If $a$ and $b$ are relatively prime integers then prove that ($a$ ,$b^2$) =1

Question

From the title I've stucked in this question for half an hour. Could anyone help me?

Answer 1

We have $$ (a,b)=1 \iff ax+by=1 \text{ for some integer $x,y$} $$ Then $$ (ax+by)^2=1 \implies a(ax^2+2bxy)+b^2 y^2=1 \implies (a,b^2)=1. $$

Answer 2

Consider the prime factors of $a$ and those of $b$. The fact that $a$ and $b$ are relatively prime means that $a$ and $b$ have no common prime factors. But then $a$ and $b^2$ have no common prime factors either; that is, $a$ and $b^2$ are relatively prime. QED.

(We have used the observation that $b^2$ has the same prime factors as $b$, with each prime factor in $b^2$ repeated twice as many times as in $b$.)

Answer 3

If some integer $d$ divides $b$, then we can write $b = dx$ for some $x \in \mathbb{Z}$. From this, we have $b^2 = (dx)^2 = d(dx^2)$, so $d$ also divides $b^2$.

Now look back to the definition for $\gcd$. What if $d|a$ and $d|b^2$ wherein $d \neq 1$?

Answer 4

Proof by contraposition.

Let $d:= \gcd (a,b^{2}) \geq 2$. Then $d \mid a$ and $d \mid b^{2}$. Note that $d \mid b^{2}$ implies $\gcd (d, b) \geq 2$. So $\gcd (a,b) \geq 2$.