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Question Number 8022 by Nayon last updated on 28/Sep/16

  prove that ∀x∈N   (2x−1)^2 +(2x^2 −2x)^2   is a proper  square number.

$$ \\ $$$${prove}\:{that}\:\forall{x}\in\boldsymbol{{N}}\: \\ $$$$\left(\mathrm{2}{x}−\mathrm{1}\right)^{\mathrm{2}} +\left(\mathrm{2}{x}^{\mathrm{2}} −\mathrm{2}{x}\right)^{\mathrm{2}} \:\:{is}\:{a}\:{proper} \\ $$$${square}\:{number}. \\ $$

Commented by prakash jain last updated on 28/Sep/16

let u=(2x−1)^2 +(2x^2 −2x)^2   =(2x^2 −2x)^2 +4x^2 −4x+1  =(2x^2 −2x)^2 +2(2x^2 −2x)+1  =(2x^2 −2x+1)^2   x∈N⇒2x^2 −2x+1∈N  or (√u)∈N

$$\mathrm{let}\:{u}=\left(\mathrm{2}{x}−\mathrm{1}\right)^{\mathrm{2}} +\left(\mathrm{2}{x}^{\mathrm{2}} −\mathrm{2}{x}\right)^{\mathrm{2}} \\ $$$$=\left(\mathrm{2}{x}^{\mathrm{2}} −\mathrm{2}{x}\right)^{\mathrm{2}} +\mathrm{4}{x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{1} \\ $$$$=\left(\mathrm{2}{x}^{\mathrm{2}} −\mathrm{2}{x}\right)^{\mathrm{2}} +\mathrm{2}\left(\mathrm{2}{x}^{\mathrm{2}} −\mathrm{2}{x}\right)+\mathrm{1} \\ $$$$=\left(\mathrm{2}{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{2}} \\ $$$${x}\in\mathbb{N}\Rightarrow\mathrm{2}{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{1}\in\mathbb{N} \\ $$$${or}\:\sqrt{{u}}\in\mathbb{N} \\ $$

Answered by prakash jain last updated on 02/Oct/16

answer in comments

$$\mathrm{answer}\:\mathrm{in}\:\mathrm{comments} \\ $$

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