Is it true that we can get zero for all $(x,y,z)\in\mathbb{N}^3$?

Question

There are three distinct positive integers $x$, $y$, and $z$.

We can choose two numbers $a,b\in\{x,y,z\}$, where $b\leq a$,

then replace $b$ by $2b$ and replace $a$ by $a-b$.

Is it true that there exists a method to repeat this process until we get a zero ?

For example, let $(x,y,z)=(3,5,10)$, then $(3,5,10)\rightarrow (6,2,10) \rightarrow (4,4,10) \rightarrow (\mathbf{0},8,10) $.

Let $(x,y,z)=(9,11,2)$, then $(9,11,2)\rightarrow (18,2,2) \rightarrow (18,\mathbf{0},4)$.

Is it true that we can get zero for all $(x,y,z)\in\mathbb{N}^3$ ?

Thanks.

thank you everyone. Now I found that this is problem from IMO 1994 and USAMT 23. Here is solution usamts.org/Solutions/Solutions_23_1.pdf

Answer 1

I wrote a Delphi program which checked that each triple $(x,y,z)$ such that $6\le x+y+z\le 100$ is reducible to a triple with a zero. You can download the program and its results here.

Answer 2

In the absence of a proof, I'll give some more numerical evidence for the conjecture. The Python code below computes the "depth" of each triple $(a,b,c)$ with a given sum, where the depth is the minimum number of moves required to make two of the three numbers equal. (A triple where no two numbers can be made equal would have depth $-1$.) I've verified the conjecture for all $a+b+c\le2000$. I also looked at the triples that set new depth records... a partial table is below.

$$ \begin{array}{c|c|l} {\text{depth}} & {\text{min }} a+b+c & {\text{triple(s)}} \\ \hline 5 & 27 & (5,9,13)^* \\ 6 & 45 & (4,15,26)^* \\ 7 & 81 & (8,27,46)^* \\ 8 & 105 & (27,35,43)^*,(8,35,62)^*,(8, 27, 70) \\ 9 & 195 & (8,57,130),(8,65,122)^*,(4,33,158),\ldots \\ 10 & 329 & (4,130,195) \\ 11 & 597 & (175,199,223)^* \\ 12 & 885 & (101,295,489)^* \\ 13 & 1425 & (206,475,744)^* \end{array} $$

At least one interesting fact stands out: almost all of these "most difficult" triples are arithmetic progressions of the form $(a-k,a,a+k)$ (these are indicated by asterisks). Also, the size of $a+b+c$ grows in a rough Fibonacci-like pattern. Looking at these examples, one might conjecture that the next record will occur around $885+1425=2310$.

Update: After running the code for a while longer, the next record comes later than expected, at

$$ \begin{array}{c|c|l} {\text{depth}} & {\text{min }} a+b+c & {\text{triple(s)}} \\ \hline 14 & 2793 & (571, 931, 1291)^*. \end{array} $$ Also, the conjecture is now verified for $a+b+c\le 3000$.

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def srt(i,j,k):
  a = min(i,min(j,k))
  c = max(i,max(j,k))
  b = (i+j+k)-(a+c)
  return (a,b,c)

def setdepth(i,j,k,val,d,queue):
  key = srt(i,j,k)
  if d.has_key(key) and d[key]>=0: return
  d[key] = val
  queue.append(key)

def addpreds(a,b,c,val,d,queue):
  if a%2==0:
    setdepth(a/2,b+a/2,c,val+1,d,queue)
    setdepth(a/2,b,c+a/2,val+1,d,queue)

def depths(n):
  d = {}
  queue = []
  for i in xrange(1,n+1):
    for j in xrange(1,n+1):
      k = n-(i+j)
      if i+j<n and i!=j and i!=k and j!=k: d[srt(i,j,k)] = -1
    if 2*i<n: setdepth(i,i,n-2*i,0,d,queue)
  ix = 0
  while ix<len(queue):
    key = (a,b,c) = queue[ix]
    val = d[key]
    addpreds(a,b,c,val,d,queue)
    addpreds(b,c,a,val,d,queue)
    addpreds(c,a,b,val,d,queue)
    ix += 1
  return d