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I went about trying to prove above statement but need a little help. I don't know if my reasoning for why $gcd(a*c, b*c)= c*gcd(a, b)$ is correct. Could I get help on clarifying why this theorem is so?

Proof:

Let $a$ and $b$ be even integers. Then, $a=2k_{1}$ and $b=2k_{2}$, where $k_{1}, k_{2}\in\mathbb{Z}$ by the definition of even integers.

The $gcd(a, b)$ is then, $gcd(2k_{1}, 2k_{2})$. We know the gcd is the product of the prime factorizations that divide both $a$ and $b$, taken with their minimum multiplicities. Thus, multiplying $a$ and $b$ by a third integer increases the multiplicities identically and increases the minimum in the same way. Therefore, $gcd(2k_{1}, 2k_{2})= 2* gcd(k_{1}, k_{2})$.

We know that $a=2k_{1}$ and $b=2k_{2}$, so solving for $k_{1}$ and $k_{2}$ gives us: $k_{1}=\frac{a}{2}$ and $k_{2}=\frac{b}{2}$.

Substituting $k_{1}$ and $k_{2}$ back into $2*gcd(k_{1}, k_{2})$, we get $gcd(a, b)=2*gcd(\frac{a}{2}, \frac{b}{2})$. Therefore, when both $a$ and $b$ are even integers, the $gcd(a, b)= 2*gcd(\frac{a}{2}, \frac{b}{2})$.

Bill Dubuque
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Frances Lee
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  • To do it that way it boils down to proving $,\min(j,k) = 1+\min(j-1,k-1),,$ where $j,k$ are the exponents of $2$ in the prime factorizations of $,a,b.,$ See the linked dupe for other ways. – Bill Dubuque Dec 10 '20 at 22:17
  • "Thus, multiplying a and b by a third integer increases the multiplicities identically and increases the minimum in the same way. Therefore, gcd(2k1,2k2)=2∗gcd(k1,k2)." Clean this up and say it in a way that is mathematically sound. This will be the heart of your argument but right now it is incoherent. (Hint: This isn't just true of $2$. $\gcd(ka, kb) = k\gcd(a,b)$. for any non zero integers. Try to prove that. That's the heart of it all.) – fleablood Dec 10 '20 at 22:25
  • would it be correct to say: The gcd(a, b) can be expressed as $gcd(a, b)= 2^{min(a_{1}, b_{1})}$. Where $a_{1}=1$ and $b_{1}=1$ since both a and b are even integers. When a/2 and b/2, then min(a1-1, b1-1). and multiplying 2 to the gcd, $$2gcd(a/2, b/2)= 22^{min(a_{1}, b_{1})}= 2^{min(a_{1}-1, b_{1}-1)+1}$$ $$min(a_{1}, b_{1})=min(a_{1}-1, b_{1}-1)+1$$ $$min(1, 1)=min(1-1, 1-1)+1=1$$ – Frances Lee Dec 10 '20 at 22:50
  • I think what Bill Dubuque is getting at is if $a = 2^{k}\prod p_i^{m_i}$ and $b=2^j\prod p_i^{n_i}$ then $\gcd(a,b) = 2^{\min(k,j)}\prod p_i^{\min(m_i,n_i)}$. So if $\frac a2= 2^{k-1}\prod p_i^{m_i}$ and $\frac b2=2^{j-1}\prod p_i^{n_i}$, then $\gcd(\frac a2,\frac b2) = .........$ – fleablood Dec 10 '20 at 23:01
  • Thank you so much!! – Frances Lee Dec 10 '20 at 23:19

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