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I am tying to disprove it but I feel my proof is lacking, please let me know what I can improve on from here, thanks!:

  • If $a ≡ b$ mod $m$, then $m|a-b$ i.e. $a-b=km$, $k$ is an int.
  • Square both sides we get: $(a-b)^2=k^2m^2\implies a^2-2ab+b^2=k^2m^2$
  • Then $a^2+b^2=2ab+k^2m^2$
  • This means that $m^2|a^2+b^2$ and not $m^2|a^2-b^2$ i.e. $a^2≡b^2$ mod $m^2$
  • Therefore the statement If $a ≡ b$ mod $m$, then $a^2 ≡ b^2$ mod $m^2$, is false.
Bill Dubuque
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quadingle the 2nd
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    $3\equiv 8\pmod{5}$, but $9\not\equiv 64\pmod{25}$ – Steven Creech Apr 03 '23 at 13:53
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    @StevenCreech, You're right! but i was trying to disprove without using a counterexample, – quadingle the 2nd Apr 03 '23 at 13:57
  • I do not see why your fourth bullet point is true - you would need that $m^2$ divides $2ab$ for that to hold. Moreover, to disprove the claim you just need to present a single counterexample, like the one @StevenCreech pointed out. – ABanerjee Apr 03 '23 at 13:57
  • @ABanerjee Is there a another way to disprove it than a counterexample? – quadingle the 2nd Apr 03 '23 at 13:59
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    Teasing out a system of equations to understand why there may be counterexamples is a good way to find counterexamples, but it's not a good proof form. Just provide the simplest counterexample you can find. – Matthew Leingang Apr 03 '23 at 13:59
  • @quadinglethe2nd Philosophically, a counter-example is a sufficient and necessary condition for the claim to be disproved. If the claim is $P(a, b, m)$ is true for all possible values of $a, b,m$, then all you need to establish that it doesn't hold is to present a single instance where $P(a, b, m)$ is false. Trying to "disprove" it for arbitrary values of $a, b, m$ is much harder - as now you are arguing that $P(a, b, m)$ is not true for all values of $a, b, m$! In fact, if you take $(a, b, m) = (1, 3, 2)$, the claim above is true. All of this is to say a counterexample is ideal in this case. – ABanerjee Apr 03 '23 at 14:09
  • That said, if you're curious, then it can be beneficial to explore the reason counterexamples CAN exist by studying the equations obtained above, as @MatthewLeingang suggested, and you could argue the claim doesn't hold using them. It would be a valuable exercise to carry out perhaps, but certainly not a good form of proof in general, as the much simpler method of presenting a counter-example exists! – ABanerjee Apr 03 '23 at 14:13
  • More generally, see LTE = Lifting The Exponent – Bill Dubuque Apr 03 '23 at 19:07
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    @BillDubuque In connection with the new tag ([tag:lifting-the-exponent]), I just wanted to mention that there is a related tag ([tag:hensels-lemma]) - but very likely you're aware of that. (I have posted some related stats in the Tagging chatroom.) – Martin Sleziak Apr 05 '23 at 05:41

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Rather than directly prove the statement false, we can use the relations to look for counterexamples.

Suppose there exist $a$, $b$, and $m$ such that $a \equiv b \pmod{m}$ and $a^2 \equiv b^2 \pmod{m^2}$. Then there exist integers $k$ and $\ell$ such that $a-b=km$ and $a^2-b^2 = \ell m^2$. Since $a^2-b^2 = (a-b)(a+b)$, we have $km(a+b) = \ell m^2$, which means $k(a+b) = \ell m$. So $a+b$ is a zero divisor modulo $m$. Since $a+b \equiv 2b \pmod{m}$, we see that $2b$ is a zero divisor modulo $m$.

To find a counterexample, look for $b$ and $m$ such that $2b$ and $m$ are relatively prime, and set $a=b+km$ for any nonzero integer $k$. For instance, choose $b=1$ and $m=3$, so $a=4$. Then $16 \not\equiv 1 \pmod{9}$, so we have a counterexample. Steven Creech's counterexample comes from the choice $b=3$ and $m=5$.

Matthew Leingang
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