South African Mathematics Olympiad

Let us consider the expression:
$$5^{2015} + 5^{2016} + 5^{2017} + 31^{2016}$$
First, factor $5^{2015}$ from the first three terms:
$$5^{2015} + 5^{2016} + 5^{2017} = 5^{2015}(1 + 5 + 5^2) = 5^{2015}(1 + 5 + 25) = 5^{2015} \times 31$$
So the expression becomes:
$$5^{2015} \times 31 + 31^{2016}$$
Now, factor $31$:
$$= 31(5^{2015} + 31^{2015})$$
So $31$ is a prime factor.

Now, consider $5^{2015} + 31^{2015}$.
Notice that both $5$ and $31$ are odd, so their powers are odd, and their sum is even. Thus, $2$ is a factor.

Let us check if $2$ is a factor:
- $5^{2015}$ is odd
- $31^{2015}$ is odd
- Odd + Odd = Even
So $2$ is a factor.

Now, check if $5 + 31 = 36$ is divisible by $3$.
Let us check modulo $3$:
- $5 \equiv 2 \pmod{3}$
- $31 \equiv 1 \pmod{3}$
So $5^{2015} \equiv 2^{2015} \pmod{3}$
$31^{2015} \equiv 1^{2015} = 1 \pmod{3}$
Now, $2^{2015}$ is $2$ if $2015$ is odd, which it is.
So $2^{2015} \equiv 2 \pmod{3}$
Thus,
$$5^{2015} + 31^{2015} \equiv 2 + 1 = 3 \equiv 0 \pmod{3}$$
So $3$ is a factor.

Therefore, the three prime factors are $2$, $3$, and $31$.