If $\gcd(a, b) = 1$ then $\gcd(ab, a+b) = 1$? In a mathematical demonstration, i saw:
If $\gcd(a, b) = 1$ Then $\gcd(ab, a+b) = 1$
I could not found a counter example, but i could not found a way to prove it too either.
Could you help me on this one ?
 A: Hint:
If a prime $p \mid ab$, then $p$ divides either $a$ or $b$.  WLOG let $p \mid a$.  If $p \mid (a+b)$ also, then $p \mid b$.
A: Hint: It is true. Indeed, if $\gcd(ab,a+b) > 1$, then there is a prime $p$ divding it, in particular, there is a prime $p$ dividing both $ab$ and $a+b$, and remember Euclid's Lemma...
A: First prove


*

*$\gcd(mn, k)=1$ if and only if $\gcd(m,k)=1$ and $\gcd(n,k)=1$.

*If $\gcd(m,k)=1$ then $\gcd(m,m+k)=1$.


The desired result follows from these like so: From $\gcd(a,b)=1$, we have $\gcd(a, a+b)=\gcd(b, a+b)=1$, implying 
$\gcd(ab, a+b)=1$.
A: Suppose that we have that some prime $p$ divides the gcd of ab and a+b. Then, we know that $p\mid a$ or $p\mid b$. WLOG suppose that p divides a. Then, it follows that a+b is divisible by by p, so b must also be divisible by p. But this means that $a$ and $b$ have a prime factor in common, contridicting the assumption that they were coprime.
A: If a prime $p$ divides $ab$, then it divides $a$ or $b$. WLOG it divides $a$. If that prime also divides $a+b$, then since it divides $a$ it must divide $b$ as well, contradicting $gcd(a,b)=1$. 
A: We can take as the definition of "relatively prime" that there are $s,t$ with $as+bt=1$ (or we can show that this is equivalent to our favorite definition.) SO given that, can we find $S,T$ with $$abS+(a+b)T=1?$$ Well we have $$a(s-t)+(a+b)t=1$$ and $$b(t-s)+(a+b)s=1$$ So multiplying gives $$ab(s-t)(t-s)+(a+b)\left[\ a(s-t)s+tb(t-s)+(a+b)st \  \right]=1 .$$ That does the job, but expanding and simplifying gives the more satisfactory $$ab(s-t)(t-s)+(a+b)[\ as^2+bt^2\ ]=1 $$
A: We use the idea of the Euclidean algorithm. I'll arbitrarily decide that "b" looks bigger than "a", so I can reduce $ab$ by subtracting off $a$ times $a+b$:
$$ \gcd(ab, a + b) = \gcd(ab - a(a+b), a+b) = \gcd(a^2, a+b)$$
Hrm. Usually this works out for homework problems. But never fear, there is something related we can still do. We know that the $\gcd$ divides $ab$, $a+b$, and $a^2$. Similarly, we can see that it also has to divide $b^2$.
In particular, it has to divide
$$ \gcd(a^2, b^2) = 1 $$
