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?

We only need to prove that
$$(5a+1)^8 + (5a+2)^8 + (5a+3)^8 +(5a+4)^8 \equiv_{25} (5a+1)^4+(5a+2)^4+(5a+3)^4+ (5a+4)^4$$
In this equations the elements of the polynomials that only matter are the elements of the form $\dbinom{8}{i}(5a)^ik^{8-i}$ with $i=0,1$, because for $i>1$, the number is $\equiv_{25} 0$. This is the same for the right side, only matters $\dbinom{4}{i}(5a)^ik^{4-i}$ with $i=0,1$. So let's check the rest of the equation and reduce it using this idea:
$$(5a+1)^8 + (5a+2)^8 + (5a+3)^8 +(5a+4)^8 -(5a+1)^4-(5a+2)^4-(5a+3)^4- (5a+4)^4 \equiv_{25} 0$$
$$\Leftrightarrow (8*5a*1+1) + (8*5a*2+2^8) + (8*5a*3+3^8) +(8*5a*4+4^8) -(4*5a*1+1^4)-(4*5a*2+2^4)-(4*5a*3+3^4)- (4*5a*4+4^4) \equiv 0$$
And using the mod 25, we get:
$$\Leftrightarrow (40a+1) + (20a+6) + (5a+11) +(10a+11) -(20a+1)-(10a+16)-(15a+6)- (5a+6) \equiv_{25} a((40+20+5+10)-(20+10+15+5))+(1+6+11+11-1-16-6-6) \equiv 29-29 \equiv 0$$
So it's done. Now, in the case of the number 99, you can divide the space in sets of 5 consecutive numbers as here: 
$$\{1,2,3,4,5\}, \{6,7,8,9,10\},\dots ,\{96, 97, 98, 99 \}$$
and as the numbers of the form $5k$ are $\equiv_{25}0$ when they're at 4th power, the other ones fulfil the identity.
