Prime Pythagorean triples

Given an arbitrary pair of integers m and n with m > n > 0. Euclid's formula states that the integers :

$$a = m^2 - n^2,\ b=2mn,\ c=m^2+n^2$$

form a Pythagorean triple: $a^2 + b^2 = c^2$

If we calculate $a^2 + b ^2$:

$$a^2 + b^2 = (m^2-n^2)^2 + (2mn)^2 = m^2 + 4m^2n^2 - 2m^2n^2 + n^2 = (m^2 + n^2)^2 = c^2$$

Which is coherent, the problem changes from finding $a$, $b$ and $c$ to finding $m$ and $n$, we have:

$$\begin{align} a+b+c &= 1000\\ 2mn + 2m^2 &= 1000\\ n &= \frac{500}{m} - m\\ \end{align}$$

We have $m > n$ since b must be positive, solving $n = m$ gives:

$$\begin{align} m &= \frac{500}{m} - m\\ 0 &= \frac{500 - 2m^2}{m}\\ 0 &= 500 - 2m^2\\ m &= \sqrt(250)\\ \end{align}$$

Resulting in:

$$m > 16 > \sqrt(250)$$

We also have $n > 0$, solving $n = 0$ gives:

$$\begin{align} 0 &= \frac{500}{m} - m\\ 0 &= 500 - m^2\\ m &= \sqrt(500)\\ \end{align}$$

Resulting in:

$$m < \sqrt(500) < 23$$

We also know that $\frac{500}{m}$ is an integer, so m must divide $500$. The only multiple of that divides $500$ with the constraint $16 > m > 23$ is 20. Having $m = 20$ result in $n = 5$. It gives $a = 200$, $b = 375$ and $c$ = 425.

The solution can also be found by using a for loop to find $m$ and $n$ with the above constraints.

From solution3.py:

def special_pythagorean_triplet():
    for m in range(16, 24):
        for n in range(1, m):
            if m * (n + m) == 500:
                return 2 * m * n * (m**4 - n**4)

    return -1