Sum of powers of a number divisible by other? I need to find the probability that $7^m+7^n$ is divisible by $5$ where $m,n\in[1,100]$. I noticed that if the remainder when m and n are divided by 7 differs by two the last digit is zero. So I did $$P=\frac{4*25^2}{100^2}=\frac14$$But I'm now unsure to prove this seemingly correct fact. On a side note I haven't studied Number Theory.
 A: The first four powers (mod 5) of 7 are $\{2,4,3,1\}.$ After this the sequence repeats. There are 16 ways of pairing these and of those 4 sum to 5: $(2,3),(3,2),(1,4),(4,1).$ So 1/4 of the possible pairings lead to a sum $7^n+7^m =0~ \text{mod}~ 5$
In general there are $(4n)^2$ pairings of which $(2n)^2$ will sum to 0 mod 5. In this case 
$$\frac{(2\cdot 25)^2}{(4\cdot 25)^2}= \frac{4\cdot25^2}{16\cdot 25^2} = 1/4. $$

Elementary probability (more than number theory) furnishes an explanation of this. If we have $n$ distinct objects and ask how many pairs can be made from the set, the number, written $nP$2, is
$$\text{Number of pairs} =\frac{n!}{(n-2)!} $$
If in addition we allow cases in which we pair two items that are indistinguishable,for example (4,4) from the set above, the number is
$$\text{Pairs including repetitions} = \frac{n!}{(n-2)!}+n = n(n-1)+ n = n^2. $$ 
Why do we include these? Because in the problem, we include cases like 
$$7^2 +7^6$$
even though both powers are congruent to 4 (mod 5). If the probability of divisibility by 5 (etc.) is defined as  
$$P =\frac{\text{number of pairs that sum to 0 (mod 5)}}{\text{number of possible pairs for} ~m,n \leq 100}$$
then this pair should be included. It also explains why the numerator grows proportionally. We treat like congruences associated with different numbers as distinct objects because their sum is still $0$ (mod 5). 

The only number theory here is I think modular arithmetic.
A: 7=2 (mod 5)
7^m=2^m (mod 5)
 Also,
2^4=1 (mod 5) =>
2^4k=1=7^4k (mod 5)
2^(4k+1)=2=7^(4k+1) (mod 5)
2^(4k+2)=4=7^(4k+2) (mod 5)
2^(4k+3)=3=7^(4k+3) (mod 5)
Thus 5|7^(4k)+ 7^(4k+2) and 5|7^(4k+1)+7^(4k+3)
Further there are 25 integers of each form.Thus, 2*25*25 satisfy case 1 and 2*25*25 satisfy case 2
Thus a total of 4*25*25 are such cases
The rest is as you did
