System again

Algebra Level 3

{ x 4 − y 4 = 0 x 2 + y 2 = 4 x + y = 4 \large \begin{cases} x^4 - y^4 = 0 \\ x^2 + y^ 2 = 4 \\ x + y = 4 \end{cases}

Consider the above system of equations above, find x 100 + 1 x^{100} +1 .

2 100 + 1 2^{100} + 1 None of the choices 1 2 100 − 1 2^{100} - 1 .

Paul Ryan Longhas by Paul Ryan Longhas , 24, Philippines

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1 solution

Paul Ryan Longhas
Jul 24, 2016

By system, ( x 2 − y 2 ) ( x 2 + y 2 ) = 0 ⟹ ( x 2 − y 2 ) 4 = 0 ⟹ x 2 − y 2 = 0 (x^2 - y^2)(x^2 + y^2) = 0 \implies (x^2 - y^2)4 = 0 \implies x^2 - y^2 = 0

So, ( x − y ) ( x + y ) = 0 ⟹ ( x − y ) ( 4 ) = 0 ⟹ x − y = 0. (x-y)(x+y) = 0 \implies (x - y)(4) = 0 \implies x-y = 0.

Note, if x − y = 0 x - y = 0 and x + y = 4 x + y = 4 then x = y = 2 x = y =2 which is not satisfy x 2 + y 2 = 4 x^2 + y^2 = 4 .

Thus, the answer is no solution.

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