For example, if we have
$x + y = z$
$d \mid x$
$d \nmid y$
Can $d \mid z$ ? I'm trying to prove that it is impossible.
To be more specific, I'm trying to prove that there exists $\phi(z/d)$ possible values for $x$ and $y$ with $x+y=z$, $x>0$, $y>0$ and $gcd(x,y) = d$. Proving the above divisibility property is the last step I need but I don't know how to do it.
If $gcd(x, y) = d$ then $gcd(x/d, y/d) = 1$. I'm trying to verify that every solution for $x/d$ correponds to a coprime of $z/d$ in $[1,z/d]$.
To prove that every coprime generates a valid pair $(x, y)$:
$a = x/d$, $b = y/d$, $c = z/d$
$gcd(a,b) = gcd(a, c-a)$
Let's say $d' \mid a$ and $d' \mid (c-a)$ with $d' > 1$
We know that $a$ and $c$ are coprime so $d' \nmid c$
If $d \nmid (c-a)$ because of the divisivility property in question, we arrive at a contradiction.
The proof for $a$ and $c$ not being coprimes is very similar.
There may be better proofs, and I would love to see them, but I think verifying that property could be helpful for future problems nontheless.
