CHECK: Show that if $b$ is an odd number, then $gcd(2a,b)=gcd(a,b)$ Show that if $b$ is an odd number, then $$gcd(2a,b)=gcd(a,b)$$
$\textbf{Proof:}$ We will prove that $(a,b)$ and $(2a,b)$ have the same set of divisors. Assume that $b$ is an odd number.
$\Leftarrow$:  Suppose $d|a$ and $d|b$. So there are integers $u$ and $v$ such that $a=u \cdot d$ and $b=v \cdot d$. We now need to check $d$ is a divisor of $(2a,b)$. We know that $d|b$. So we are left with showing $d|2a$. Well $$a=u\cdot d \iff 2a=2u\cdot d$$
Since $2u$ is an integer, $d|2a$. Therefore, $d$ is a common divisor of $(2a,b)$.
$\Rightarrow$: Now suppose $d|2a$ and $d|b$. So there are integers $r$ and $s$ such that $2a=r \cdot d$ and $b=s\cdot d$. We now need to check $d$ is a divisor of $(a,b)$. We know that $d|b$. So we are left with showing $d|a$. Since $b$ is odd then it's divisors are either 1 or its self. So if $d|2a$ and $d \not{|} 2$, then $d|a$. Therefore, $d$ is a common divisor of $(a,b)$.
Since $(a,b)$ and $(2a,b)$ have the same set of divisors, then maximum divisor of each pair must be the same. Hence $gcd(a,b)=gcd(2a,b)$.
 A: If $b$ is odd, so is the $\gcd$, and any power of $2$ in the other argument is discarded.
A: Since $\text{gcd}(x,y)=\prod_{p \in \mathbf{P}}p^{\min\{\upsilon_p(x),\upsilon_p(y)\}}$, where $\upsilon_p(x)$ represents the highest power of $p$ dividing $x$, then
$$\text{gcd}(a,b)=2^{\min\{\upsilon_2(a),0\}}\prod_{p \in \mathbf{P}\setminus \{2\}}p^{\min\{\upsilon_p(a),\upsilon_p(b)\}}=2^{\min\{\upsilon_2(2a),0\}}\prod_{p \in \mathbf{P}\setminus \{2\}}p^{\min\{\upsilon_p(a),\upsilon_p(b)\}}.$$
Since all primes in the second product are odd then
$$\text{gcd}(a,b)=2^{\min\{\upsilon_2(2a),0\}}\prod_{p \in \mathbf{P}\setminus \{2\}}p^{\min\{\upsilon_p(2a),\upsilon_p(b)\}}=\text{gcd}(2a,b).$$
A: A different proof:
Let the prime factorizations of $a$ and $b$ be
$$a = \prod_{i=1}^{k}p_i^{a_i} \\ b = \prod_{i=1}^{k}p_i^{b_i},$$
where $p_1 = 2, p_2 = 3, p_3 = 5, ...$ and the $k$th prime $p_k$ is the maximum prime in either factorization.
Then we have
$$GCD(a,b) = \prod_{i=1}^{k}p_i^{min(a_i, b_i)}.$$
Since $b$ is odd, $b_1 = 0.$  (An odd number must have only odd factors, and $2$ is the only even prime.)  Then, going from $a$ to $2a$, the only change in the prime factorization is adding one to $a_1$.
Since $min(a_1 + 1, 0) = min(a_1, 0) = 0$, $GCD(2a, b) = GCD(a,b).$
P.S.:  This also shows that the GCF is odd!
