[Guest]

Each of X,Y and Z is a positive integer satisfying: X²+Y²+1 = X*Y*Z.

Determine all possible value(s) that Z can assume.

Edited March 2, 2010 by K Sengupta

[jazzship]

3

[Guest]

x=2

y=5

z=3

Each of X,Y and Z is a positive integer satisfying: X²+Y²+1 = X*Y*Z.

Determine all possible value(s) that Z can assume.

One possible solution...

Edited March 2, 2010 by rover

[Guest]

for example, subbing x =2

5 + y^2 = z*2y

y^2 - 2zy + 5 = 0

(y - 1)(y - 5) = 0

=> -2zy = -6y

z = 3

similarly, for x=1:

y^2 -zy + 2 = 0

(y-1)(y-2) = 0

=> z=2

we can see the general solution is such that z will equal any positive integer that fits:

[sum of 2 factors of (x^2+1) ] / x, where x is a positive integer

while for x=1,2 this leads us to solutions for z by factoring (x^2 +1) into {1, (x^2+1)}, for values of x >= 3, there will be no such solution possible.

That still leaves the possibility that the sum of 2 other factors (in between) will be divisible by x. However, that is also not possible unless z = 3. Still working on trying to prove this.

for example:

x = 13:

y^2 - 13zy +170 = 0

(y-34)(y-5) = 0

=> 13z = 39

z = 3

the other factor pairs, don't produce a solution (1,170), (10, 17) , (2, 85)

Edited March 2, 2010 by jlf278

[Guest]

for example, subbing x =2

5 + y^2 = z*2y

y^2 - 2zy + 5 = 0

(y - 1)(y - 5) = 0

=> -2zy = -6y

z = 3

similarly, for x=1:

y^2 -zy + 2 = 0

(y-1)(y-2) = 0

=> z=2

we can see the general solution is such that z will equal any positive integer that fits:

[sum of 2 factors of (x^2+1) ] / x, where x is a positive integer

while for x=1,2 this leads us to solutions for z by factoring (x^2 +1) into {1, (x^2+1)}, for values of x >= 3, there will be no such solution possible.

That still leaves the possibility that the sum of 2 other factors (in between) will be divisible by x. However, that is also not possible unless z = 3. Still working on trying to prove this.

for example:

x = 13:

y^2 - 13zy +170 = 0

(y-34)(y-5) = 0

=> 13z = 39

z = 3

the other factor pairs, don't produce a solution (1,170), (10, 17) , (2, 85)

re-check your work for x=1, you'll find that z does in fact equal 3...

[Guest]

if the equation is rearranged to (x^2+y^2+1)/(x*y)=z, then z can assume an infinite number of values based on what x and y are.

[Guest]

if the equation is rearranged to (x^2+y^2+1)/(x*y)=z, then z can assume an infinite number of values based on what x and y are.

Your conclusion does not support the original requirement of the problem which is X, Y, and Z be positive intergers.

So if x=1, y=4 gives (1 + 16 + 1)/4 = 18/4 = 4.5 which means z is not an interger.

[Guest]

Your conclusion does not support the original requirement of the problem which is X, Y, and Z be positive intergers.

So if x=1, y=4 gives (1 + 16 + 1)/4 = 18/4 = 4.5 which means z is not an interger.

Not only that, but Z can never be between -2 and 3 for any real values of x and y.

[Guest]

solution.pdf

[Guest]

solution.pdf

You're proof is more complicated than necessary.

One need only consider the three cases:

1) x=1,y=1 => z=3

2) x=1,y>1 => y=2,z=3

3) x>1,y>1 => z=DNE

Thus, z=3.

[Guest]

You're proof is more complicated than necessary.

One need only consider the three cases:

1) x=1,y=1 => z=3

2) x=1,y>1 => y=2,z=3

3) x>1,y>1 => z=DNE

Thus, z=3.

You're proof is more complicated than necessary.

One need only consider the three cases:

1) x=1,y=1 => z=3

2) x=1,y>1 => y=2,z=3

3) x>1,y>1 => z=DNE

Thus, z=3.

Hi vinays84, thanks for commenting on my post but, I'm sorry, I have to disagree with your solution.

What do you mean by DNE?

In principle, I can imagine only two reasonable answers for that question: either DNE="Does Not Extist" or DNE = 3.

However, as I pointed out in my solution, the equation holds for (x, y, z) = (2, 5, 3). Hence, it's wrong to say

"3) x>1,y>1 => z Does Not Exist"

Therefore, the only possibility remaining is

"3) x>1,y>1 => z=3"

Then I ask how do you prove that? Essentially, this implication is the original problem or at least a main part of it, since the case x<1 or y<1 is trivial (as you have noticed in 1 and 2).

[Guest]

For the 2 hours I was expanding, simplifying, dividing, multiplying and all kind of stuff with x² + y² + 1 = x*y*z. I somehow found that xy^-1 - 3 = z². From here:

z² must be greater or equal to 0 (To be a real number). From the original equation we find, that z must be greater or equal to 3, so for any z greater or equal to 3 there is a xy^-1 - 3, equal to z².