If $a \equiv b\pmod{m^2}$, then prove $ a \equiv b\pmod{m}$

Question

How would I go about proving

$$ a \equiv b\pmod{m}$$ given

$$ a \equiv b\pmod{m^2} $$

I suspected this was false, so I tried to find a counterexample using a program, and there does not appear to be a counterexample for $m$ in (2, 16).

What I have tried so far: $$ a \equiv b\pmod{m^2} $$ $$ a = b + m^2k $$ $$ a - b = m^2k $$ $$ \sqrt{a - b} = \sqrt{m^2k} $$

Which would imply that $\sqrt{a-b} \equiv 0\pmod{m}$ assuming that $k$ is a perfect square. It feels like I'm barking up the wrong tree, however, and I'm not sure where to go from here.

Just use the definitions. So $m^2\mid b-a$ implies $m\mid (b-a)$. — Dietrich Burde, Apr 21 '21 at 18:12
Use \pmod{m} and \pmod{m^2}; it provides the spacing, parentheses, and correct typesetting. — Arturo Magidin, Apr 21 '21 at 18:14
Interesting exercise to consider the reverse direction; if we have $a\equiv b \bmod m$, what possible relationships are there between $a$ and $b$ considered $\bmod m^2$? This can also crystallize understanding of the answer to the question asked. — Joffan, Apr 21 '21 at 18:51

score 1 · Answer 1 · answered Apr 21 '21 at 18:13

1

$a\equiv b\pmod{m^{2}}$ means that:

$$a-b=pm^{2}$$

For some integer $p$. However, letting $q = pm$, we see:

$$a-b=qm$$

Which is the definition of $\boxed{a\equiv b\ \ (\text{mod}\ m).}$

Note that the converse is certainly false; just let $a = 0$ and $b = m$.

answered Apr 21 '21 at 18:13

Joshua Wang

6,103

That really clarifies it! Thank you! – bren Apr 21 '21 at 18:20
Please strive not to add more dupe answers to dupes of FAQs. – Bill Dubuque Apr 21 '21 at 18:33

score 1 · Answer 2 · answered Apr 21 '21 at 18:13

1

$a-b=km^2=(km)m$. This same argument can be used to show that if $a\equiv b \pmod{mn}$ then $a\equiv b \pmod m$.

answered Apr 21 '21 at 18:13

Aaron

24,207

Please strive not to add more dupe answers to dupes of FAQs. – Bill Dubuque Apr 21 '21 at 18:33

If $a \equiv b\pmod{m^2}$, then prove $ a \equiv b\pmod{m}$

2 Answers2