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I have a question as follow:

Suppose I have two positive integers $a, b$ (both not equal to $1$) such that $a$ divides $b^2.$ Show that $a$ and $b$ have common divisors.

It is straightforward to use prime factorisation to prove. However, I am seeking for alternative explanation.

Anne Bauval
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Novice
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    You can use that there are integers $x,y$ with $xa + yb = \gcd(a,b)$. – Daniel Fischer Nov 26 '16 at 14:13
  • Here is my attempt using your hint: Suppose the $gcd(a,b)=1$. With $xa+yb=1$, we have $xab+yb^2=b$. Since $a|b^2$, $b^2=ka$ for some integer $k$. Therefore, $xab+yka=b$ which means $a(xb+yk)=b$ which means $a|b$. This means $a|gcd(a,b)=1$, which is a contradiction. – Novice Nov 26 '16 at 15:20
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    That's one way to use it. Another is to note that $a \mid (xa + yb)^2$, so $a \mid \gcd(a,b)^2$, whence $\gcd(a,b) \geqslant \sqrt{a} > 1$. – Daniel Fischer Nov 26 '16 at 15:22

3 Answers3

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Because of $$a|b^2$$ there is a positive integer $k$ with $$ak=b^2$$ Hence, we have $$\frac{k}{b}=\frac{b}{a}$$

If we assume $gcd(a,b)=1$, we can conclude $a|b$ because every fraction equal to $\frac{b}{a}$ is of the form $\frac{mb}{ma}$, hence $ma=b$ for some $m$.

Hence $a\ne 1$ is a common divisor of $a$ and $b$. This is a contradiction to $gcd(a,b)=1$, hence $gcd(a,b)>1$ completing the proof.

Peter
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Let $a,b$ be two coprime integers such that $a\mid b^2.$ By Gauss's generalization of Euclid's lemma (used twice), $a\mid1.$

Anne Bauval
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If $\gcd(a,b)=1,$ then $\exists m,n\in \Bbb Z$ such that $ma+nb=1$ which implies that $mab+nb^2=b \tag1$ Since $a|b^2$, $a$ divides the left hand side of the equation $(1)$. Hence, $a|b$ and $\gcd(a,b)=a=1$.

Bob Dobbs
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    This is the best answer posted to date. I would have concluded a bit more fully: Having established $a\mid b$, it is the case that $(a\mid b) \land (\gcd(a,b)=1) \Rightarrow a=1$. But the problem specifies $a,b>1$. Therefore the starting assumption in the proof that $\gcd(a,b)=1$ is false, whence $\gcd(a,b)>1$, which is what OP particularly asks for. – Keith Backman Jul 23 '23 at 21:13
  • This is a usual proof of Gauss's generalization of Euclid's lemma in a particular case. Note that this generalization actually does not need Bézout identity. It can be proved directly in any GCD domain. – Anne Bauval Jul 23 '23 at 21:38