<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] If $\gcd(x, 25) \neq 1$, then $x$ is a multiple of $5$, so $x^4$ and $x^8$ are multiples of $25$ and don't contribute to the sum. Thus, we will focus on the cases where $\gcd(x, 25)=1$.
Also, $x \equiv x+25 \pmod{25}$, so we really have the following:
$$4(1^4+2^4+3^4+[...]+24^4) \equiv 4(1^8+2^8+3^8+[...]+24^8) \pmod{25}$$
Now, your work shows that there is a bijective mapping between the fourth powers of the group of units $25$ and the eighth powers of the group of unts of $25$, so for any fourth power on the left side, we can find an eighth power on the right side and then subtract the two from both sides. We can do this repeatedly until we've subtracted all of the powers and gotten to $0 \equiv 0 \pmod{25}$, concluding the proof.
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