Prove that $1^8+2^8+\cdots+99^8 \equiv 1^4+2^4+\cdots+99^4 \pmod{25}.$ 
Prove that $$1^8+2^8+\cdots+99^8 \equiv 1^4+2^4+\cdots+99^4 \pmod{25}.$$

Attempt:
We can easily show that an eighth power can be expressed as a fourth power since $x^8 = (x^2)^4$. Conversely, by Fermat's Little Theorem, $x^{\phi(25)} = x^{20} \equiv 1 \pmod{25}$ if $\gcd(x,25) = 1$ and thus $x^4 = x \cdot x^3 \equiv x^{21} \cdot x^3 \equiv x^{24} \equiv (x^3)^8 \pmod{25}$. $\square$
I am not sure if the above proves the result, but it does show that if we have an $8$th power of an integer modulo $25$ and it is relatively prime to $25$, then it is equal to a fourth power of an integer modulo $25$ and vice-versa. Does that therefore mean they are equivalent?

$\mathbb{Z}/(25\mathbb{Z})^*$ is a cyclic group generated by $2$, hence
$$ \sum_{\substack{1\leq n \leq 25\\ 5\nmid n}}n^4 \equiv \sum_{r=1}^{20}2^{4r}\equiv \frac{16^{24}-16^4}{16-1}\equiv 20\pmod{25} $$
and
$$ \sum_{\substack{1\leq n \leq 25\\ 5\nmid n}}n^8 \equiv \sum_{r=1}^{20}2^{8r}\equiv \frac{256^{168}-256^8}{256-1}\equiv 20\pmod{25} $$
prove our claim. 
To compute $2^N\pmod{125}$ is not difficult since $2$ is a generator of $\mathbb{Z}/(125\mathbb{Z})^*$, too.
