find all the positive integers triples (x, y, z) that Satisify the equation x^4+y^4+z^4=2x^2*y^2+2y^2*z^2+2z^2+x^2-63
Answers
2 , 2 , 3 & 4 , 4 , 1 are two triplets Satisfy the equation x⁴ + y⁴ + z⁴ = 2x²y² + 2y²z² + 2z²x² - 63
Step-by-step explanation:
x⁴ + y⁴ + z⁴ = 2x²y² + 2y²z² + 2z²x² - 63
=> x⁴ + y⁴ + z⁴ - 2x²y² - 2y²z² - 2z²x² = - 63
=> (x² - y² - z²)² - (2yz)² = -63
=> (x² - y² - z² + 2yz) (x² - y² - z² - 2yz) = -63
=> (x² - (y -z)²)(x² -(y + z)²) = -63
=> ( x + y - z)(x - y + z) ( x + y + z)(x - y - z) = -63
=> ( x + y - z)(x - y + z) ( x + y + z)(-x + y + z) = 63
=> ( x + y + z) ( x + y - z)(x - y + z)(-x + y + z) = 63
=> A . B . C . D = 63
A , B , C D can be 1 , 3 , 7 , 9 , 21 , 63 only
B + C + D = x + y - z + x - y + z -x + y + z = x + y + z = A
so possible Factor of B C D are
3 , 3 , 1 & A = 7 => 3 * 3 * 1 * 7 = 63
7 , 1 , 1 & A = 9 => 7 * 1 * 1 * 9 = 63
Case 1
x + y + z = 7
we get x , y z as 2 , 2 , 3 ( not respectively)
Case 2
x + y + z = 9
we get x , y z as 4 , 4 , 1 ( not respectively)