Least Common Multiple and Greatest Common Divisor

Question

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.

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. — ccorn, Jul 27 '15 at 03:08
ccorn's hint is: show that $\gcd(a,b)$ and $\text{lcm}(a,b)$ are roots of $x^2-(a+b)x+ab$. — Batominovski, Jul 27 '15 at 03:27

score 3 · Accepted Answer · edited Jul 27 '15 at 04:43

3

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 $a$.

edited Jul 27 '15 at 04:43

Michael Hardy

1

answered Jul 27 '15 at 03:25

marty cohen

107,799

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. – marty cohen Jul 27 '15 at 05:21

Least Common Multiple and Greatest Common Divisor

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