Question

find all the positive integers triples (x, y, z) that satisfy the equation: x power 4 + y power 4 + y power 4 = 2x power 2 y power 2 + 2y power 2 z power2 + 2z power 2 x power 2 - 63​

Answer 1

Answer:

${x}^{4} + {y}^{4} + {z}^{4} = 2 {x}^{2} {y}^{2} + 2 {y}^{2} {z}^{2} + 2 {z}^{2} {x}^{2} - 63$

Make it quadratic in z^2

$( {z}^{2} ) {}^{2} - 2( {y}^{2} + {x}^{2} ) {z}^{2} + {x}^{4} + {y}^{4} - 2 {x}^{2} {y}^{2} + 63 = 0 \\ \\ {z}^{2} = \frac{2( {y}^{2} + {x}^{2}) \: \pm \: \sqrt{4( {x}^{2} + {y}^{2} ) {}^{2} - 4( {x}^{4} + {y}^{4} - 2 {x}^{2} {y}^{2} + 63) } }{2} \\ \\ {z}^{2} = {x}^{2} + {y}^{2} \pm \: \sqrt{4 {x}^{2} {y}^{2} - 63 } \\ z = \pm \sqrt{ {x}^{2} + {y}^{2} \pm \: \sqrt{4 {x}^{2} {y}^{2} - 63 } }$

Now for integer solutions

$4 {x}^{2} {y}^{2} - 63 \: \: \: \: must \: be \: \: a \: \: perfect \: \: square$

Possible values , of

$4 {x}^{2} {y}^{2} = 64 \: \: and \: \: 144 \\ {x}^{2} {y}^{2} = 16 \: \: and \: \: 36\\xy= \pm 4\: , \:\pm6$

【 Because 64 - 63 = 1 and 144 - 63 = 81

are perfect squares 】

So possible values of x , y

x = 4 , 3 , , 2 , 1

y = 1 , 2 , 3 , 4

Note for 6 , 1 z = √(36+1 +/- √144-63

z= √(37+/-9) is not an integer

And find corresponding values of z .