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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

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    @Ross: I see that I failed to make it clear, but I was talking there about $2a\bmod 5$, not about the original problem. Unfortunately, I forgot that when I wrote the second sentence.2012-11-30

2 Answers 2

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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 $\{(a, 5k): a, k\in \mathbb{Z}\}$.

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    if you subtract (a - b) - (a + b) = 0 (mod 5), you get a-a - b - b= -2b = 0 (mod 5). That happens if and only if b is a multiple of 5: $b\in \{...-10, -5, 0, 5, 10, ...\}$.2012-11-30
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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.