Prove that when $n$ is square free, then $a^2b = a^2c \text{ mod }n$ implies that $ab = ac \text{ mod } n$

Question

I am trying to prove that when $n$ is a square-free composite number ($n = p_1\dots p_r$ where $p_i$ are primes), then $a^2b = a^2c \text{ mod }n$ implies that $ab = ac \text{ mod } n$. I have tried considering the identity $a^2(b - c) = 0 \text{ mod } n$, then we have: $$a^2(b - c) = qn \text{ for some } q$$ Now to prove that $ab = ac \text{ mod } n$, I must prove that $a | q$. But I am not sure how to proceed from here. Any ideas?

Answer 1

You are given that $p_1 \cdots p_n | a^2(b-c)$, so for each $i$ you have that $p_i|a^2(b-c)$. This means that either $p_i|a^2$ or $p_i | (b-c)$.

In the first case you have $$p_i|a^2 \implies p_i| a \implies p_i | a(b-c).$$ In the second case you have $$p_i | (b-c) \implies p_i | a(b-c).$$ Either way, you conclude that $p_i | a(b-c)$.

Now can use the fact that if $p$ and $q$ are distinct primes, then $p|k \wedge q|k \implies pq|k$. Since $p_i | a(b-c)$ for each $i$ you have $$ \left( \prod_{i=1}^n p_i \right) \bigg| \,a(b-c).$$

Answer 2

Note that $\ n\mid a^2d\mid (ad)^2\Rightarrow\, n\mid ad,\,$ by $\,n\,$ squarefree. $ $ QED $\ $ [$d = b\!-\!c\,$ in OP]

Remark $ $ The above arrow is $(4)$ in a handful of common characterizations of squarefree integers.

$\begin{align} {\rm Similarly}\ \ &n\mid a_1^{\large \color{#c00}{e_1}}\cdots a_k^{\color{#c00}{\large e_k}}\mid (a_1\cdots a_k)^{\large e}\\ \Rightarrow\ \ & n\mid a_1\,\cdots\ a_k\ \ {\rm using}\ \ e = \max e_i,\ \ \text{i.e. we can erase the }\color{#c00}{\rm exponents}\end{align}$

Answer 3

Note: $n|a^2b -a^2c = a^2(b-c)$.

If we let $n= n_1*n_2$ where $n_1|a^2$ and $n_2|b-c$ [1]

Then we note $n_1$ is squarefree [2].

And so as $n_1$ is squarefree and $n_1|a^2$ then $n_1|a$ [3]

So $n_1|a$ and $n_2|b-c$ so $n=n_1n_2|a(b-c)=ab-ac$ so $ac\equiv bc\pmod n$.

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[1] Lemma 1: If $n|jk$ then there are $n_1,n_2$ so that $n=n_1n_2$ and $n_1|j$ and $n_2|k$.

Proof: Let $n'=\frac n{\gcd(n,j)}$ and $j'=\frac j{\gcd(n,j)}$. It's standard knowledge that $n'$ and $j'$ are relatively prime (if they had any nontrivial factors in common then that factor times $\gcd(n,j)$ would be a common factor larger than $\gcd(n,j)$).

$n|jk$ so $n'|j'k$. As $n'$ and $j'$ are relatively prime $n'|k$. So $n = \gcd(n,j)*n'$ and $\gcd(n,j)|j$ and $n'|k$.

[2] Lemma 2: If $k|n$ and $n$ is squarefree then $k$ is squarefree.

Proof: Oh, come on! All square factors of $k$ are also square factors of any multiple of $k$ and as $n$ is squarefree and doesn't have any neither does $k$.

[3] Lemma 3: If a squarefree $n$ divides a square $a^2$ then $n|a$.

Pf: This is the heart of the question Let $n = \prod p_i$ be the prime factorization of $n$. (Note: Each $p_i$ are distinct and to a single power because... $n$ is squarefree.) So each $p_i|a^2$ and by Euclids Lemma $p_i|a$.

So $n = \prod p_i|a$.

(Note: Had any of the prime factors of $n$ had a power more than $1$ we could not reach that result.)

...............

Just read Bill Dubuques solution.

I could have skipped Lemma 1 and 2: entirely and just used Lemma 3, plus

Corolary to Lemma 3: If $n$ is square free and $n|a^2d$ then $n|ad$.

Pf: $a^2d|a^2d^2=(ad)^2$ so $n|(ad)^2$ and by Lemma 3: $n|ad$.

.... D'oh....

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