Find the solutions for the diophantine equation x^2 + y^2 + z^2 = 2xyz

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

Solution x = y = z = 0 is obvious.

Let's prove that we don't have other solutions, so we'll suppose the opposite.

x^2 + y^2 + z^2 is an even number and at least one of the numbers x, y, z is even. Considering the symmetry of the equation, then ca x is even, if x = 2x1. Then 4|y2 + z2, and this is only when y and z are even. It's true if y is even and z is odd, then 4 is not divisible by y2 + z2. If both are odd then:

y^2 + z^2 = (2u + 1)^2 + (2v + 1)^2 =

=4(u^2 + v^2 + u + v) + 2

So 4 is not divisible by y^2+z^2.

So that x = 2x1, y = 2y1, z = 2z1 and considering the equation, we'll determine:

x1^2 + y1^2 + z1^2 = 2^2x1y1z1.

By taking the same arguments from the equality above, we'll have 2|x1, 2|y1, 2|z1, so that 2^2|x, 2^2|y, 2^2|z. We can show that 2^n|x, 2^n|y, 2^n|z for any n belongs to N. It contradicts the hypothesis.

So, the equation does have a single solution and this is (0,0,0).

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neela | High School Teacher | (Level 3) Valedictorian

Posted on

x^2+y^2+z^2 = 2xyz.


Since this is an equation, with 3 variables, there are many solutions.

There is no constant term , which is free from x, y, z.

So  the curve passes through the origin.

 So (0 ,0 ,0) is an obvious solution.

Now we rewrite the equation as:

x^2 + (-2yz)x +(y^2+z^2) = 0

This is  a quadratic in x and the solutions are:

x1 = {2yz + sqrt{(2yz)^2 - 4(y^2+z^2)}/2

x1 ={yz +sqrt{(yz)^2 - (x^2+y^2)}/2 and

x2 = {(yz)^2-(x^2+y^2)}/2

Example :

let y = 2, z=3. then x = {2*3 +sqrt[(2*3)^2 - 2^2-3^2]}

x = 10.7958.

Therefore x =10.7958 (nearly ), y = 2 and z = 3 is a Diophantine solution.

Similarly take any integer yand z such that (yz)^2 > x^2+z^2, then we can get a real Diophantine solution for x. 

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