Proving if the sum of 5 ints (each raised to the power of 5) is divisible by 25 then the product of those 5 ints is divisible by 5 So I'm trying to prove the following:
For any 5 ints, if the sum of the numbers (each raised to the power of 5) is divisible by 25 then the product of those original 5 ints is divisible by 5.
What I did first was notice that any integer raised to the power of 5 is congruent to 0,1,7,18 or 24 mod 25. And if we choose 5 numbers, and the sum of these powers of 5 is divisible by 25, then it seems like 0 must be one of those 5 numbers (It doesn't seem like any 5 numbers as a combination of 1,7,18,24 summed would be divisible by 25) But if we choose 0 as one of the 5 numbers, it'll work. Then if must always include a 0 for the sum to be divisible by 25, then the product will be divisible by 5 since the product will be 0.
That's what I noticed but I don't know how to prove this. Any advice?
 A: You are basically done already. You just have to write out your argument in detail as for why it is the case that the only way to get back $0\bmod 25$ from the sum of five fifth powers is to have one of those powers be congruent to $0\bmod 25$.
If this is for your own edification, you could simply exhaust over the possible sums that exclude $0\bmod 25$ (there are 1024 such sums if you naively count them, but if you are really careful there are only 56 important cases). If you don't want to do that, you could make a series of arguments like the following...
The only residues $\bmod 25$ that are fifth powers are $0,\pm 1$, and $\pm 7$. If exactly one of the five numbers in the sum has a residue of $7$ and no numbers have a residue of $-7$, then the entire sum can only be $3,4,5,6,7,8,9,10$, or $11\bmod 25$. The same argument applies to the case of exactly one number having residue $-7$ and no numbers of residue $7$. Similar arguments will handle sums in which there are only $7$'s or $-7$'s. So the only way to get back $0\bmod 25$ is to have a mixture of $7$'s and $-7$'s. Now keep on handling cases...
While this is much less effort than enumerating all possible sums, you can see that the entire argument will still be long and boring. Is there a more clever way? I hope so, but it is late, and I can't think very well.
At the end, you will see that the only sums that work are those that have at least one number with a residue of $0\bmod 25$. But wait, which numbers have fifth powers that are congruent to $0$? Those that are divisible by 5. Thus, when you do to take the product of the terms in your sum, it too will be divisible by $5$.
