Algebraic Identities Come In Handy!

Number Theory Level 2

x + y x 3 + y 3 = 1 10 \large \dfrac { x+y }{ { x }^{ 3 }+{ y }^{ 3 } } =\frac { 1 }{ 10 }

How many integral solutions does the above equation have?


The answer is 0.

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Hamza A by Hamza A , 19, USA

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1 solution

Tran Quoc Dat
Mar 31, 2016

Condition: x 3 + y 3 ≠ 0 ⇔ x ≠ − y x^3+y^3 \neq 0 \Leftrightarrow x \neq -y

x + y x 3 + y 3 = 1 10 \frac{x+y}{x^3+y^3}=\frac{1}{10} ⇔ 1 x 2 − x y + y 2 = 1 10 \Leftrightarrow \frac{1}{x^2-xy+y^2}=\frac{1}{10} ⇔ 3 4 ( x − y ) 2 + 1 4 ( x + y ) 2 = 10 \Leftrightarrow \frac{3}{4}(x-y)^2+\frac{1}{4}(x+y)^2=10 ⇔ ( x + y ) 2 + 3 ( x − y ) 2 = 40 \Leftrightarrow (x+y)^2+3(x-y)^2=40 .

There's no pair of integers x + y x+y and x − y x-y satisfy the above equation. Hence, the equation has no integral solutions.

9 Helpful 0 Interesting 0 Brilliant 0 Confused

how did you go from 1/(x^2 - xy + y^2) = 1/10 --> (3/4)(x - y)^2 + (1/4)(x + y)^2 = 10 ?

Karish Thangarajah - 5 years, 1 month ago

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It's simple ..simplify you can easily see that 3/4(x - y)^2 + 1/4(x + y)^2 = 10

Bhupendra Jangir - 5 years, 1 month ago

Cross multiply and complete the square

Gambler Ho - 5 years, 1 month ago

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