So complicated

Algebra Level 4

Given x x and y y are distinct real numbers such that x 2 + y 2 x 2 − y 2 + x 2 − y 2 x 2 + y 2 = 4 \frac{x^{2}+y^{2}}{x^{2}-y^{2}}+\frac{x^{2}-y^{2}}{x^{2}+y^{2}}=4 If x 8 + y 8 x 8 − y 8 + x 8 − y 8 x 8 + y 8 \frac{x^{8}+y^{8}}{x^{8}-y^{8}}+\frac{x^{8}-y^{8}}{x^{8}+y^{8}} can be expressed as a b \frac{a}{b} , where a a and b b are positive numbers and gcd ( a , b ) = 1 \text{gcd}(a,b)=1 , find a + b a+b .


The answer is 61.

Skye Rzym by SKYE RZYM , 20, Indonesia

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2 solutions

Chew-Seong Cheong
Jul 17, 2017

x 2 + y 2 x 2 − y 2 + x 2 − y 2 x 2 + y 2 = 4 ( x 2 + y 2 ) 2 + ( x 2 − y 2 ) 2 = 4 ( x 2 + y 2 ) ( x 2 − y 2 ) 2 ( x 4 + y 4 ) = 4 ( x 4 − y 4 ) x 4 + y 4 = 2 x 4 − 2 y 4 ⟹ x 4 = 3 y 4 \begin{aligned} \frac {x^2+y^2}{x^2-y^2} + \frac {x^2-y^2}{x^2+y^2} & = 4 \\ \left(x^2+y^2 \right)^2 + \left(x^2-y^2 \right)^2 & = 4\left(x^2+y^2 \right)\left(x^2-y^2 \right) \\ 2 \left(x^4+y^4 \right) & = 4 \left(x^4-y^4 \right) \\ x^4+y^4 & = 2 x^4- 2 y^4 \\ \implies x^4 & = 3y^4 \end{aligned}

Then, we have:

x 8 + y 8 x 8 − y 8 + x 8 − y 8 x 8 + y 8 = 9 y 8 + y 8 9 y 8 − y 8 + 9 y 8 − y 8 9 y 8 + y 8 = 10 8 + 8 10 = 5 4 + 4 5 = 41 20 \begin{aligned} \frac {x^8+y^8}{x^8-y^8} + \frac {x^8-y^8}{x^8+y^8} & = \frac {9y^8+y^8}{9y^8-y^8} + \frac {9y^8-y^8}{9y^8+y^8} \\ & = \frac {10}8+\frac 8{10} = \frac 54 + \frac 45 = \frac {41}{20} \end{aligned}

⟹ a + b = 41 + 20 = 61 \implies a+b=41+20 = \boxed{61}

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I also solved like this brother.

Bibhor Singh - 3 years, 2 months ago

How do you explain it to someone

Abby Pulver - 3 years, 1 month ago

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I don't know what you mean.

Chew-Seong Cheong - 3 years, 1 month ago
Jd Money
Jul 17, 2017

x 2 + y 2 x 2 − y 2 + x 2 − y 2 x 2 + y 2 = 4 \frac {x^2+y^2}{x^2-y^2}+\frac {x^2-y^2}{x^2+y^2}=4

( x 2 + y 2 ) 2 ( x 2 − y 2 ) ( x 2 + y 2 ) + ( x 2 − y 2 ) 2 ( x 2 + y 2 ) ( x 2 − y 2 ) = 4 \frac {(x^2+y^2)^2}{(x^2-y^2)(x^2+y^2)}+\frac {(x^2-y^2)^2}{(x^2+y^2)(x^2-y^2)}=4

x 4 + 2 x 2 y 2 + y 4 + x 4 − 2 x 2 y 2 + y 4 x 4 − y 4 = 4 \frac {x^4+2x^2y^2+y^4+x^4-2x^2y^2+y^4}{x^4-y^4}=4

2 x 4 + 2 y 4 x 4 − y 4 = 4 \frac {2x^4+2y^4}{x^4-y^4}=4

x 4 + y 4 x 4 − y 4 = 2 \frac {x^4+y^4}{x^4-y^4}=2

x 4 + y 4 x 4 − y 4 + x 4 − y 4 x 4 + y 4 = x 4 + y 4 x 4 − y 4 + 1 x 4 + y 4 x 4 − y 4 = 2 + 1 2 = 5 2 \frac {x^4+y^4}{x^4-y^4}+\frac {x^4-y^4}{x^4+y^4}=\frac {x^4+y^4}{x^4-y^4}+\frac{1}{\frac {x^4+y^4}{x^4-y^4}}=2+\frac{1}{2}=\frac{5}{2}

( x 4 + y 4 ) 2 ( x 4 − y 4 ) ( x 4 + y 4 ) + ( x 4 − y 4 ) 2 ( x 4 + y 4 ) ( x 4 − y 4 ) = 5 2 \frac {(x^4+y^4)^2}{(x^4-y^4)(x^4+y^4)}+\frac {(x^4-y^4)^2}{(x^4+y^4)(x^4-y^4)}=\frac{5}{2}

x 8 + 2 x 4 y 4 + y 8 + x 8 − 2 x 4 y 4 + y 8 x 8 − y 8 = 5 2 \frac {x^8+2x^4y^4+y^8+x^8-2x^4y^4+y^8}{x^8-y^8}=\frac{5}{2}

2 x 8 + 2 y 8 x 8 − y 8 = 5 2 \frac {2x^8+2y^8}{x^8-y^8}=\frac{5}{2}

x 8 + y 8 x 8 − y 8 = 5 4 \frac {x^8+y^8}{x^8-y^8}=\frac{5}{4}

x 8 + y 8 x 8 − y 8 + x 8 − y 8 x 8 + y 8 = x 8 + y 8 x 8 − y 8 + 1 x 8 + y 8 x 8 − y 8 = 5 4 + 4 5 = 41 20 \frac {x^8+y^8}{x^8-y^8}+\frac {x^8-y^8}{x^8+y^8}=\frac {x^8+y^8}{x^8-y^8}+\frac {1}{\frac{x^8+y^8}{x^8-y^8}}=\frac{5}{4}+\frac{4}{5}=\frac{41}{20}

a + b = 41 + 20 = 61 a+b=41+20=\boxed{61}

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