Need help proving/disproving

I'm not sure how to best phrase this, so bear with me:
If you're given
abc = xyz
and
a+b+c = x+y+z
where a, b, c, x, y, z are all positive integers, is there a way to prove or disprove a = x, b = y, c = z?

I want to know if an integer shown in this way is unique.

Since abc = xyz, both sides are composed of the same prime factors.

Let a := p_a1 * ... * p_an, b := p_b1 * ... * p_bn, c := p_c1 * ... * p_cn, with all p_ai, p_bi, p_ci being prime.

Then for a+b+c = x+y+z with {a,b,c} = {x,y,z} we get

p_a1 * ... * p_an + p_b1 * ... * p_bn + p_c1 * ... * p_cn =
p_a1 * ... * p_an + p_b1 * ... * p_bn + p_c1 * ... * p_cn

with the sums both sides being composed of the same products of prime numbers (not necessarily a = x, b = y, c = z of course).

We now need to show somehow that {a,b,c} != {x,y,z} is not possible.

If we switched any two prime factors that aren't equal on one side, i.e. changing the prime factorisation, we'd obviously get two different numbers.

Let us try it with p_ai != p_bi (without loss of generality) first.

a + b + c = p_a1 * ... * p_ai-1 * p_ai+1 * ... * p_an + p_b1 * ... * p_bi-1 * p_bi+1 * ... * p_bn + c

<=> (subtracting the left from the right side)

0 = (p_bi - p_ai)(p_a1 * ... * p_ai-1 * p_ai+1 * ... * p_an) + (p_ai - p_bi)(p_b1 * ... * p_bi-1 * p_bi+1 * ... * p_bn)

Set d := |p_bi - p_ai| and let (p_ai - p_bi) =: -d, without loss of generality.

Then

0 = d(p_a1 * ... * p_ai-1 * p_ai+1 * ... * p_an) - d(p_b1 * ... * p_bi-1 * p_bi+1 * ... * p_bn)

But this can only be true if all prime factors p_a1*...*p_an and p_b1*...*p_bn, except for p_ai and p_bi are equal, which means that a and b merely switched places, but we didn't get two new numbers.

Let us prove by induction, no matter how often we switch prime factors, we'll always keep sums of the same products on both sides.

We've already proven that it's true for one switch, so let us assume it was true for n.

By our assumption the products stay the same after n-switches, thus after n-switches we still have a + b + c = a + b + c on both sides, and we can easily refer to our first case about one switch, which proves it for n+1.


I don't know if I'm right, but that's how I'd try it. Obviously, you need to know about the uniqueness of prime factorisation for that.