0
$\begingroup$

Given $a, b$ positive integers, such that $\gcd(a, b) = 2$, $\gcd(a, 5) = 1$ and $a|5b$, show $a = 2$.

My solution:

Once for $d = gcd(a, b) \implies d|a$ and $d|b$, therefore for $\gcd(a, b) = 2$, $2|a$ and $2|b$, this implies $a, b$ are even numbers. Once the greatest common factor of $a$ and $b$ is $2$, therefore $a$ or $b$ must be equal to $2$.

According to Bézout's Theorem, for $a, b \in\mathbb{Z}$, such that $d = gcd(a, b)$. Then there exist $x, y \in\mathbb{Z} : ax + by = d$. Therefore, $2 = ax+by$ and $1 = ax + 5y$. So,

$$2-by = 1 - 5y \implies 1 = by - 5y \implies 1 = y(b-5)$$

Considering $b = 2$, so $1 = y(2-5) \implies 1 = -3y \implies y = -\dfrac{1}{3}$, but $-\dfrac{1}{3}\not\in \mathbb{Z}$, therefore $a = 2$.

I am not sure if this solution is correct. I even not considered $a|5b$, how can I use it in the proof?

I would like to know a better solution or what would make my attempt better.

$\endgroup$
1
  • $\begingroup$ The issue with your proof is that you use Bézout's Theorem two times but use the same numbers twice. From Bézout's Theorem you know that there must exist $x_0,y_0,x_1,y_1\in \mathbb{Z}$ such that $ax_0 + by_0 = 2$ and $ax_1 + 5y_1 = 1$; it's not the same $x$ and $y$ for both relations. $\endgroup$
    – Peluso
    Commented Feb 8, 2021 at 19:34

1 Answer 1

2
$\begingroup$

Once the greatest common factor of a and b is 2, therefore a or b must be equal to 2.

Wrong : counter example is $a = 4, b = 6.$

From Number Theory, if $r,s$ relatively prime and $r|(ks)$, then $r|k$.

$\gcd(a, b) = 2$, $\gcd(a, 5) = 1$ and $a|5b$, show $a = 2$

From the premises, you therefore have that $a|b$.

Therefore, $\gcd(a,b) = a = 2.$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .