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 $3$: - $5\equiv 2(\mathrm{mod}3)$ - $31\equiv 1(\mathrm{mod}3)$ So $5^{2015}\equiv 2^{2015}(\mathrm{mod}3)$ $31^{2015}\equiv 1^{2015}=1(\mathrm{mod}3)$ Now, $2^{2015}$ is $2$ if $2015$ is odd, which it is. So $2^{2015}\equiv 2(\mathrm{mod}3)$ Thus, $$5^{2015}+31^{2015}\equiv 2+1=3\equiv 0(\mathrm{mod}3)$$ So $3$ is a factor.

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