I can, can you?

Find the sum of all integral values of κ \kappa such that there exists positive integers x x , y y and z z satisfying x 2 + y 2 + z 2 = κ x y z x^2+y^2+z^2=\kappa xyz

3 0 \infty 4

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1 solution

Sudhanshu Mishra
Jun 21, 2015

Let us first consider the case when x 2 = y 2 = z 2 x^{2}=y^{2}=z^{2}

Hence, x = y = z x=y=z

Then, 3 x 2 = k x 3 3 x^{2}=k x^{3}

As x 0 x\neq0 Therefore 3 x = k \frac{3}{x}=k

x = 1 , 3 \therefore x={1,3} k = 3 , 1 \therefore k={3,1}

And we can easily prove by divisibility that there is no solution for x y z x\neq y\neq z k = 1 + 3 = 4 \therefore \sum k= 1+3 = 4

what do you mean prove by divisibility? good solution otherwise

Silver Vice - 5 years, 2 months ago

How do you know that there are no solutions for κ \kappa when x 2 y 2 z 2 x^2\neq y^2\neq z^2 ?

Abdur Rehman Zahid - 5 years, 1 month ago

"And we can easily prove by divisibility that there is no solution for distinct x,y,z."

Consider k=3 and (x,y,z)=(2,5,29)

Ananda Bhaduri - 1 year, 11 months ago

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