[Guest]

Determine all possible triplet(s) (x, y, z) of positive integers, with x ≤ y ≤ z, that satisfy the equation: xy + yz+ zx - xyz = 2.

Edited September 30, 2009 by K Sengupta

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(1,1,1). Beyond that, "xy + yz + xz" grows faster than "xyz" so the difference will be more than 2.

Edited September 30, 2009 by Carl

[superprismatic]

  On 9/30/2009 at 7:27 PM, Carl said:

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(1,1,1). Beyond that, "xy + yz + xz" grows faster than "xyz" so the difference will be more than 2.

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(2,3,4) ?

[Guest]

You're right, Superprismatic. I was in too much of a hurry, I guess!

[Guest]

  On 9/30/2009 at 7:34 PM, superprismatic said:

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(2,3,4) ?

I believe that these two answers comprise the entire solution.

[superprismatic]

  On 9/30/2009 at 8:46 PM, ogden_tbsa said:

I believe that these two answers comprise the entire solution.

So do I, but I'm having difficulty proving it.

[Guest]

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0,1,2

Edited October 1, 2009 by Pieter :-?

[Guest]

  On 10/1/2009 at 5:54 AM, Pieter :-? said:

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0,1,2

X, Y and Z must be positive integers. You have 0 in your solution

[Guest]

I believe this also works

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1,1,2

[Guest]

  On 9/30/2009 at 8:46 PM, ogden_tbsa said:

I believe that these two answers comprise the entire solution.

(4,2,3)

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(1,1,1), (4,2,3), (2,3,4)

  On 9/30/2009 at 6:19 PM, K Sengupta said:

Determine all possible triplet(s) (x, y, z) of positive integers, with x ≤ y ≤ z, that satisfy the equation: xy + yz+ zx - xyz = 2.

[Guest]

I was wrong with my answer.

There are only 2 answers so far. 4,2,3 does not conform to x ≤ y ≤ z.

[Guest]

  On 10/1/2009 at 3:42 PM, victorules said:

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(1,1,1), (4,2,3), (2,3,4)

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(4,2,3), while equivalent to (2,3,4) anyway, is not valid since KS specifies that x <= y <= z.

Edited October 1, 2009 by jerbil

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Given that x <= y <= z

xy + xz + yz <= 3zy

rearranging: xy + yz + zx - xyz = 2

to get: xy + yz + zx = 2 - xyz

and the substitute: 2 - xyz <= 3zy

(2/zy) + x <= 3

From this we can see that x must be less than 3. I can't put an upper limit on y or z yet.

Edited October 1, 2009 by psychic_mind

[Guest]

(0,1,2) and (1,1,1)

[superprismatic]

  On 10/1/2009 at 4:10 PM, psychic_mind said:

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Given that x <= y <= z

xy + xz + yz <= 3zy

rearranging: xy + yz + zx - xyz = 2

to get: xy + yz + zx = 2 - xyz

and the substitute: 2 - xyz <= 3zy

(2/zy) + x <= 3

From this we can see that x must be less than 3. I can't put an upper limit on y or z yet.

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Thanks for the start, psychic-mind. As you

have showed, x=1 or 2. If x=1, we plop it

into the original formula and get,

y+z+yz-yz=2. So, we have to solve

y+z=2 in the positive integers. Hence, y=z=1.

Hence, the solution (1,1,1).

Now, when x=2, we get 2y+2z+yz-2yz=2 which

is 2y+2z-yz=2. Solving for z, we get

z=(2y-2)/(y-2). Now, y<=z so

y<=(2y-2)/(y-2) from which we get

y²-4y+2<=0. This is true when

2-sqrt(2)<=y<=2+sqrt(2). The only positive

integer greater than x in this range is 3. Plugging x=2

and y=3 into the original, we get z=4.

Hence, the (only) other solution (2,3,4).

Edited October 1, 2009 by superprismatic