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I can't get to get a good proof of this, any help?

What I thought was:

$$b^n = a^nk$$ then, by the Fundamental theorem of arithmetic, decompose $b$ such:

$$b=p_1^{q_1}p_2^{q_2}...p_m^{q_m}$$

with $p_1...p_m$ primes and $q_1...q_n$ integers.

then

$$b^n=(p_1^{q_1}p_2^{q_2}...p_m^{q_m})^n= p_1^{q_1n}p_2^{q_2n}...p_m^{q_mn}$$

but here i get stucked, and i can't seem to find a good satisfactory way to associate $a$ and $b$...

Any help will be appreciated

FranckN
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  • This is pretty similar to this question: http://math.stackexchange.com/questions/182988/if-a2-divides-b2-then-a-divides-b – Ben West Feb 09 '14 at 19:07
  • http://math.stackexchange.com/questions/607937/let-a-b-n-be-elements-of-mathbb-n-such-that-an-mid-bn-show-that?rq=1 – user26486 Feb 09 '14 at 19:39

6 Answers6

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Note: I am making one assumption you did not state. I assume $n \in \{1, 2, ... \}$.

Without loss of generality, let $ a = p_1^{\alpha_1} p_2 ^{\alpha_2} ... p_m^{\alpha_m}$ and $b = p_1^{\beta_1} p_2 ^{\beta_2} ... p_m^{\beta_m}.$ Note this means that some $\alpha_k, \beta_k, k \in \{1, 2, ... m\}$ may be zero. Now assume $a^n | b^n$. Then,

$a^n = p_1^{n \alpha_1} ... p_m^{ n \alpha_m}$ and $b^n = p_1^{n \alpha_1} ... p_m^{n \alpha_m}$ are such that $n \alpha_k \le n \beta_k, \forall k \in \{1, ... , m\}$ (because $a^n$ divides $b^n$). This is true if and only if $\alpha_k \le \beta_k \forall k \in \{1, 2, ... , m\}$ (because $n > 0$). However, this is true if and only if $a | b$.

mlg4080
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  • can somebody please just explain why from $a^n | b^n$ follows that $n\alpha_k \leq n\beta_k$ ? – syphracos Jan 16 '22 at 18:10
  • In words: $a \mid b$ is equivalent to saying that all the primes in the prime factorization of $a$ have a smaller power than the same prime numbers in the prime factorization of $b$. In the notation of the problem, that is to say, $a \mid b$ if and only if $\alpha_{k} \le \beta_{k}$ for all $k$. Equivalently, applying the previous statement to $a^{n}$ and $b^{n}$ in place of $a$ and $b$, we have $a^{n} \mid b^{n}$ if and only if $n \alpha_{k} \le n \beta_{k}$ for all $k$. – mlg4080 Jan 19 '22 at 08:50
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Start with a single prime $p$.

$p^a|p^b \Leftrightarrow a\leq b$.

So if $p^{na}|p^{nb}$ then $na \leq nb$ which means that $a \leq b$, so $p^a | p^b$. Use prime decomposition for general case.

Arek Krawczyk
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0

Show that if $p$ is prime and $p \mid a$, then $p \mid b$.

Then note that $({ a \over p})^n \mid ({b \over p})^n$. This is the induction step.

copper.hat
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Assume $p^r\|b$ (i.e. $p^r\mid b$ and $p^{r+1}\not\mid b$) and $p^s\|a$. Then $a^n\mid b^n$ gives aus $ns\le nr$. Provided $n>0$ this implies $s\le r$. Do so for all primes to conclude $a\mid b$

Hagen von Eitzen
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If you know unique factorisation, you can do this by contradiction.

Suppose $a$ is not a factor of $b$. Then there must be a prime $p$ with $p^r$ a factor of $a$, but $p^r$ is not a factor of $b$ (otherwise $a$ is a factor of $b$). Let $p^s$ be the maximum power of $p$ which divides $b$ - and note that $s\lt r$.

Then $a^n=p^{rn}a_1^n$ and $b^n=p^{sn}b_1^n$ with $a_1$ and $b_1$ coprime to $p$. Hence $a^n $ cannot be a factor of $b^n$. Contradiction.

Mark Bennet
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Hint $\ p^{n\alpha}\!\mid p^{n\beta}\! \iff n \alpha \le n\beta \iff \alpha \le \beta \iff p^\alpha\mid p^\beta.\,$ Apply it to prime factorizations of a,b.

Simpler: $\, (b/a)^n = k\in\Bbb Z\Rightarrow b/a\in \Bbb Z\,$ by the Rational Root Test applied to $\,x^n - k.$

Bill Dubuque
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