I got mixed up by the use of triple bar vs equals sign in linear congruences. Is it worth asking the professor for a regrade?

Asked May 17 '21 at 01:10

Active May 19 '21 at 23:58

Viewed 55 times

On an exam I recently took, I was asked the following question:

What are the number of solutions to the equation:

$ax = b$ (mod $m$) when gcd($b,m$) = 1 and gcd($a,m$) = $d$ > 1

I know this is entirely sematic, but I was confused when I saw them use = instead of ≡ in the equation. For this reason, I interpreted the right-hand side to be a specific value $(b\%m)$ and stated there was one solution. In reality, there are no solutions as the expression ax + km is always a multiple of d and b is not.

What I really want to know is if my interpretation was valid at all, and if it is worth asking the professor for a regrade in the context of ≡ vs = or if I just take the L.

edited May 19 '21 at 23:58

Bill Dubuque

272,048

asked May 17 '21 at 01:10

William Minnis

4

In any case there is no solution... – nir shahar May 17 '21 at 01:13
2

$ax = (b\bmod m) = b - qm,$ so $,d\mid a,m\Rightarrow d\mid b,,$ contra hypothesis. So that too has no solution. Or use $,ax = (b\bmod m)\Rightarrow ax\equiv b\pmod{m}$ – Bill Dubuque May 17 '21 at 01:51
@BillDubuque What about for the case where gcd(a,m) = d and d|b? – William Minnis May 17 '21 at 04:24
$ax\equiv b\pmod{m}$ is solvable $\iff \gcd(a,m)\mid b,,$ see here. – Bill Dubuque May 17 '21 at 07:40

I got mixed up by the use of triple bar vs equals sign in linear congruences. Is it worth asking the professor for a regrade?

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