What did I do wrong in this proof? I have this answer marked partially correct. I'd like to get an idea of where I went wrong
Prove that for all integers $a,b \in \mathbb{Z}$, $(a+b)^5 \equiv a^5+b^5 \mod 5$.
My answer:
Suppose that $a,b \in \mathbb{Z}$. Then, by definition of congruence modulo, $5|(a+b)^5-(a^5+b^5)$ meaning that there exists an integer $k$ such that $$(a+b)^5-(a^5+b^5)=5k$$ Then, $$a^5+5a^4b+10a^3b^2+10a^2b^3+5b^4a+b^5-a^5-b^5 = 5k$$ $$= 5(a^4b+2a^3b^2+2a^2b^3+b^4)=5k$$
This shows that the deduced result is divisible by 5, hence, for all integer $a,b,\in\mathbb{Z}$, $(a+b)^5 \equiv a^5+b^5 \mod 5$.
 A: Your idea is correct, but your proof looks backwards. I suggest the following modifications. Observe how now the proof looks like a tale of deduction rather than one of back-engineering, if you may.

Suppose that $a,b \in \mathbb{Z}$. Then, by definition of congruence
  modulo, we must show that $5|(a+b)^5-(a^5+b^5)$. This means
  that there exists an integer $k$ such that $$(a+b)^5-(a^5+b^5)=5k$$ 
  Let us show this holds. To see this, note that, $$(a+b)^5 = a^5+5a^4b+10a^3b^2+10a^2b^3+5b^4a+b^5 = a^5+b^5+5k$$ 
where $a^4b+2a^3b^2+2a^2b^3+b^4=k$ is obtained by collecting terms. This shows that the difference $(a+b)^5 -a^5-b^5$ is divisible by 5, hence,
  for all integer $a,b,\in\mathbb{Z}$, $(a+b)^5 \equiv a^5+b^5 \mod 5$.

A: It looks as though you're starting from what you want to prove, although this is not the case. I think it would be clearer if you started with just the LHS of the expression and showing that you can find a factorization of $5k$ where $k \in \mathbb{Z}$. You should also argue that the expression $a^4b+2a^3b^2+2a^2b^3+b^4$ is an integer.
