attempted solution:
a - b(mod5) = a + b(mod5)
2a(mod5) = 0
a = 5
5 + b mod 5 = 5 - b mod 5
b mod 5 = -b mod 5
b = 0
attempted solution:
a - b(mod5) = a + b(mod5)
2a(mod5) = 0
a = 5
5 + b mod 5 = 5 - b mod 5
b mod 5 = -b mod 5
b = 0
If you're working in the usual residue system of $\{0, 1, 2, 3, 4\}$, and $b = 0$, $a$ can be any number $n$ in the set, that is, $a-0 = a+0 \pmod 5 \iff a = a \pmod 5$ which is true for any $a$. Solutions of the form $(a, b)$ are given by $\{(a, 0): a \in \{0, 1, 2, 3, 4\}\}$
Similarly, if you're working over the integers, then provided $b$ is any integer multiple of $5$ (that is, provided $b\equiv 0 \bmod 5),$ then $a-b\equiv a+b\pmod 5 \iff a - 5k \equiv a + 5k \pmod 5$, which is true for all integer values of $a$, $k \in \mathbb{Z}$. Solutions of the form $(a, b)$ are given by $\{(a, 5k): a, k\in \mathbb{Z}\}$.
From $a-b\equiv a+b \pmod 5$ you can derive $2b \equiv 0 \pmod 5$, so $b \equiv 0 \pmod 5$, but $a$ can be anything-it cancels. Try $a=3, b=5$, for example.