Show-that-2x-7-4x-4-4x-2-6x-6-3-Has-no-solution-in-N-

Question Number 89906 by Ar Brandon last updated on 19/Apr/20

S h o w t h a t

2 x^{7} - 4 x^{4} + 4 x^{2} = 6 x^{6} + 3

H a s n o s o l u t i o n i n N

Answered by MJS last updated on 20/Apr/20

transform to

x^{7} - 3 x^{6} - 2 x^{4} + 2 x^{2} - \frac{3}{2} = 0

p = x^{7} - 3 x^{6} - 2 x^{4} + 2 x^{2} \land q = - \frac{3}{2}

x \in N \Rightarrow p \in N but q \notin N

or

x, m, n \in N

2 x^{7} - 4 x^{4} + 4 x^{2} = 2 (x^{7} - 2 x^{4} + 2 x^{2}) = 2 m

6 x^{6} + 3 = 3 (2 x^{6} + 1) = 3 (2 n + 1)

2 m = 3 (2 n + 1)

\frac{2}{3} m = 2 n + 1

let m = 3 k; k \in N

2 k = 2 n + 1

even = uneven

impossible

Commented by Ar Brandon last updated on 20/Apr/20

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