This is my proof

Let $m$ be $\gcd(ka,kb)$ then exists $s$ and $t$ such that

$ska+ tkb = m => k(sa+tb) = m $

$=> sa + tb = m/k$ (1)

since (1) is linear combination of $a$ and $b$ and $m$ is smallest positive linear combination of $ka$ and $kb$ => $m/k$ is smallest positive then $\gcd(a,b) = m/k$.

Is this proof is good enough to prove $\gcd(ka,kb)=k\cdot \gcd(a,b)$

  • 1
    $\begingroup$ Why would m being the smallest linear combination of ka and kb imply that m/k is the smallest linear combination of a and b? Isn't that what you are proving. Also, in using Bezout's lemma, are you sure that you haven't assumed the result to begin with? Maybe the proof of Bezout's lemma doesn't require knowing this, but are you sure it doesn't? $\endgroup$
    – fleablood
    Commented Sep 28, 2017 at 14:59
  • 1
    $\begingroup$ you are right.This proof not good enough $\endgroup$
    – xuoimai
    Commented Oct 3, 2017 at 9:56

1 Answer 1


Let $a=p_{1}^{q_{1}}p_{2}^{q_{2}}...p_{k}^{q_{k}}$ and $b=p_{1}^{r_{1}}p_{2}^{r_{2}}...p_{k}^{r_{k}}$ be their respective prime decompositions.

Thus, $gcd(a,b)=p_{1}^{\min(q_{1},r_{1})}p_{2}^{\min(q_{2},r_{2})}...p_{k}^{\min(q_{k},r_{k})}$

Now, if $k=p_{1}^{s_{1}}p_{2}^{s_{2}}...p_{k}^{s_{k}}$ , then

$ka=p_{1}^{s_{1}+q_{1}}p_{2}^{s_{2}+q_{2}}...p_{k}^{s_{k}+q_{k}}$ and $kb=p_{1}^{s_{1}+r_{1}}p_{2}^{s_{1}+r_{1}}...p_{k}^{s_{k}+r_{k}}$

Now, note that $\min(i+x,j+x)=x+\min(i,j)$. Hence,

$$gcd(ka,kb)=p_{1}^{\min(s_{1}+q_{1},s_{1}+r_{1})}p_{2}^{\min(s_{2}+q_{2},s_{1}+r_{1})}...p_{k}^{\min(s_{k}+q_{k},s_{k}+r_{k})}$$ $$=(p_{1}^{s_{1}}p_{2}^{s_{2}}...p_{k}^{s_{k}})(p_{1}^{\min(q_{1},r_{1})}p_{2}^{\min(q_{2},r_{2})}...p_{k}^{\min(q_{k},r_{k})})$$ $$=k \cdot gcd(a,b)$$ Hence proved.

Basically, this has the simple intuition that by the definition of gcd, the gcd is the collection of the common factors. So, when we replace $a$ by $ka$ and $b$ by $kb$, there is one extra factor added to this collection, and that factor is nothing but $k$. Hence, we get the result.


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