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

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.
