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What about the converse: Given that $a^2 \mid b^2$ is it necessarily the case that $a \mid b$? Justify.

My train of thought:

Proof: Direct

Assume: $a$ and $b$ are integers. And $a^2$ divides $b^2$

so,

$b^2 = ka^2$ where $k$ is some integer square: $k = j^2$

then, $b = \sqrt{k}\cdot a \implies b = ja$

This implies $a \mid b$. Done.

ForgotALot
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Loc
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    Welcome to Stackexchange. You'll find that simple "Here's the statement of my exercise, solve it for me" posts will be poorly received. What is better is for you to add context: What you understand about the problem, what you've tried so far, etc. Something to both show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance. –  Mar 20 '16 at 01:59
  • It is correct.${}$ –  Mar 20 '16 at 02:02
  • Yes. I appreciate the look out, Shahab. I was editing while you wrote that. Is my proof enough for this? – Loc Mar 20 '16 at 02:02
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    It is true that $k$ is a square, but that requires proof. – André Nicolas Mar 20 '16 at 02:02
  • Ah. I see. How do I make it more explicit or proof that k is actually squared? – Loc Mar 20 '16 at 02:04
  • You use the prime decomposition. –  Mar 20 '16 at 02:04
  • For which integer? So basically j are two prime numbers? – Loc Mar 20 '16 at 02:05
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    $b$ and $a$, so you recover the ones of $b^2$ and $a^2$, from which you recover that of $k$. I don't understand what you mean by "$j$ is two prime numbers", but there are istances of $a$ and $b$ for each $j\in \Bbb Z$. –  Mar 20 '16 at 02:06
  • At a loss, with the prime decomposition. I thought for some reason it would make sense with the Js. So if its with a and b, wouldn't that just make it (bb) = k (aa) – Loc Mar 20 '16 at 02:14
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    @b5er: No problem. Your proof is correct in essence. Aternatively note that every single prime $p$ going in $a$ must also be going in $b$ due to $b^2=ka^2$ because of unique prime factorization of both sides. –  Mar 20 '16 at 02:21

1 Answers1

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The clearest way is to use the Unique Factorization Theorem. We give a proof that bypasses it.

Assume (it makes no difference) that $a$ and $b$ are positive. Let $d=\gcd(a,b)$. We want to show that $d=a$.

Let $a=da_0$ and $b=db_0$. Then $a_0$ and $b_0$ are relatively prime. From $a^2\mid b^2$ we conclude that $a_0^2\mid b_0^2$. Since $a_0$ and $b_0$ are relatively prime, from $a_0\mid b_0\cdot b_0$ we conclude that if $a_0$ does not divide (the first) $b_0$, it must divide the second. So $a_0\mid b_0$, and therefore $a_0=1$. Thus $a=d$, and we are finished.

Added: For completeness, we write down the Unique Factorization Theorem argument that shows that your $k$ is a perfect square. Let $b=p_1^{b_1}\cdots p_t^{b_t}$, where the $p_i$ are distinct primes. Then $b^2=p_1^{2b_1}\cdots p_t^{2b_t}$. Since $a^2$ divides $b^2$, all the prime divisors of $a^2$, and hence of $a$, are among the $p_i$. So $a=p_1^{a_1}\cdots p_t^{a_t}$, where some of the $a_i$ may be $0$. It follows that $a^2=p_1^{2a_1}\cdots p_t^{2a_t}$.

Divide. We find that $k=\frac{b^2}{a^2}=p_1^{2b_1-2a_1}\cdots p_t^{2b_t-2a_t}$. Since all the exponents of the $p_i$ in the factorization of $k$ are even, it follows that $k$ is a perfect square.

André Nicolas
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  • Ah. Alright. I'm new at proofs. So this is the entire proof? We don't need to essential proof k in this case, right? And they are relative primes because they have the same gcd? – Loc Mar 20 '16 at 05:41
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    The numbers $a_0$ and $b_0$ are relatively prime because if some $e\gt 1$ divided both, then $de$ would divide $a$ and $b$, contradicting the fact that $d$ is the greatest common divisor of $a$ and $b$. Please note that as I pointed out the proof that uses the Unique Factorization Theorem is more natural. I was hoping someone would post it. It is late now, but perhaps tomorrow if no one has I will add a version of it. – André Nicolas Mar 20 '16 at 05:47