Integers simply!

Find all the integers x , y , z x,y,z satisfying

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

Evaluate the sum \textbf{sum} of all integers of each pair.

Details \textbf{Details}

For e.g. If you get your integers solution as

( 1 , 2 , 3 ) (1,2,3) and ( 2 , 1 , 5 ) (2,-1,5)

Then enter your answer as 1 + 2 + 3 + 2 1 + 5 = 12 1+2+3+2-1+5=\boxed{12}


The answer is 0.

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1 solution

広務 山下
Apr 4, 2018

right side is even.so one of them is only even or every is even.

If one of them is even,right side is multiple of four , left side is not multiple of four.*[(2q-1)^2=4q^2+4q+1]

so every is even…①

If every is even,x=2p y=2q z=2r 4p^2+4q^2+4r^2=16pqr

p^2+q^2+r^2=4pqr

so p,q,r are even(①)………

that will never end.

so answer is 0.

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