<s>[INST] 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?
 [/INST] $100 = 2^2 \times 5^2$. It suffices to show that the two sums are equal mod $4$ and mod $25$.
If $x$ is even, $x^4 \equiv x^8 \equiv 0 \mod 2^2$, while if $x$ is odd,
$x^4 \equiv x^8 \equiv 1 \mod 2^2$.  So $\sum_{i=1}^{99} x^4 \equiv \sum_{i=1}^{99} x^8 \mod 2^2$.
Similarly, if $x$ is divisible by $5$, $x^4 \equiv x^8 \equiv 0 \mod 5^2$.
On the other hand, the multiplicative group $U$ of units in $\mathbb Z / 25 \mathbb Z$ (i.e. the numbers mod $25$ that are not divisible by $5$) is cyclic of order $\phi(5^2) = 20$.  The maps $t \to t^4$ and $t \to t^8$ are homomorphisms of this group, with the same range (namely the $5$ elements $y$ such that $y^5 = 1$).  Thus $x^4$ for $x \in U$ and $x^8$ for $x \in U$ run over this same set, each element occurring $4$ times.  If we take the 
numbers from $1$ to $99$ not divisible by $5$, each element occurs $20$ times.  Thus the two sums mod $25$ are the same.
EDIT:  Somewhat more generally, $\sum_{x=1}^{N-1} x^4 \equiv \sum_{x=1}^{N-1} x^8 \mod N$ for all positive integers $N$.  One way to see this is that
$$ \sum_{x=1}^{N-1} (x^8 - x^4) = \dfrac{N\; (2\,{N}^{8}-9\,{N}^{7}+12\,{N}^{6}-12\,{N}^{4}+9\,{N}^{3}-2\,{N}^{2})}{18}$$ and $2\,{N}^{8}-9\,{N}^{7}+12\,{N}^{6}-12\,{N}^{4}+9\,{N}^{3}-2\,{N}^{2} \equiv 0 \mod 18$ for all $N$.
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