Proof Using Modular Arithmetic - Divisibility using co-prime numbers

Question

If a number is divisible by two or more co-prime numbers, then that number is also divisible by the product of the co-prime numbers also. How to prove this using modular arithmetic?

Let $N$ be an integer
Let $p_1,p_2,p_3$ be co-prime numbers.

$N \equiv 0 \pmod{p_1}\cdots(1)\\ N \equiv 0 \pmod{p_2}\cdots(2)\\ N \equiv 0 \pmod{p_3}\cdots(3)$

Then, $N \equiv 0 \pmod{p_1p_2p_3}\cdots(4)$

i.e., if first three equations are true, 4th equation must be true.

I hope, I have written this correctly. But, how to prove this. please help

I tried to rewrite the first first three equations as follows, to reach the fourth, but it did not help

$(1) \implies N = k_1p_1\\ (2) \implies N = k_2p_2\\ (3) \implies N = k_3p_3$

Note: modular arithmetic is not in my syllabus. But I am learning as it is a very interesting topic. appreciated if anybody can provide links to any online material or book to understand these kind of concepts.

Essentially this is a case of the Chinese remainder theorem... — Parcly Taxel, Oct 23 '16 at 14:29
oh, I wanted to learn Chinese remainder theorem also, so i was learning basics using the facts I know. Surprised, I have come to this:) — Kiran, Oct 23 '16 at 14:31

score 2 · Accepted Answer · answered Oct 23 '16 at 14:43

2

For $2$ coprime divisors:

Write $n=p_1q_1=p_2q_2$, and use a Bézout's relation between $p_1$ and $p_2$: $\;u_1p_1+u_2p_2=1$. We deduce $$n=nu_1p_1+nu_2p_2=p_2q_2u_1p_1+p_1q_1u_2p_2=p_1p_2(q_2u_1+q_1u_2).$$

For any number of coprime divisors: use an easy induction, knowing that if $p_1,p_2, \dots,p_r$ are pairwise coprime, then $p_r$ is coprime with $p_1\dotsm p_{r-1}$.

answered Oct 23 '16 at 14:43

Bernard

175,478

Beautiful applying Bézout's lemma. – Piquito Oct 23 '16 at 14:47

score 1 · Answer 2 · answered Oct 23 '16 at 14:31

1

If $N\equiv0\pmod{p_1}$ and $N\equiv0\pmod{p_2}$

Then

$N=0+kp_1\equiv0\pmod{p_2}$

$\Rightarrow k\equiv 0 \pmod{p_2}$ (since $\gcd(p_1,p_2)=1$)

So $N=0+k'p_2p_1\equiv0 \pmod {p_1p_2}$

Now repeat with $p_1'=p_1p_2$ and $p_2'=p_3$...

answered Oct 23 '16 at 14:31

Evariste

2,511

score 1 · Answer 3 · answered Oct 23 '16 at 14:46

For all $N$ one has $$N=\prod_{1\le i\le n} p_i=p_1p_2.....p_n$$ therefore $$p\text { divides }N\iff N\equiv0\pmod p\iff p=p_i \text { for some }i\in\{1,2,...,n\}$$ It follows $$p_1\text { and } p_2\text { divides }N\iff p_1=p_i\text { and } p_2=p_j\text { for some }i,j\in\{1,2,...,n\}$$ This is equivalent to $$ N\equiv0\pmod {p_1p_2}$$ and this extend easily to a finite number of primes $p_1,p_2,p_3,...,p_k$

Proof Using Modular Arithmetic - Divisibility using co-prime numbers

3 Answers3