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Question Number 204349 by Abdullahrussell last updated on 14/Feb/24

Commented by mr W last updated on 14/Feb/24

$$you{can}^{\prime }tprovesomethingwrong!$$

Commented by mr W last updated on 14/Feb/24

$$m=\frac{\sqrt{6}+\sqrt{5}}{2}<\frac{\sqrt{9}+\sqrt{9}}{2}=3$$$$\Rightarrow m^5<3^5=243$$$$m=\frac{\sqrt{6}+\sqrt{5}}{2}>\frac{\sqrt{4}+\sqrt{4}}{2}=2$$$$m^5>2^5=32$$$$\Rightarrow \frac{32}{m^5}<1$$$$\Rightarrow m^5+\frac{32}{m^5}<243+1=244$$$$but1048\sqrt{6}>1048\sqrt{4}=2096$$$$\Rightarrow m^5+\frac{32}{m^5}canneverbe=1048\sqrt{6}!$$

Answered by Rasheed.Sindhi last updated on 14/Feb/24

$$m+n=\sqrt{6}\mathrm{\&}m-n=\sqrt{5}$$$$\Rightarrow m=\frac{\sqrt{6}+\sqrt{5}}{2}$$$$\Rightarrow \frac{2}{m}=\frac{2}{\frac{\sqrt{6}+\sqrt{5}}{2}}=\frac{4}{\sqrt{6}+\sqrt{5}}\cdot \frac{\sqrt{6}-\sqrt{5}}{\sqrt{6}-\sqrt{5}}$$$$=\frac{4(\sqrt{6}-\sqrt{5})}{1}=4(\sqrt{6}-\sqrt{5})$$$$\Rightarrow m+\frac{2}{m}=\frac{\sqrt{6}+\sqrt{5}}{2}+4(\sqrt{6}-\sqrt{5})$$$$=\frac{5\sqrt{6}-3\sqrt{5}}{2}$$$$\Rightarrow {(m+\frac{2}{m})}^2={\left(\frac{5\sqrt{6}-3\sqrt{5}}{2}\right)}^2$$$$m^2+\frac{4}{m^2}={\left(\frac{5\sqrt{6}-3\sqrt{5}}{2}\right)}^2-4=\frac{25\left(6\right)+9\left(5\right)-30\sqrt{30}-16}{4}=\frac{179-30\sqrt{30}}{4}$$$${(m+\frac{2}{m})}^3={\left(\frac{5\sqrt{6}-3\sqrt{5}}{2}\right)}^3$$$$m^3+\frac{8}{m^3}=\frac{1425\sqrt{6}-1485\sqrt{5}}{8}-6\left(\frac{5\sqrt{6}-3\sqrt{5}}{2}\right)$$$$=\frac{1425\sqrt{6}-1485\sqrt{5}-24(5\sqrt{6}-3\sqrt{5})}{8}$$$$=\frac{1425\sqrt{6}-1485\sqrt{5}-120\sqrt{6}+72\sqrt{5})}{8}$$$$=\frac{1305\sqrt{6}-13\sqrt{5}}{8}$$$$(m^2+\frac{4}{m^2})(m^3+\frac{8}{m^3})=\left(\frac{179-30\sqrt{30}}{4}\right)\left(\frac{1305\sqrt{6}-13\sqrt{5}}{8}\right)$$$$m^5+\frac{32}{m^3}+4(m+\frac{2}{m})=\frac{(179-30\sqrt{30})(1305\sqrt{6}-13\sqrt{5})}{32}$$$$m^5+\frac{32}{m^3}+4\left(\frac{5\sqrt{6}-3\sqrt{5}}{2}\right)=\frac{(179-30\sqrt{30})(1305\sqrt{6}-13\sqrt{5})}{32}$$$$m^5+\frac{32}{m^3}=\frac{(179-30\sqrt{30})(1305\sqrt{6}-13\sqrt{5})}{32}-4\left(\frac{5\sqrt{6}-3\sqrt{5}}{2}\right)$$$$Continue$$

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