent.

**Corollary** (to the proof) There are no integers $a,b,c$ satisfying $a^4-b^4=c^2$.

**Proof** We may assume $a,b,c$ are all relatively prime. We have
$$
(a^2-b^2)(a^2+b^2)=c^2
$$
and hence
$$
a^2+b^2=n^2\text{ and }a^2-b^2=m^2
$$
Let $u=(n-m)$ and $v=(n+m)$, so $uv=n^2-m^2=2b^2$ whereas, $u^2+v^2=2n^2+2m^2=4a^2$. Thus $(u,v,2a)$ is an integer triangle with square area.

**Corollary**: There are no positive integers satisfying $x^4+y^4=z^4$.

## Euler

### Infinitude of primes

Consider $\zeta(s)=\sum_{n=1}^\infty \frac{1}{n^s}$. For $s=1$, this gives the harmonic series
$$
\zeta(1)=1+\frac{1}{2}+\frac{1}{3}+\cdots,
$$
which diverges. (Proof: Group terms by powers of 2.) The fundamental theorem of algebra tells us that $n^s= (p_1^{m_1})^s\cdot (p_2^{m_2})^s\cdots (p_k^{m_k})^s$ So we get
$$
\zeta(s)=\prod_{p\text{ prime}}\left(1+\frac{1}{p^s}+\left(\frac{1}{p^s}\right)^2+\cdots\right)=\prod_{p\text{ prime}}\left(\frac{1}{1-p^{-s}}\right)
$$
If there were only finitely many primes, then $\zeta(1)$ would converge.

### The Basel Problem

The goal here was to compute $\zeta(2)$, in fact, Euler computed $\zeta(2n)$ for all positive n. Here is the proof for
$$
\frac{\pi^2}{6}=\zeta(2)=1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots
$$
The trick here is to use the Taylor series for $\sin(x)$ in a tricky way.
$$
\sin(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots
$$
so
$$
\frac{sin(x)}{x} = 1-\frac{x^2}{3!}+\frac{x^4}{5!}-\cdots
$$
The roots of this are $\pm n\pi$ for $n$ a positive integer. The leading coefficient is $1$, so we have
$$
\begin{align}
\frac{\sin(x)}{x} = &1-\frac{x^2}{3!}+\frac{x^4}{5!}-\cdots=\\
&\left(1-\frac{x}{\pi}\right)\left(1+\frac{x}{\pi}\right)\left(1-\frac{x}{2\pi}\right)\left(1+\frac{x}{2\pi}\right)\cdots=\\
&\left(1-\frac{x^2}{\pi^2}\right)\left(1-\frac{x^2}{2^2\pi^2}\right)\cdots=\\
&1+\left(-\frac{x^2}{\pi^2}-\frac{x^2}{2^2\pi^2}-\cdots\right)+\cdots
\end{align}
$$
So in particular
$$
\frac{x^2}{3!}=\frac{x^2}{\pi^2}\sum_{n=1}^\infty\frac{1}{n^2}
$$
A little algebra gets the result from this.