Fermat's Last Theorem states that for integers $p > 2$, there are are no natural numbers $a, b, c$ such that

$$a^p + b^p = c^p.$$

Note that, for example, for $p=2$, there are such $a, b, c$ (e.g., $3^2+4^2=5^2.$

Let's write a program that tries to find a, b, and c, such that $a^p + b^p = c^p.$ The strategy is as follows: first, we'll try all possible a, b, and c in the range 1..1. Then we'll try all possible a, b, and c in the range 1..2. Then we'll try all possible a, b, and c in the range 1..3, and so on. If we find a solution at any point, we return.

def fermat(p):
    n = 1
    while True:
        for i in range(1, n):
            for j in range(1, n):
                for k in range(1, n):
                    if i**p + j**p == k**p:
                        return i, j, k
        n += 1

We can call fermat(2) to make sure that this works:

fermat(2)

(3, 4, 5)

According to Fermat's Last Theorem, fermat(p) will never return for p > 2. (So that we could claim that the complexity is infinity.) However, this is not at all obvious a priori (it took hundreds of years to figure out!)