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For this problem would it be $7$ for both $m$ and $n$ because you will get $49$ on both side and then it is divisible by $7$. Or am I thinking of how to do this completely wrong?

Alekos Robotis
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AliceFasca
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  • Did you mean $n^2$? Here is a MathJax tutorial – J. W. Tanner Oct 06 '20 at 21:58
  • I'm not at all sure what you mean by any of this. Would what be $7$ for both $m$ and $n$. What does "getting $49$ on both sides mean. And then what is divisible by $7$? And ... then what? What would that imply? why is that a concern? What is your question? – fleablood Oct 06 '20 at 22:44
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    I've posted an answer but I've hidden it. I'll show it if you could make it more clear what you are asking. It sounds like you are on the right track. If $7m=n^2$ then $n^2$ must be a multiple of $49$ (why?) and therefore $7$ divides $m$ (why?). And that's enough to prove $7|\gcd(m,n)$ (why?). If I can be more clear what you are asking I can try to answer more specifically (and not just write a blanket solution to the problem for you). – fleablood Oct 06 '20 at 22:48
  • @Alice The proof is essentially the same as in the classical proofs showing square-roots irrational, by applying Euclid's lemma to show $,7\mid n^2\Rightarrow, 7\mid n,,$ then $,7^2\mid 7m\Rightarrow7\mid m.\ $ Alternatively, by unique prime factorization $7$ occurs to an even power $k$ in $7m = n^2$ (by RHS) and $k \ge 1$ by LHS, so $k \ge 2$, so $,7\mid m,n,$ (compare here and here) – Bill Dubuque Oct 06 '20 at 23:51

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I'm assuming you are looking to prove $7m=n^2$ implies that $7\mid \gcd(m,n)$.

Hint 1: Note that it suffices to show that $7\mid m$ and $7\mid n$.

Hint 2: Compare the prime factorizations of both sides. Can a squared number $n^2$ satisfy $p\mid n^2$ but $p^2\nmid n^2$ for $p$ a prime?

Alekos Robotis
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