Euclid’s formula says that, \((a,b,c)\) are a Pythagorean triple, i.e., \(a^2+b^2=c^2\) for \(a,b,c\) are integers, if and only if \(a=2mn\), \(b=m^2-n^2\), \(c=m^2+n^2\) for some integers \(m,n\).

The proof is as follows: Since \(a^2+b^2=c^2\), we have

\[\big(\frac{a}{c}\big)^2 + \big(\frac{b}{c}\big)^2 = 1\]

i.e., the point \(P=(x,y)=(a/c,b/c)\) is on the unit circle. Assume \(a,b,c\) are all non-negative. Hence P is in the first quadrant. Consider the point \(Q=(0,-1)\) on the unit circle, we have line QP intersect with the \(x\)-axis on \(R=(k,0)\), which, by section formula, we have

\[k = \frac{x}{1+y}.\]

Since \(x\) and \(y\) are rational numbers, so does \(k\). Now we assume \(k=n/m\) for some non-negative \(n,m\). Obviously, point \(R\) is inside the unit circle, so \(n\le m\). Now consider the line QR, it is given by the equation

\[\begin{aligned} (y-0)(0-k)&=(x-k)(-1-0) \\ x-ky-k &= 0, \end{aligned}\]

and the unit circle is given by

\[x^2+y^2 = 1.\]

Solving the two equations simultaneously, it gives the solution \(x=0,y=-1\), i.e. point \(Q\), and the other solution is

\[\begin{aligned} x&=\frac{2k}{1+k^2} & y&=\frac{1-k^2}{1+k^2}. \end{aligned}\]

Subsititute \(k=n/m\) we have

\[\begin{aligned} x&=\frac{2\frac{n}{m}}{1+\frac{n^2}{m^2}} & y&=\frac{1-\frac{n^2}{m^2}}{1+\frac{n^2}{m^2}} \\ x&=\frac{2mn}{m^2+n^2} & y&=\frac{m^2-n^2}{m^2+n^2} \\ x&=\frac{a}{c} & y&=\frac{b}{c}. \end{aligned}\]

Therefore, we have

\[\begin{aligned} a&=2mn & b&=m^2-n^2 & c&=m^2+n^2. \end{aligned}\]