# Gcd(a,b,c)lcm(a,b,c)=abc => a,b,c relatively prime in pairs

1. Apr 10, 2010

### kingwinner

Claim: If gcd(a,b,c)lcm(a,b,c) = abc, then gcd(a,b)=gcd(b,c)=gcd(a,c)=1.

I'm trying to understand why this is true...
How can we prove it?

Any help is appreciated!

2. Apr 10, 2010

### JSuarez

Start by the prime factorizations of a,b and c, then use the expressions for the gcd and lcm in terms of these.

3. Apr 12, 2010

### kingwinner

What do you mean?
I've written out the prime factorizations of a,b, and c. But I don't know what to do next...

4. Apr 12, 2010

### JSuarez

If the prime factorizations of a, b and c are:

$$a=2^{e_1}3^{e_2}\cdots p^{e_i}$$

$$b=2^{f_1}3^{f_2}\cdots p^{f_i}$$

$$c=2^{g_1}3^{g_2}\cdots p^{g_i}$$

(If a particular prime factor doesn't appear in the factorization, its exponent is zero)

Then you should know that:

$$gcd\left(a,b,c\right)=2^{min \left\{e_1,f_1,g_1\right\}}3^{min \left\{e_2,f_2,g_2\right\}}\cdots p^{min\left\{e_i,f_i,g_i\right\}}$$

And:

$$lcm\left(a,b,c\right)=2^{max \left\{e_1,f_1,g_1\right\}}3^{max \left\{e_2,f_2,g_2\right\}}\cdots p^{max\left\{e_i,f_i,g_i\right\}}$$

Now plug these in your equality and see what must happen for the exponents to agree.

5. Apr 12, 2010

### kingwinner

I think we'll then have min{ei,fi,gi}+max{ei,fi,gi}=ei+fi+gi, but why does this imply gcd(a,b)=gcd(b,c)=gcd(a,c)=1?

6. Apr 13, 2010

### JSuarez

For a given i, what must happen to the ei's, fi's and gi's for that equality to be true? For example, can they all be > 0?

7. Apr 25, 2010

### disregardthat

Use that $$\text{gcd}(a,b,c)=\text{gcd}(\text{gcd}(a,b),c)$$, and likewise for $$\text{lcm}$$. Also, the fact that $$\text{gcd}(a,b)\text{lcm}(a,b)=ab$$ might come in handy. You can extract a lot of information from the equation using this, and you do not have to go the way through their respective prime factorizations.