$\gcd(a,b) = \gcd(a + b, \mathrm{lcm}[a,b])$

Question

Show that if $a$, $b$ are positive integers, then we have: $\gcd(a,b) = \gcd(a + b, \mathrm{lcm}[a,b])$.

Answer 1

Below are a few proofs. First, here's a couple from my sci.math post on 2001/11/10
For variety here is yet another using $\rm\ (a,b)\ \ [a,b]\ =\ ab\ \ $ and basic gcd laws:

$\rm\quad (a,b)\ (a+b, [a,b])\ =\ (aa+ab,\ ab+bb,\ ab)\ =\ (aa,\:bb,\:ab)\ =\ (a,b)^2\quad $ QED

By the way, recall that the key identity in the second proof arose the other day in our discussion of Stieltjes $\rm\ 4\:n+3\ $ generalization of Euclid's proof of infinitely many primes. Here's a slicker proof:

LEMMA $\rm\ \ (a+b,\:ab) = 1\ \iff\ (a,\:b) = 1$

Proof $\rm\ \ \ (a,\:b)^2 \subset (a+b,\:ab) \subset (a,\:b)\ \ $ since e.g. $\rm\ \ a^2 = a\:(a+b)-ab\in (a+b,\:ab)$

Thus $\rm\ 1\in (a+b,\ ab)\ \Rightarrow\ 1\in (a,\:b)\:.\:$ Conversely $\rm\ 1 \in (a,\:b)\ \Rightarrow\ 1 \in (a,\:b)^2 \subset (a+b,\:ab)\quad$ QED

Answer 2

Another Dubuquesque attempt; for legibility, write $d=\gcd(a,b)$: \begin{align*} \gcd\Bigl(d(a+b), ab\Bigr) &= \gcd\Bigl(d(a+b), ab, ab\Bigr)\\ &=\gcd\Bigl(d(a+b),\ ab-a(a+b),\ ab-b(a+b)\Bigr)\\ &=\gcd\Bigl(d(a+b),\ a^2,\ b^2\Bigr)\\ &=\gcd\Bigl(d(a+b),\ \gcd(a^2,b^2)\Bigr)\\ &=\gcd\Bigl(d(a+b),\ \gcd(a,b)^2\Bigr)\\ &=\gcd\Bigl(d(a+b),\ d^2\Bigr)\\ &= d\gcd\Bigl(a+b,d\Bigr)\\ &= d\gcd\Bigl(a+b,\gcd(a,b)\Bigr)\\ &= d\gcd(a,b)\\ &= \gcd(a,b)\gcd(a,b). \end{align*}

(Second line uses the fact that $a(a+b)$ and $b(a+b)$ are both multiples of $d(a+b)$).

Now divide through by $\gcd(a,b)$ to get the desired result.

Answer 3

Start by writing $a=d a'$, $b=d b'$, where $d=(a,b)$.

Answer 4

Perhaps overkill, but if you accept the 'distribution law' $(x,[y,z])=[(x,y),(x,z)]$ stated at Wikipedia, then it is easy:

$(a+b,[a,b])=[(a+b,a),(a+b,b)]=[(b,a),(a,b)]=(a,b)$

where in the second equality I used the easy fact $(x,y)=(x,y \mod x)$.