Geometric proof : infinite dissimilar right triangles with integral sides Wha is the proof that there are infinitely many right angled (non-similar) triangles whose sides have integral lengths? I know that this is equivalent to showing that there are infinite pythagorean triples, which can be proven easily, but I would like to know about any purely geometrical proof.
 A: Consider the following diagram:

So the area of the green part must be equal to the area of the orange part by the pythagorean theorem. However, the area of the green part is $2X+1$. Hence, we can make the area of the orange part any odd number with a positive integer choice of $X$. In particular we can look at the sequence where $2X+1=p_i^2$ where $p_i$ is the $i$-th odd prime. Hence we have an infinite sequence of right triangles with of side lengths $p_i,X,X+1$ where $2X+1=p_i^2$. Note that none of these triangles are similar because one sidelength of each is prime.
A: The wikipedia page on Pythagorean triples describes the geometric proof of the enumeration of Pythagorean triples. In brief, there are two steps. First, there is a one-to-one correspondence between primitive Pythagorean triples $(a,b,c)$ and rational points $(r,s)=(a/c,b/c)$ on the unit circle $S^1$. Second, sterographic projection, radiating away from the north pole $N$ and projecting $S^1 - N$ to the $x$-axis, defines a one-to-one correspondence $S^1 - N \leftrightarrow \mathbb{R}$. Furthermore, stereographic projection restricts to a one-to-one correspondence between rational points $(r,s)$ on the unit circle and the set $\mathbb{Q}$ of rational numbers on the $x$-axis (this is where you need some formulas, which can be found in the wikipedia link provided). 
Here are the formulas for stereographic projection, which I am copying from the wikipedia page. If $P = (r,s)$ is a rational point on the unit circle and if $P' = (m/n,0)$ is a rational point on the real line, so that stereographic projection relates $P$ to $P'$, then the formulas relating the coordinates of $P$ and of $P'$, are:
$$r = \frac{2mn}{m^2+n^2}, \,\,\, s = \frac{m^2-n^2}{m^2+n^2}
$$
and 
$$\frac{m}{n} = \frac{r}{1-s}
$$
These formulas can be derived using elementary geometry, starting from the picture of stereographic projection.
