2 equations and 3 unknowns

Algebra Level 4

Real numbers x x , y y and z z are such that x + y = 2 x+y=2 and x y − z 2 = 1 xy-z^2=1 . Find x 5 + y 5 + z 5 x^5+y^5+z^5 .


The answer is 2.

Sathvik Acharya by Sathvik Acharya , 17, India

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3 solutions

Sathvik Acharya
Apr 2, 2017

( x − y ) 2 = ( x + y ) 2 − 4 x y = 2 2 − 4 ( z 2 + 1 ) (x-y)^2=(x+y)^2-4xy=2^2-4(z^2+1) ⟹ ( x − y ) 2 = − 4 z 2 \implies (x-y)^2=-4z^2 Since x , y x,y are real, the LHS is non-negative, so the RHS must be non-negative. This is possible only when z = 0 z=0 and x = y x=y . Solving the equations, we get x = 1 , y = 1 x=1, y=1 .

Therefore, x 5 + y 5 + z 5 = 1 5 + 1 5 + 0 5 = 2 \boxed{x^5+y^5+z^5=1^5+1^5+0^5=2}

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Same method

Aditya Kumar - 3 years, 11 months ago
Hana Wehbi
Apr 9, 2017

x + y = 2 ⟹ x = 1 , y = 1 x+y=2 \implies x=1 , y=1

x y − z 2 = 1 ⟹ z = 0 xy -z^2 = 1\implies z=0

Thus, x 5 + y 5 + z 5 = 1 + 1 + 0 = 2 x^5+y^5+z^5= 1+1+0=2

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Ankit Kumar Jain
Apr 6, 2017

See this

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