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Question Number 197843 by a.lgnaoui last updated on 30/Sep/23

$$\mathbf{Exercice}2$$

Commented by a.lgnaoui last updated on 30/Sep/23

Commented by witcher3 last updated on 07/Oct/23

$$1,\mathrm{tout}\mathrm{en}\mathrm{puissance}2$$$$\left(\frac{2\mathrm{a}+2\sqrt{{\mathrm{a}}^2-\mathrm{b}}}{2(\mathrm{a}+1)}\right)=\frac{2\mathrm{a}+2\sqrt{1}}{2(\mathrm{a}+1)}=1\iff 0⩽\frac{\sqrt{\mathrm{a}-\sqrt{\mathrm{b}}}+\sqrt{\mathrm{a}+\sqrt{\mathrm{b}}}}{\sqrt{2\mathrm{a}+2}}=1$$$${\mathrm{n}}^2\equiv 0\left[3\right]$$$$3\mid {\mathrm{n}}^2,3\mathrm{est}\mathrm{premier}\Rightarrow 3\mid \mathrm{n}‘‘\mathrm{p}\mid \mathrm{n}.\mathrm{m}\mathrm{avec}\mathrm{n}\mathrm{premier}\mathrm{alors}$$$$\mathrm{p}\mid \mathrm{n}\mathrm{ou}\mathrm{p}\mid {\mathrm{m}}^{″}\mathrm{Theorem}\mathrm{de}\mathrm{gause}\mathrm{ic}3\mid \mathrm{n}.\mathrm{n}\Rightarrow 3\mid \mathrm{n}$$$${\mathrm{x}}^2-\mathrm{x}+1=(\mathrm{x}-\frac{1}{2})+\frac{3}{4}⩾\frac{3}{4}$$$$\Rightarrow \frac{1}{({\mathrm{x}}^2-\mathrm{x}+1)}⩽\frac{4}{3}\Rightarrow \frac{\sqrt{\mathrm{x}}}{{\mathrm{x}}^2-\mathrm{x}+1}⩽\frac{4}{3}\sqrt{\mathrm{x}}$$$$\sqrt{{4\mathrm{n}}^2+5\mathrm{n}+3}\in \mathbb{N}\iff \mathrm{\exists }\mathrm{a}\in \mathbb{N}\mid {\mathrm{a}}^2={4\mathrm{n}}^2+5\mathrm{n}+3$$$$\iff {(2\mathrm{n}+1)}^2+\mathrm{n}+2={\mathrm{a}}^2$$$$\Rightarrow (\mathrm{a}-2\mathrm{n}-1)(\mathrm{a}+2\mathrm{n}+1)=\mathrm{n}+2\Rightarrow$$$$\mathrm{a}>2\mathrm{n}+1,\mathrm{car}\mathrm{n}+2⩾0$$$$\Rightarrow \mathrm{a}+2\mathrm{n}+1⩾4\mathrm{n}+2$$$$\Rightarrow (\mathrm{a}-2\mathrm{n}-1)(\mathrm{a}+2\mathrm{n}+1)⩾4\mathrm{n}+2>\mathrm{n}+2\mathrm{\forall }\mathrm{n}>1$$$$\mathrm{pour}\mathrm{n}=0\sqrt{3}\notin \mathbb{N}$$$$$$$$$$$$$$$$\Rightarrow$$

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