diophantine equation

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Joined: Aug 4, 2011

by hemangsarkar, Nov 12, 2015, 5:11 PM

Q) Solve over integers $x^{2} + y^{2} + z^{2} = 2xyz$ .

Solution: The left hand side is even. Hence of $x, y$ and $z$ we must have either $0$ or $2$ odd numbers.
If we have $2$ odd numbers says $x$ and $y$ then

$x^{2} + y^{2}$ will be of form $4k+2$ and then LHS will be of form $4m + 2$ while RHS is divisible by $4$ .

Hence none of them is odd.

$x = 2u, y = 2v, z = 2w$

$u^{2} + v^{2} + w^{2} = 4uvw$

Continuing this way we see that $x, y, z$ must be divisible by all powers of $2$ .

The only solution possible here is $x = 0, y = 0 , z = 0$