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How to prove the following?: Let $m,n\in \mathbb{N}$ ; $ \;m^{2}\mid n^{2}\Longrightarrow \;\;m\mid n$

Just a hint please. I tried two ways but did not work.

Martin Sleziak
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SonicFancy
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    prove every prime factor p of m divide n – delta Mar 06 '14 at 03:39
  • Please list which ways you tried so we can help. What does $m^2\mid n^2$ mean at a definition level? – abiessu Mar 06 '14 at 03:40
  • @delta OK I will try it now – SonicFancy Mar 06 '14 at 03:40
  • @abiessu exact division – SonicFancy Mar 06 '14 at 03:50
  • @delta sorry can I get some more hints... No improvement... – SonicFancy Mar 06 '14 at 03:53
  • express $m=\prod_i p_i^{k_i}$, here we try to prove for every $p_i^{k_i}$ divide n.the basic idea is if $p^2|n^2$, then $p|n$, if not, then $p^2 \nmid n^2$. – delta Mar 06 '14 at 04:01
  • If $p$ is a prime factor of $m$ and $p|n^2$, does that mean $p|n$? Why or why not? – Nick D. Mar 06 '14 at 04:02
  • @NickD. Yes, if $p \nmid n$, then in the factorization of n, it does not have factor p, so does $n^2$, which lead to $p \nmid n^2$. – delta Mar 06 '14 at 04:05
  • @NickD. Oh, I come up with a better explaination. because $p$ is prime, then gcd(p,n) is either p or 1, if it is p, then $p|n$, if not, then we have gcd(p,n)=1, as we know, there must exist integer a and b s.t. ap+bn=1, then bn=1-ap, $(bn)^2=(1-ap)^2=a^2p^2-2ap+1$, because $p|n^2,p|p^2,p|2ap$, so $p|1$, which is not possible. – delta Mar 06 '14 at 04:14

3 Answers3

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Prime factorization is the natural way to go. Another way is to let $d$ be the greatest common divisor of $m$ and $n$, and let $m=ad$, $n=bd$. Note that $a$ and $b$ are relatively prime.

Then from $m^2$ divides $n^2$ we conclude that $a^2$ divides $b^2$. We now show that $a^2=1$. To do this, note that if $a^2\gt 1$ then there is a prime $p$ that divides $a^2$. Use this to contradict the fact that $a$ and $b$ are relatively prime.

André Nicolas
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Assume that $m^2 \mid n^2$ but $m\not\mid n$. Decompose $m$ and $n$ into their prime factors and exponents, i.e. $m = p_1^{a_1}\ldots p_N^{a_N}$, $n = p_1^{b_1}\ldots p_N^{b_N}$. If $m\not\mid n$, what can you say about these prime factors? And what can you say about the prime factors of $m^2$ and $n^2$? And then infer from these answers whether it is indeed possible that $m^2 \mid n^2$ but $m\not\mid n$.

fgp
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Another way is to note that the integer $n^2/m^2$ is the square of a rational (namely, $n/m$), so we need to show that if $k$ is a positive integer and $\sqrt k$ is rational, then it is actually an integer. There are many proofs of this fact. Here is one that does not use prime factorizations:

If $\displaystyle \sqrt k=\frac nm$, we may start by assuming that $n,m$ have no common factors (else, divide both of them by their gcd. This does not change their quotient). Now, also $$ \sqrt k=\frac k{\sqrt k}=\frac k{\frac nm}=\frac {mk}n. $$ Since $n,m$ have no common factors, then there are integers $x,y$ with $nx+my=1$. Simple algebra verifies that since $$ \frac nm=\frac{km}n, $$ then also $$ \frac nm=\frac{km}n=\frac{ny+kmx}{my+nx}, $$ but the latter is an integer, since $nx+my=1$.

Andrés E. Caicedo
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