Assume $a$, $b$, $c$ are positive numbers, $a \le b$ and $b \le c$.
If $a^2\equiv0\pmod c$ and $b^2\equiv0\pmod c$, and $a$ is the minimal number satisfy $a^2\equiv0\pmod c$, do we always have $a\mid b$?
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Updated the question, sorry, forgot to mention that $a$ is the minimal number that satisfy the condition – wddd Nov 26 '23 at 11:27
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Hint: the set of square multiples of $c$ is closed under gcd: $c\mid a^2,b^2\Rightarrow c\mid\gcd(a^2,b^2)!=!\gcd(a,b)^2,,$ so classical Euclidean / gcd descent applies as here in the dupe, i.e. if $,a\nmid b,$ then $\gcd(a,b) < a,$ so $,\gcd(a,b)^2< a^2$ contra minimality of $,a.\ \ $ – Bill Dubuque Nov 26 '23 at 18:08
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@wddd I deleted my comment giving the smallest counterexample to your prior version, but the rest seems still relevant: "for next time, please read Quick beginner guide for asking a well-received question and avoid "no clue" questions (show some work), and use a more descriptive title". – Anne Bauval Nov 26 '23 at 18:35