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Let $m$ and $n$ be two positive integers relatively prime to each other. Prove that for every positive integer $k$, the following statements are equivalent:
1. $n + m$ is a divisor of $n^2 + km^2$;
2. $n+m$ is a divisor of $k+1$.

I couldn't do too much of this question, I feel that we should find a integer $q$ such that $n^2+km^2-q(n+m)=k+1$, but I can't go on and even see where $\gcd(m,n)=1$ applies. I'll be thankful for help.

Ian Barquette
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    Have you tried using $q = n + km$ (although that would give a multiple of $k + 1$ on the right side instead of just $k + 1$)? – John Omielan Jun 02 '22 at 22:29
  • Let $q = k+1$__ – DanielV Jun 02 '22 at 23:11
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    @JohnOmielan thank you for the insight. I think I can complete the proof with this. Just asking about the choice of $q$: is $q=n+km$ used with first degree to homogenize the exponents? – Ian Barquette Jun 03 '22 at 00:08
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    @IanBarquette You're welcome. As for the choice of $q = n + km$, it's just that this removes the $n^2$ and $km^2$ terms, leaving just only one term, in particular $-(k + 1)mn$, which is a multiple of $k + 1$ (although, if I understand you correctly, it also does "homogenous" the exponents). In addition, since you can complete the proof with this idea, even though you've already accepted the current answer, you may wish to add your own answer detailing the procedure. – John Omielan Jun 03 '22 at 00:15
  • By here in dupe: $,n!+!m\mid f(n)\iff n!+!m\mid f(-m),,$ for $f(x)$ any poly with integer coef's. So in OP: $,n!+!m\mid n^2!+!km^2\iff n+m\mid m^2(1!+!k)\iff n+m\mid 1!+!k,,$ by Euclid & $(n!+!m,m)=(n,m)=1$ – Bill Dubuque Jun 03 '22 at 01:41
  • @BillDubuque why is this a duplicate? It's a relativity different question, basically one can generalize the other... – Ian Barquette Jun 03 '22 at 10:08
  • @Ian Because it is a special case of the general method described in the linked answer in the dupe. We already have hundreds of variants on such and this adds nothing novel. – Bill Dubuque Jun 17 '22 at 15:54

1 Answers1

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Denote your two conditions by (1) and (2) respectively.

First observe that $m+n|(m-n)(m+n)=m^2-n^2$. Therefore $(1) \iff m+n|n^2+km^2 + m^2-n^2 = m^2(k+1)$. This condition denote by (3)

Now, observe that $\mathrm{GCD}(m+n,m^2)=1$, since if a prime number $p$ satisfies $p|m^2$ and $p|m+n$ then $p|m,n$, which contradicts the assumption.

Therefore $(3)\iff (2)$.

Mateo
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