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I have seen this question a couple of times before on this website with very different, yet correct, proofs of this proposition. I came up with the following proof, and would like to know if it is correct.

Using that if $(a,b)=d$ then $d=ma+nb$, we have

if $(a+b,a-b) =d$, then

$$d=a+b+a-b=2a$$ $$d=a+b-(a-b)=2b$$

Thus,

$d | 2a$ and $d| 2b$

And,

$d=1$ or $d=2$

Basem Fouda
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2 Answers2

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$$d=gcd(a+b,a-b)\implies \begin{cases}d|(a+b)\\ d|(a-b)\end{cases}$$ So $$d|(a+b+a-b)\iff d|2a$$ and $$d|(a+b-(a-b))\iff d|2b$$ That is $$d|gcd(2a,2b)=2gcd(a,b)=2\implies d\in\{1,2\}.$$

mfl
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  • I am new to number theory, can you say that if $a|b$ and $a|c$ then $a|mb + nc$ ?Also the book from which I am studying has the theorem that I used above: The greatest common divisor of two integers $a$ and $b$, not both $0$ is the least positive integer such that $ma + nb =d$ for some integers $m$ and $n$. Am I misunderstanding it or misusing it? – Basem Fouda Nov 05 '16 at 13:43
  • $a|b\iff \exists k: b=ka$ and $a|c\iff \exists l: c=la.$ Then we have that if $a|b$ and $a|c$ then $mb+nc=mka+nla=(mk+nl)a,$ which means that $a|(mb+nc).$ The other part of your question is Bézout's identity https://en.wikipedia.org/wiki/B%C3%A9zout%27s_identity – mfl Nov 05 '16 at 16:13
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By Bezout $\ d := (a,b) = ma + nb\, $ for some $\,m,n,\, $ not any $\,m,n.\, $ So you cannot simply assume $\,m,n = 1,\pm1.\ $ However, since $\,(a,b)\mid a,b\,$ it is true that $\, (a,b)\mid ma+nb\,$ for all $\,m,n.\,$ Thus if you change $\, d =\ldots\, $ to $\,d\mid\ldots\,$ in your displayed equations then they are correct.

Finally you need to justify your claim that $\,d\mid 2a,2b\,\Rightarrow\, d\mid 2.\ $

Bill Dubuque
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  • So it is not true to say that if $a|b$ and $a|c$ then $a|mb+nc$. However it is only true if $gcd(b,c)=a$. Am I correct? Also the book from which I am studying seems to be missing on a lot of the basic rules. Would you recommend a book for beginners in number theory? Thanks. – Basem Fouda Nov 05 '16 at 14:14
  • That divisibility statement is always true. But you wrote it incorrectly as an equality $\ d = ,\ldots,$ in the displayed equations in your question. Hence the explanation above. – Bill Dubuque Nov 05 '16 at 14:18
  • I don't think you got my question. I am asking if it is true to say that : if $x|y$ and $x|z$ then $x|my+nz$ without $x$ being the gcd of $y$ and $z$. – Basem Fouda Nov 05 '16 at 14:22
  • @BasemFouda Yes, as I said, it is always true, since $\ m(b/a) + n(c/a)\ $ is an integer since $, b/a, c/a,$ are integers. – Bill Dubuque Nov 05 '16 at 14:23
  • Thanks. Would you have any book recommendations? – Basem Fouda Nov 05 '16 at 14:25
  • Any textbook on elementary number theory will cover these basics. I cannot speculate what will work best for you without knowing much more about your background. It may help to browse various textbooks at a university library. Also iirc there are previous questions on book recommendations. – Bill Dubuque Nov 05 '16 at 14:28
  • Let us continue this discussion in chat. – Basem Fouda Nov 05 '16 at 14:30