Infinitely many Proofs of Pythagoras' Theorem

This post is inspired by an MSE question and the brilliant answers that it received. However, it can be understood independently. Everything here is joint work with Satvik Saha.

Consider the following diagram made of 16 right triangles of sides $a\le b<c$ and four squares of side $b-a$. This being a tiling of the outer square (of side length $2c$), we have

$$(2c{)}^2=4(b-a{)}^2+16\times \frac{1}{2}ab$$

hence implying

$$c^2=a^2+b^2$$

thus proving the Pythagoras' Theorem.

Outer square of side $2c$

This is similar to previously known proofs (eg. see proof #3 and #116 here), but it is hard to say whether this exact tiling appears anywhere before.

Now, consider the following diagram made up of right triangles of sides $a\le b<c$, (pink) sqaures of side $a$, (yellow) squares of side $b$, and (green) squares of side $(b-a)$. The reader is welcome to check that this also gives a completely valid proof of the Pythagoras' Theorem.

Outer square of side $3c$

Some of the readers might have started asking the right question. And the answer is yes, it can be generalised! With the outer sqaure of side $nc$, we get $4(n-2)$ of the red and yellow squares, $n^2-4n+8$ of the green squares, and $4(n^2-4n+8)$ of the triangles. This gives

$$n^2{c}^2=4(n-2)(a^2+b^2)+(n^2-4n+8)(a-b{)}^2+2(n^2-4n+8)ab$$

hence proving the Pythagoras' Theorem. For your viewing pleasure, here's the $16\times 16$ big square. Feel free to print it out and put it on your wall.

Outer square of side $16c$

Of course, the keen ones among you would ask whether this is truly a tiling for any values of $a$ and $b$ or is it a coincindence of the value chosen in the diagrams. To clear this doubt, here's an animated version of the diagrams. Notice that the animation also shows that it doesn't matter whether you call the larger side of the triangle $a$ or $b$.

Outer square of side $6c$

The keener ones would of course ask for a proof that this is indeed a tiling. For that, one only needs to note that the diagram is made of four spiky triangles (made of a layer of blue triangles followed by a layer of yellow and a layer of pink squares followed by a number of layers of blue triangles) - this is almost a triangle except two spikes on two sides. All one needs to convince oneself is that these spiky triangles fit with each other properly and leave a space in the middle which can be nicely filled by the green squares.

Feel free to play with the animation here (p5js code here). Also, feel free to check out my answer in the aforementioned MSE post.

Worst Way to Calculate Area of a Triangle

A useful technique from Complex Analysis helps us to identify whether a real number $u$ is positive or negative by checking a discontinuous integral. We have

$$\frac{1}{2\pi i}{\int }_{c-i\mathrm{\infty }}^{c+i\mathrm{\infty }}{e}^{su}\frac{\text{d}s}{s}=\{\begin{array}{ll}0& \text{ if }u<0\\ \frac{1}{2}& \text{ if }u=0\\ 1& \text{ if }u>0\end{array}$$

for $c>0$.

We will apply this to calculate the area of the triangle

$$T(N):=\{(x,y)\in {\mathbb{R}}_{\ge 0}:ax+by\le N\}$$

with $(a,b)=1$. This area is given by

$$\begin{array}{rl}& {\int }_{x,y\ge 0}{\mathbf{1}}_{N-ax-by\ge 0}\text{d}x\text{d}y\\ =& {\int }_{x,y\ge 0}\frac{1}{2\pi i}{\int }_{s=c-i\mathrm{\infty }}^{c+i\mathrm{\infty }}{e}^{s(N-ax-by)}\frac{\text{d}s}{s}\text{d}x\text{d}y\\ =& \frac{1}{2\pi i}{\int }_{s=c-i\mathrm{\infty }}^{c+i\mathrm{\infty }}\left({\int }_{x\ge 0}{e}^{-asx}\text{d}x\right)\left({\int }_{y\ge 0}{e}^{-bsy}\text{d}y\right)\frac{e^{sN}}{s}\text{d}s\\ =& \frac{1}{ab}\cdot \frac{1}{2\pi i}{\int }_{s=c-i\mathrm{\infty }}^{c+i\mathrm{\infty }}\frac{e^{sN}}{s^3}\text{d}s\end{array}$$

by the above mentioned trick. Notice that the integral with $N-ax-by=0$ is zero, and hence was ignored.

Now, moving the contour far to the left we pick up the residue from the pole of order $3$ at $s=0$ and hence the area comes out to be

$$\frac{1}{ab}\cdot \frac{N^2}{2}=\frac{N^2}{2ab}$$

which matches with the area calculated by other known methods!

Acknowledgement : I learnt this from lectures by Prof. Andrew Granville for the course Distribution of Prime Numbers.