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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
Show that   2x^7 −4x^4 +4x^2 =6x^6 +3  Has no solution in N
$${Show}\:{that}\: \\ $$$$\mathrm{2}{x}^{\mathrm{7}} −\mathrm{4}{x}^{\mathrm{4}} +\mathrm{4}{x}^{\mathrm{2}} =\mathrm{6}{x}^{\mathrm{6}} +\mathrm{3} \\ $$$${Has}\:{no}\:{solution}\:{in}\:\mathbb{N} \\ $$
Answered by MJS last updated on 20/Apr/20
transform to  x^7 −3x^6 −2x^4 +2x^2 −(3/2)=0  p=x^7 −3x^6 −2x^4 +2x^2 ∧q=−(3/2)  x∈N ⇒ p∈N but q∉N  or  x, m, n ∈N  2x^7 −4x^4 +4x^2 =2(x^7 −2x^4 +2x^2 )=2m  6x^6 +3=3(2x^6 +1)=3(2n+1)  2m=3(2n+1)  (2/3)m=2n+1  let m=3k; k∈N  2k=2n+1  even=uneven  impossible
$$\mathrm{transform}\:\mathrm{to} \\ $$$${x}^{\mathrm{7}} −\mathrm{3}{x}^{\mathrm{6}} −\mathrm{2}{x}^{\mathrm{4}} +\mathrm{2}{x}^{\mathrm{2}} −\frac{\mathrm{3}}{\mathrm{2}}=\mathrm{0} \\ $$$${p}={x}^{\mathrm{7}} −\mathrm{3}{x}^{\mathrm{6}} −\mathrm{2}{x}^{\mathrm{4}} +\mathrm{2}{x}^{\mathrm{2}} \wedge{q}=−\frac{\mathrm{3}}{\mathrm{2}} \\ $$$${x}\in\mathbb{N}\:\Rightarrow\:{p}\in\mathbb{N}\:\mathrm{but}\:{q}\notin\mathbb{N} \\ $$$$\mathrm{or} \\ $$$${x},\:{m},\:{n}\:\in\mathbb{N} \\ $$$$\mathrm{2}{x}^{\mathrm{7}} −\mathrm{4}{x}^{\mathrm{4}} +\mathrm{4}{x}^{\mathrm{2}} =\mathrm{2}\left({x}^{\mathrm{7}} −\mathrm{2}{x}^{\mathrm{4}} +\mathrm{2}{x}^{\mathrm{2}} \right)=\mathrm{2}{m} \\ $$$$\mathrm{6}{x}^{\mathrm{6}} +\mathrm{3}=\mathrm{3}\left(\mathrm{2}{x}^{\mathrm{6}} +\mathrm{1}\right)=\mathrm{3}\left(\mathrm{2}{n}+\mathrm{1}\right) \\ $$$$\mathrm{2}{m}=\mathrm{3}\left(\mathrm{2}{n}+\mathrm{1}\right) \\ $$$$\frac{\mathrm{2}}{\mathrm{3}}{m}=\mathrm{2}{n}+\mathrm{1} \\ $$$$\mathrm{let}\:{m}=\mathrm{3}{k};\:{k}\in\mathbb{N} \\ $$$$\mathrm{2}{k}=\mathrm{2}{n}+\mathrm{1} \\ $$$$\mathrm{even}=\mathrm{uneven} \\ $$$$\mathrm{impossible} \\ $$
Commented by Ar Brandon last updated on 20/Apr/20
Thank you Sir
$${Thank}\:{you}\:{Sir} \\ $$

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