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Prove that if $\mathop{\mathrm{lcm}}( a, b) + \gcd(a, b) = a+b$, $a$ divides $b$ or $b$ divides $a$.

This problem seemed simple at first, however I cannot figure out a way to prove this. If I assume both parts of the statement, I can show it is true, but that is not a valid proof method.

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  • $\begingroup$ ... or $b$ divides $a$. $\endgroup$ – ccorn Jul 27 '15 at 3:00
  • $\begingroup$ Yes, sorry, I have updated it $\endgroup$ – Young Padawan Jul 27 '15 at 3:01
  • $\begingroup$ Please use LaTeX notation like this: $\gcd(a,b)\operatorname{lcm}(a,b)=ab$ which typesets as $\gcd(a,b)\operatorname{lcm}(a,b)=ab$. Oh, and this example could be understood as a hint. $\endgroup$ – ccorn Jul 27 '15 at 3:08
  • $\begingroup$ @ccorn It hasn't seemed to help me $\endgroup$ – Young Padawan Jul 27 '15 at 3:10
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    $\begingroup$ ccorn's hint is: show that $\gcd(a,b)$ and $\text{lcm}(a,b)$ are roots of $x^2-(a+b)x+ab$. $\endgroup$ – Batominovski Jul 27 '15 at 3:27
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Let $d = \gcd(a, b)$, and let $a = dA$ and $b = dB$.

Then, since $\operatorname{lcm}(a, b) =\frac{ab}{\gcd(a, b)} =\frac{ABd^2}{d} =ABd $, $d+ABd =Ad+Bd $, so $1+AB =A+B $ or $0 =AB-A-B+1 =(A-1)(B-1) $.

Therefore either $A=1$ or $B=1$. If $A=1$, then $a = dA = d$ divides $b = dB$. Similarly, if $B = 1$, $b$ divides $a$.

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  • $\begingroup$ Ah, Mr. (or, more likely, Dr. Prof.) Hardy. You are definitely a smooth operator(name). Me, I'm too lazy - I will use $a, b, c$ for $\alpha, \beta, \gamma$ and, in general, anything to minimize keystrokes. $\endgroup$ – marty cohen Jul 27 '15 at 5:21

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