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.

Let me correct that: does the set {a, b, c} contain all the elements in the set {x, y, z} (and vice versa)?

abc = three digit integer, or a.b.c?

2+2+2=2x3x1

false by contradiction

>>4
But 2*2*2 != 2*3*1

I think it's probably false, but I don't have a contradiction yet.  I do know that either a,b,c and x,y,z are the same sets, or they're totally disjoint, but that's not very helpful.

1+5+8 = 2+2+10 = 14
1*5*8 = 2*2*10 = 40

Moar counterexamples:

1 7 15
4 17 1
 
1 7 18
2 21 3
 
1 7 18
3 21 2
 
1 8 12
2 16 3
 
1 8 12
3 16 2
 
1 9 10
2 15 3
 
1 9 10
3 15 2
 
1 10 20
2 25 4
 
1 10 20
4 25 2
 
1 11 16
2 22 4
 
1 11 16
4 22 2
 
1 12 14
2 21 4
 
1 12 14
4 21 2
 
1 13 18
3 26 3
 
1 14 20
2 28 5
 
1 14 20
5 28 2
 
1 15 18
2 27 5
 
1 15 18
5 27 2

How did you come up with all of those?

>>8
Edit: That first set doesn't work.  There's some bug in my program or something that's giving wrong solutions. :/

>>9
OK, suppose a \le b \le c and x\le y\le z are *different* sets of numbers with the property you want.  Think of the two cubic polynomials with roots a,b,c and x,y,z:

f_1(t) = (t-a)(t-b)(t-c) = t^3 - p t^2+qt-r
f_2(t) = (t-x)(t-y)(t-z) = t^3 - p t^2+q't-r

Where p = a+b+c, r = abc, and q=ab+bc+ca and q'=xy+yz+zx are two different numbers (if they were the same, then a,b,c and x,y,z would have to be the same).  If you set them equal to each other, you get (q-q')t = 0, so the graphs only cross on the y-axis.  Now if you sketch out what the graphs have to look like, keeping in mind how the graph of a cubic is shaped, you have to have

x < a < b < y < z < c

Now if we have a cubic with roots x,y,z, then for any other integer a, we can find another cubic with the same p and r that has a as a root by plugging in a and solving for q.  Then the other two roots are the solutions of a quadratic, and if you're lucky, they're integers.  Now the idea is just to write a program to loop through a crapload of possible values for x,y,z and a, and find some where b and c come out to be integers.

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.

>>11 here
Scratch that.

Next time I should probably read the whole thread before posting.

>>11
>>12
lol I think there might be a hole in your proof dude.

The initial conditions are too vague to come up with any concise solution.  For example, if we say

a=x+y+z-b-c and substitute it into the first equation, we get

bc(x+y+z-b-c)=xyz, which is an equation of 5 variables and has no simple form to its solutions.

Further, from a1*a2*a3=b1*b2*b3 we note the gcd(ai,bi) needn't be greater than 1.  This fact implies simplification is not necessarily possible.

>>14
What? It's not vague at all. 

Problem: If a,b,c,x,y,z are positive integers such that a+b+c = x+y+z and abc = xyz, does it follow that {a,b,c} = {x,y,z}?

And >>7 is about as concise a solution as you can get.

Here's a generalization:

What if all of the sums of products of k-combinations with 0 < k <= n match up? So if

a + b + c = x + y + z
ab + ac + bc = xy + xz + yz
abc = xyz

then up to permutation, a = x;b=y;c=z.

>>16

Yes, just use polynomials.

in the case of just 3 variables. If they were not the same, up to re-ordering, then consider this polynomial.

f(t)=(t-a)(t-b)(t-c)= t^3 - (a+b+c)t^2 + (ab+ac+bc)t - abc
    =t^3 - (x+y+z)t^2 + (xy+xz+z)t - xyz [By the equalities]
    =(t-x)(t-y)(t-z)


so as long the base field K we're working is a UFD, this implies K[x] is also a UFD, and so we have (without loss of generality) a=x,b=y,c=z.

>>17
Ah, what a nice proof. I knew you'd probably use the fact that the symmetric polynomials are the coefficients, but I didn't expect it to flow so nicely.