When is $\gcd(a+b,c)=\gcd(a,c-b)$? Let $a,b,c$ be three positive integers. Under which conditions do we have $\gcd(a+b,c)=\gcd(a,c-b)$ ?
Set $d= a+b$. Then the question becomes: When is $\gcd(d,c) = \gcd(d-b,c-b)$?

Let $g=\gcd(d,c)$. Then $d=sg$ and $c=tg$ (with $s,t$ coprime). So there are two cases when the sought-for equality will not hold:


$g$ does not divide $b$ (equivalently, $b$ is not a multiple of $g$). 
$b=rg$ such that $s-r$ and $t-r$ have a common factor, so the resulting $\gcd$ is a multiple of $g$. This is harder to test for.


Sample case: $(a,b,c) = (40, 6, 16)$. $\gcd(a+b,c) \ne \gcd(a,b) \ne (gcd(a,b-c)$
Set $d= a+b$. Then the question becomes: When is $\gcd(d,c) = \gcd(d-b,c-b)$?

Let $g=\gcd(d,c)$. Then $d=sg$ and $c=tg$ (with $s,t$ coprime). So there are two cases when the sought-for equality will not hold:


$g$ does not divide $b$ (equivalently, $b$ is not a multiple of $g$). 
$b=rg$ such that $s-r$ and $t-r$ have a common factor, so the resulting $\gcd$ is a multiple of $g$. This is harder to test for.


Sample case: $(a,b,c) = (40, 6, 16)$. $\gcd(a+b,c) \ne \gcd(a,b) \ne (gcd(a,b-c)$