Diophantine Equations (Problem 3, Version 2)

Number Theory Level pending

x n − y n = 1001 x^n - y^n = 1001

Are there are solutions for n = 1 , n = 2 n = 1, n = 2 where x , y x,y are integers?

Yes No

A Former Brilliant Member by A Former Brilliant Member

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1 solution

When n = 1 , x − y = 1001 : x ≥ 1001 , y ≥ 0 n = 1, x - y = 1001: x \geq 1001, y \geq 0

When n = 2 , x 2 − y 2 = 1001 : x ≥ 32 : x = 32 , y = 45 n = 2, x^2 - y^2 = 1001: x \geq 32 : x = 32, y = 45

Therefore, the answer is Yes

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n = 3 n = 3 is also true

x 3 − y 3 ⟹ { ( 10 ) 3 − ( − 1 ) 3 = 1001 ( 1 ) 3 − ( − 10 ) 3 = 1001 x^3 - y^3 \implies \begin{cases}(10)^3 - (-1)^3 = 1001 \\ (1)^3 - (-10)^3= 1001 \end{cases}

Mahdi Raza - 1 year ago

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Oh, thanks - the guy who gave me the idea said he couldn't solve it for n = 3 n = 3

A Former Brilliant Member - 1 year ago

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