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Question Number 196770 by Amidip last updated on 31/Aug/23

Commented by Frix last updated on 31/Aug/23

$$‘‘\mathbb{N}\mathrm{atural}\mathrm{numbers}{\left(\mathbb{Z}\right)}^{″}-\mathrm{really}?!?$$

Commented by Frix last updated on 31/Aug/23

$$\mathrm{This}\mathrm{has}\mathrm{been}\mathrm{answered}.$$$$\mathrm{Question}196637$$

Answered by mr W last updated on 31/Aug/23

$$\sqrt{12+\sqrt{m}}+\sqrt{12-\sqrt{m}}=n...\left(i\right)$$$$(\sqrt{12+\sqrt{m}}-\sqrt{12-\sqrt{m}})(\sqrt{12+\sqrt{m}}+\sqrt{12-\sqrt{m}})=2\sqrt{m}$$$$\sqrt{12+\sqrt{m}}-\sqrt{12-\sqrt{m}}=\frac{2\sqrt{m}}{n}...\left(ii\right)$$$$\left(i\right)+\left(ii\right)/2:$$$$\sqrt{12+\sqrt{m}}=\frac{\sqrt{m}}{n}+\frac{n}{2}$$$$\Rightarrow 12=\frac{m}{n^2}+\frac{n^2}{4}$$$$\Rightarrow m=\frac{n^2(48-n^2)}{4}$$$$nmustbeeven,sayn=2k$$$$m=4k^2(12-k^2)>0\Rightarrow k^2⩽12\Rightarrow k⩽3$$$$m=4k^2(12-k^2)⩽144\Rightarrow k^2⩾6\Rightarrow k⩾3$$$$\Rightarrow onlypossibilityisk=3$$$$\Rightarrow n=6,m=4\times 9\times 3=108$$$$m+n=108+6=114✓$$

Commented by Frix last updated on 01/Sep/23

$$\mathrm{Yes}.$$$$\mathrm{But}\mathrm{if}m,n\in \mathbb{Z}\mathrm{we}\mathrm{have}\mathrm{infinite}\mathrm{solutions}$$$$\mathrm{for}m=\frac{n^2(48-n^2)}{4}\wedge n=2k\wedge k⩾3\mathrm{because}$$$$\mathrm{for}m<0\mathrm{the}\mathrm{imaginary}\mathrm{parts}\mathrm{cancel}\mathrm{out}:$$$$\sqrt{12-\sqrt{m}}+\sqrt{12+\sqrt{m}}=\sqrt{24+2\sqrt{144-m}};m<0$$

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