For each of the following, decide whether or not the equality holds for all $a,b$. Prove each of the true statements and provide counterexamples for the false ones.
$$\gcd(a+b,a-b)=\gcd(a,b)$$
$$\gcd(a+b,2a-b)=\gcd(a,b)$$
$$\gcd(a+b,2a+b)=\gcd(a,b)$$
$a$ and $b$ have to be positive integers, where $a>b$.
I said the first statement is not true. Let $a=5$ and $b=3$. $a+b=8$, and $a-b=2$. Their gcd is $2$, but $5$ and $3$ are relativ3ly prime, so their gcd is $1$, which is not correct using the statement.
Statement 2 is also false for $a=7$ and $b=2$. $gcd(9,12)=3$ and $gcd(7,2)=1$, and $3$ doesn't equal $1$
I said the third statement is true because if $a$ divides $b$, then a+b and 2a+b both will sum to something that has the gcd. Is my reasoning flawed?
Notes on a proof of the third one.
It is clear that $\gcd(a,b)$ divides both $a+b,\,2a+b$. We need to argue that any common divisor of $a+b,\,2a+b$ divides both $a,b$.
We will use the fact that if $d$ divides $m,n$ then $d$ divides $Am+Bn$ for all $A,B\in \mathbb Z$.
So, suppose that $d$ divides both $a+b,\,2a+b$. Then:
$d\,|\,a$ Proof: $d$ must divide $2a+b-(a+b)=a$
$d\,|\,b$ Proof: $d$ must divide $2(a+b)-(2a+b)=b$
and we are done.