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Let $a,b\in \mathbb{Z}$. It is a known fact that if there exists $x,y\in\mathbb{Z}$ such that $ax+by=1$, then 1 is the $\gcd(a,b)$. Let $d\in\mathbb{Z}$. Suppose there exists $x,y\in\mathbb{Z}$ such that $ax+by=d$, Prove that $d=1$ is the only integer for what this implies $\gcd(a,b)=d$.

An exercise in an introductory Algebra book just asks you to JUSTIFY that 1 is the only integer that satisfies this. This IS in one of the first chapters, before introducing Diophantine equations or Congruences. I solve it like this: Let $d$ be a positive integer such that there exists $x,y\in\mathbb{Z}$ with $ax+by=d$. If $m$ is $\gcd(a,b)$, then $m|d$, because $d$ is a linear integer combination of $a$ and $b$. Therefore $d$ is a multiple of $m$, which constitutes an infinite set. The only case in which we can conclude that $m=d$ is when $d=1$, because the only positive divisor of 1 is 1 itself. Other numbers have at least two divisors (itself and 1).

I think it was a good justification and answered the exercise in the book. But, on the other hand, this argument is not a proof, because, at first glance, nothing prevents other facts from causing some specific $d$ to have the desired property. A counterexample doesn't work here. How to formally prove this?

Benjamim Júnior
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  • "Prove that $1$ is the only integer satisfying this." Satisfying what exactly? – Dietrich Burde Feb 03 '24 at 16:24
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    If $ax+by=1$, then $axd+byd=d$. – Geoffrey Trang Feb 03 '24 at 16:24
  • @Dietrich (...) Satisfying the proposition right before, of course. It means that replacing 1 in that proposition for any other positive integer makes the proposition not valid. – Benjamim Júnior Feb 03 '24 at 16:32
  • Above you are not saying "other positive integer", only "other integer". So $-1$, for example? – Dietrich Burde Feb 03 '24 at 16:34
  • But this is not necessary, because the gcd is always positive, by definition. – Benjamim Júnior Feb 03 '24 at 16:41
  • I'll edit the question to make it clearer. – Benjamim Júnior Feb 03 '24 at 16:43
  • Theorems don't have reciprocals. Maybe the word you want is "converse"? – Gerry Myerson Feb 04 '24 at 10:21
  • @Gerry Why not? Aren't theorems condicional sentences? Every condicional sentences hás a reciprocal. At least, we use an equivalente word in portuguese... Anyway, It is not a reciprocal formally speaking, as it only applies to the case d=1. But in the literature I found this term being used informally for this theorem. – Benjamim Júnior Feb 04 '24 at 12:38
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    Maybe It was just a translation problem. – Benjamim Júnior Feb 04 '24 at 13:10
  • By here in the dupe, $,ax+by=d,$ is solvable $\iff \gcd(a,b)\mid d,,$ so you question is equivalent to showing that $,\gcd(a,b)\mid d\iff \gcd(a,b)=d,$ is true iff $,d=1$, i.e. that $,d=1,$ is the only natural whose only divisor is itself - whose proof is obvious. – Bill Dubuque Feb 04 '24 at 13:19
  • @Bill. Thank's for your contribuition. But I don't think my question is answered in the other question indicated. They are different questions, and the fact that one of the answers contains information from which it is easy to deduce the answer to my question does not make it redundant on the site. Someone who is directly interested in the converse of Bezout's theorem will probably not research the relationship between the inverse element in a congruence ring and Diophantine equations. – Benjamim Júnior Feb 04 '24 at 14:00
  • @Bill Furthermore, different people, at different levels of study, have different perceptions of what is obvious or trivial. You, for example, may think this is obvious, but others may not, and I believe that this community aims to be more inclusive. – Benjamim Júnior Feb 04 '24 at 14:00
  • Do you really think it is nontrivial to prove that a natural $>1$ has a divisor other than itself? – Bill Dubuque Feb 04 '24 at 14:09
  • @Bill Not this fact isolated, but the sequence of equivalences you use until stop on It. And, like I've said, beeing trivial is something relative. It depends on the point of view and the student level. – Benjamim Júnior Feb 04 '24 at 14:13
  • There is no "sequence". Rather there is only the fundamental equivalence I linked to in the dupe. It follows immediately from that. – Bill Dubuque Feb 04 '24 at 14:16
  • @Bill You think so because of your experience. Students in initial levels do not see in this way. The more we learn, more dificult is to us understanding the peculiarities and difficulties in more basic content. – Benjamim Júnior Feb 04 '24 at 14:21

1 Answers1

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We will not demonstrate that the implication is valid for $d=1$, since, as stated, it is a well-known fact.

Let $d\in \mathbb{Z}$ be such that $d$ is different from $1$. Consider any $a,b\in \mathbb{Z}$ such that $a$ and $b$ are relatively prime numbers, that is, $\gcd(a,b)=1$. It follows, by Bezout's theorem, that there are $x,y\in \mathbb{Z}$ such that $xa+yb=1$. This implies $dxa+dyb=d$. Therefore, there are $x',y'\in \mathbb{Z}$ such that $x'a+y'b=d$, namely: $x'=dx$ and $y'=dy$. Thus, we find a linear combination in the integers of $a$ and $b$ resulting in $d$ without $d$ being $\gcd(a,b)$, as $a$ and $b$ were taken so that $\gcd(a,b)=1$ and $d$ was taken as different from $1$. Thus, the intended implication is not valid for any $d\in\mathbb{Z}$ that is different from $1$. As we wanted to demonstrate.

Benjamim Júnior
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