# Number Theory gcd problems

## Homework Statement

Let a,b,c be integers. If (b,c)=1, then (a,bc)=(a,b)(a,c)

## Homework Equations

This is difficult to answer because some theorems that we haven't proven yet, we can't use.

## The Attempt at a Solution

Let g=(a,b) and h=(a,c), g and h are integers.
That means g|a and g|b, h|a and h|c.
That means:
a=gk, for some integer k
b=gq, for some integer q
a=hp for some integer p
c=ht for some integer t.

Multiply b and c together to get bc=(a,b)(a,c)qt. Then (a,bc)= (gk, (a,b)(a,c)qt), but that's as far as I got.

## Answers and Replies

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matt grime
Homework Helper
Sometimes it is just easier to think what things mean. Suppose that d divides a and bc. Think about the prime factors of d. Can you split them into two types? (Or is uniqueness of prime factorisation something you can't use?)

Yeah, we can use uniqueness of prime factorization. We proved it last month, so that's why we can use it. I don't know how to do correct mathematical notation on the computer here, so hopefully you'll understand what I'm typing.

So d=(a,bc). d=(p1^r1)(p2^r2)(p3^r3)... or we can say d=(2^r2)(3^r3)(5^r5)... But I don't see how that's going to help me.

matt grime
Homework Helper
So you know that gcd(a,b) is the product of the primes that divide a and b (with the right powers), similarly for gcd(a,c). And b and c are coprime so no primes divide both of them. Do you see where this is going?

Theorem: If (b,c)=1, then (a,bc)=(a,b)(a,c).

Let g=(a,b) and h=(a,c), g and h are integers.
That means g|a and g|b, h|a and h|c.
Therefore:
b=qg, for some integer q.
c=th for some integer t.
g|b and h|c, and (b,c)=1, so (g,h)=1.
Therefore a=pgh, for some integer p.
(a,b)=g => (a,b)/g=1 => (pgh,qg)/g=1 => g(ph,q)/g=1 => (ph,q)=1 => (p,q)=1
(a,c)=h => (a,c)/h=1 => (pgh,th)/h=1 => h(pg,t)/h=1 => (pg,t)=1 => (p,t)=1
Therefore (p,qt)=1
(a,bc)=(pgh,qgth)=gh(p,qt)=gh*1=gh=(a,b)(a,c)
Therefore (a,bc)=(a,b)(a,c)
Q.E.D.

I've never done work like this before, so I'm sure I have several form issues, but I went ahead and showed as much of my thought process as I could in hopes that you understand. To the best of my knowledge, this proves the theorem, but if I took any liberties, please, feel free to correct me.