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Question Number 211954 by RojaTaniya last updated on 25/Sep/24

x, y a r e r a t i o n a l n u m b e r s w h e r e

x \neq 0, y \neq 0, x \neq y, t h e n i s i t

p o s s i b l e : x^{5} + y^{5} = 2 x^{2} y^{2} ?

Answered by Frix last updated on 25/Sep/24

x^{5} - 2 x^{2} y^{2} + y^{5} = 0

y = p x

(p^{5} + 1) x^{5} - 2 p^{2} x^{4} = 0

(p^{5} + 1) x - 2 p^{2} = 0

x = \frac{2 p^{2}}{p^{5} + 1} \Rightarrow y = \frac{2 p^{3}}{p^{5} + 1}

p \in Q \Rightarrow (x, y) \in Q^{2}

Commented by RojaTaniya last updated on 25/Sep/24

S i r p e r f e c t .