if-q-is-prime-number-fixed-then-solve-for-natural-numbers-the-equation-1-q-1-x-1-y-1-z-

Question Number 149567 by mathdanisur last updated on 06/Aug/21

$${if}\:\:\boldsymbol{{q}}\:\:{is}\:{prime}\:{number}\:{fixed},\:{then} \\ $$$${solve}\:{for}\:{natural}\:{numbers}\:{the}\:{equation}: \\ $$$$\frac{\mathrm{1}}{{q}}\:=\:\frac{\mathrm{1}}{{x}}\:+\:\frac{\mathrm{1}}{{y}}\:-\:\frac{\mathrm{1}}{{z}} \\ $$

Commented by Rasheed.Sindhi last updated on 07/Aug/21

My answer is too lengthy. The question is waiting for a better solution.

$$\mathrm{My}\:\mathrm{answer}\:\mathrm{is}\:\mathrm{too}\:\mathrm{lengthy}. \\ $$$$\mathrm{The}\:\mathrm{question}\:\mathrm{is}\:\mathrm{waiting}\:\mathrm{for}\:\mathrm{a}\:\mathrm{better} \\ $$$$\mathrm{solution}. \\ $$

Commented by mathdanisur last updated on 07/Aug/21

Dear Ser We hawe: (1/n) = (1/(2(n-1))) + (1/(2(n+1))) - (1/((n-1)(n+1))) ⇒ n=q .... Ser, this true or wrong.?

$$\mathrm{Dear}\:\boldsymbol{\mathrm{S}}\mathrm{er} \\ $$$$\mathrm{We}\:\mathrm{hawe}:\:\frac{\mathrm{1}}{\mathrm{n}}\:=\:\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{n}-\mathrm{1}\right)}\:+\:\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{n}+\mathrm{1}\right)}\:-\:\frac{\mathrm{1}}{\left(\mathrm{n}-\mathrm{1}\right)\left(\mathrm{n}+\mathrm{1}\right)} \\ $$$$\Rightarrow\:\mathrm{n}=\mathrm{q}\:…. \\ $$$$\mathrm{Ser},\:\mathrm{this}\:\mathrm{true}\:\mathrm{or}\:\mathrm{wrong}.? \\ $$

Commented by Rasheed.Sindhi last updated on 07/Aug/21

(1/(2(q−1)))+(1/(2(q+1)))−(1/((q−1)(q+1))) =((q+1+q−1−2)/(2(q−1)(q+1)))=((2q−2)/(2(q−1)(q+1))) =((2(q−1))/(2(q−1)(q+1)))=(1/(q+1))≠(1/q) Not correct ser.

$$\frac{\mathrm{1}}{\mathrm{2}\left({q}−\mathrm{1}\right)}+\frac{\mathrm{1}}{\mathrm{2}\left({q}+\mathrm{1}\right)}−\frac{\mathrm{1}}{\left({q}−\mathrm{1}\right)\left({q}+\mathrm{1}\right)} \\ $$$$\:\:\:=\frac{{q}+\mathrm{1}+{q}−\mathrm{1}−\mathrm{2}}{\mathrm{2}\left({q}−\mathrm{1}\right)\left({q}+\mathrm{1}\right)}=\frac{\mathrm{2}{q}−\mathrm{2}}{\mathrm{2}\left({q}−\mathrm{1}\right)\left({q}+\mathrm{1}\right)} \\ $$$$=\frac{\cancel{\mathrm{2}\left({q}−\mathrm{1}\right)}}{\cancel{\mathrm{2}\left({q}−\mathrm{1}\right)}\left({q}+\mathrm{1}\right)}=\frac{\mathrm{1}}{{q}+\mathrm{1}}\neq\frac{\mathrm{1}}{{q}} \\ $$$${Not}\:{correct}\:{ser}. \\ $$

Commented by mathdanisur last updated on 07/Aug/21

$$\mathrm{Sorry}\:\boldsymbol{\mathrm{S}}\mathrm{er},\:\:…\:\frac{\mathrm{1}}{\mathrm{n}\left(\mathrm{n}-\mathrm{1}\right)\left(\mathrm{n}+\mathrm{1}\right)} \\ $$

Commented by Rasheed.Sindhi last updated on 07/Aug/21

(1/(2(n-1))) + (1/(2(n+1))) - (1/(n(n-1)(n+1))) ((n(n+1)+n(n−1)−2)/(2n(n-1)(n+1))) ((n^2 +n+n^2 −n−2)/(2n(n-1)(n+1)))=((2n^2 −2)/(2n(n-1)(n+1))) ((2(n−1)(n+1))/(2n(n-1)(n+1)))=(1/n) ✓ It means that this is the solution.

$$\:\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{n}-\mathrm{1}\right)}\:+\:\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{n}+\mathrm{1}\right)}\:-\:\frac{\mathrm{1}}{\mathrm{n}\left(\mathrm{n}-\mathrm{1}\right)\left(\mathrm{n}+\mathrm{1}\right)} \\ $$$$\frac{\mathrm{n}\left(\mathrm{n}+\mathrm{1}\right)+\mathrm{n}\left(\mathrm{n}−\mathrm{1}\right)−\mathrm{2}}{\mathrm{2n}\left(\mathrm{n}-\mathrm{1}\right)\left(\mathrm{n}+\mathrm{1}\right)} \\ $$$$\frac{\mathrm{n}^{\mathrm{2}} +\mathrm{n}+\mathrm{n}^{\mathrm{2}} −\mathrm{n}−\mathrm{2}}{\mathrm{2n}\left(\mathrm{n}-\mathrm{1}\right)\left(\mathrm{n}+\mathrm{1}\right)}=\frac{\mathrm{2n}^{\mathrm{2}} −\mathrm{2}}{\mathrm{2n}\left(\mathrm{n}-\mathrm{1}\right)\left(\mathrm{n}+\mathrm{1}\right)} \\ $$$$\frac{\mathrm{2}\left(\mathrm{n}−\mathrm{1}\right)\left(\mathrm{n}+\mathrm{1}\right)}{\mathrm{2n}\left(\mathrm{n}-\mathrm{1}\right)\left(\mathrm{n}+\mathrm{1}\right)}=\frac{\mathrm{1}}{\mathrm{n}}\:\checkmark \\ $$$$\mathrm{It}\:\mathrm{means}\:\mathrm{that}\:\mathrm{this}\:\mathrm{is}\:\mathrm{the}\:\mathrm{solution}. \\ $$

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