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

Commented by Frix last updated on 31/Aug/23

“Natural numbers (Z)” − really?!?

$$``\mathbb{N}\mathrm{atural}\:\mathrm{numbers}\:\left(\mathbb{Z}\right)''\:−\:\mathrm{really}?!? \\ $$

Commented by Frix last updated on 31/Aug/23

This has been answered. Question 196637

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

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

(√(12+(√m)))+(√(12−(√m)))=n ...(i) ((√(12+(√m)))−(√(12−(√m))))((√(12+(√m)))+(√(12−(√m))))=2(√m) (√(12+(√m)))−(√(12−(√m)))=((2(√m))/n) ...(ii) (i)+(ii) /2: (√(12+(√m)))=((√m)/n)+(n/2) ⇒12=(m/n^2 )+(n^2 /4) ⇒m=((n^2 (48−n^2 ))/4) n must be even, say n=2k m=4k^2 (12−k^2 )>0 ⇒k^2 ≤12 ⇒k≤3 m=4k^2 (12−k^2 )≤144 ⇒k^2 ≥6 ⇒k≥3 ⇒only possibility is k=3 ⇒n=6, m=4×9×3=108 m+n=108+6=114 ✓

$$\sqrt{\mathrm{12}+\sqrt{{m}}}+\sqrt{\mathrm{12}−\sqrt{{m}}}={n}\:\:\:\:...\left({i}\right) \\ $$$$\left(\sqrt{\mathrm{12}+\sqrt{{m}}}−\sqrt{\mathrm{12}−\sqrt{{m}}}\right)\left(\sqrt{\mathrm{12}+\sqrt{{m}}}+\sqrt{\mathrm{12}−\sqrt{{m}}}\right)=\mathrm{2}\sqrt{{m}} \\ $$$$\sqrt{\mathrm{12}+\sqrt{{m}}}−\sqrt{\mathrm{12}−\sqrt{{m}}}=\frac{\mathrm{2}\sqrt{{m}}}{{n}}\:\:\:...\left({ii}\right) \\ $$$$\left({i}\right)+\left({ii}\right)\:/\mathrm{2}: \\ $$$$\sqrt{\mathrm{12}+\sqrt{{m}}}=\frac{\sqrt{{m}}}{{n}}+\frac{{n}}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{12}=\frac{{m}}{{n}^{\mathrm{2}} }+\frac{{n}^{\mathrm{2}} }{\mathrm{4}} \\ $$$$\Rightarrow{m}=\frac{{n}^{\mathrm{2}} \left(\mathrm{48}−{n}^{\mathrm{2}} \right)}{\mathrm{4}} \\ $$$${n}\:{must}\:{be}\:{even},\:{say}\:{n}=\mathrm{2}{k} \\ $$$${m}=\mathrm{4}{k}^{\mathrm{2}} \left(\mathrm{12}−{k}^{\mathrm{2}} \right)>\mathrm{0}\:\Rightarrow{k}^{\mathrm{2}} \leqslant\mathrm{12}\:\Rightarrow{k}\leqslant\mathrm{3} \\ $$$${m}=\mathrm{4}{k}^{\mathrm{2}} \left(\mathrm{12}−{k}^{\mathrm{2}} \right)\leqslant\mathrm{144}\:\Rightarrow{k}^{\mathrm{2}} \geqslant\mathrm{6}\:\Rightarrow{k}\geqslant\mathrm{3} \\ $$$$\Rightarrow{only}\:{possibility}\:{is}\:{k}=\mathrm{3} \\ $$$$\Rightarrow{n}=\mathrm{6},\:{m}=\mathrm{4}×\mathrm{9}×\mathrm{3}=\mathrm{108} \\ $$$${m}+{n}=\mathrm{108}+\mathrm{6}=\mathrm{114}\:\checkmark \\ $$

Commented by Frix last updated on 01/Sep/23

Yes. But if m, n ∈Z we have infinite solutions for m=((n^2 (48−n^2 ))/4)∧n=2k∧k≥3 because for m<0 the imaginary parts cancel out: (√(12−(√m)))+(√(12+(√m)))=(√(24+2(√(144−m)))); m<0

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

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