Prove or Disprove $∀a, b ∈ Z$, if $a ≡ b$ (mod 4) then the average of $a$ and $b$ has the same parity as $a$. Prove or Disprove  $∀a, b ∈ Z$, if $a ≡ b$ (mod 4) then the average of $a$ and $b$ has the same parity as $a$. 
Classic case that I have an idea of how this works but not sure how to "prove" it. Was thinking a counterexample of the contrapositive might be easier too so if I come up with nothing down this avenue I'll move onto these. 
Congruence implies $\frac{a-b}{4}$ is perfectly divisible and 4 divides into only even numbers. 
I think only an odd pair of a and b or an even pair of a and b will prove to be congruent but that doesn't get me anywhere.
I know the average of two even's or two odd's can be even or odd but that doesn't seem to lead me anywhere. 
I feel like I'm missing a key part about basic number theory around odds and evens or congruence that make the proof trivial but alas I'm in the rut. Any suggestions/hints/solutions will be studied hard! 
 A: Saying that the average of $a,b$ and $a$ have the same parity, is the same as saying $\frac{a+b}{2} \equiv a \mod 2$. 
Suppose $a \equiv k \mod 4$. Then, note that $b \equiv k \mod 4$, obviously. We also have that $a+b \equiv 2k \mod 4$.
Suppose $k=1,3$. Then, $a+b \equiv 2 \mod 4$, so that $\frac{a+b}2$ is odd, of the same parity as $a$.
Suppose $k = 0,2$. Then, $a+b \equiv 0 \mod 4$, so that $\frac{a+b}{2}$ is even, of the same parity as $a$. Hence, since $k=0,1,2,3$ are the only values $k$ can take, we have that $a$ has the same parity as $\frac {a+b}2$. 

This kind of analysis helps. Suppose that we only assume that $a-b$ is divisible by $2$, and not $4$. Then, this does not work out, since if $a$ and $b$ are odd numbers summing to a multiple of $4$, then $\frac{a+b}2$ will be even, but $a$ is odd. For example, $1$ and $7$ sum to $8$, so their average $4$ is even while $1$ is odd.
A: Since $a\equiv b \pmod{4}$, you have $b = a +4k$ for some integer $k$.  Then the average is:
$$\frac{a+b}{2} = \frac{a+(a+4k)}{2} = a+2k.$$
Adding an even number to $a$ won't change its parity.  Or you could
continue the above calculation:
$$ a+2k \equiv a \pmod{2}.$$
A: You want to show that $\frac{a+b}2$ has the same parity as $a,$ that is, that $\frac{a+b}2-a$ is an even integer; in other words, you want to show that
$$\frac{\frac{a+b}2-a}2$$ is an integer. But that complex fraction simplifies to $$\frac{b-a}4$$ which is an integer, because that's what $a\equiv b\pmod4$ means.
