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Kindly mention solution-techniques along with solution

Cameron Buie
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Arnab Dutta
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1 Answers1

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Hint: if $({\frac{1}{2}})^p + ({\frac{1}{4}})^p + ({\frac{1}{8}})^p = 1 $, let $ x=({\frac{1}{2}})^p$, then $$x+x^2+x^3=1\to \frac{x^4-1}{x-1}=2$$ we can find all Roots of a cubic function by using the discriminant (see here http://en.wikipedia.org/wiki/Cubic_function)

M.H
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  • How did you conclude that last bit? – Tim Vermeulen Jun 01 '13 at 19:54
  • It comes from the formula for a geometric sum. – Cameron Williams Jun 01 '13 at 19:57
  • @timjver $1+x+\cdots+x^n=\dfrac{x^{n+1}-1}{x-1}$ – Pedro Jun 01 '13 at 19:57
  • Please don't assume they will be in relations like x , x^2 , x^3 Then how to solve ? – Arnab Dutta Jun 01 '13 at 20:15
  • Arnab Dutta:$(\frac{1}{2})^p+(\frac{1}{2})^{p^2}+(\frac{1}{2})^{p^3}=1 $$\to $$\frac{({\frac{1}{2}})^{p^4}-1}{({\frac{1}{2}})^p-1}=2 $ then use discriminant – M.H Jun 01 '13 at 20:22
  • Hi everybody: If you put values of p(like $\frac{1}{2}$, 2) back in the euation it doesn't satisfy! Moreover what - if the equation stands like?: $ ({\frac{1}{2}})^p + ({\frac{1}{3}})^p + ({\frac{1}{7}})^p - 1 = 0. $ (So please dont assume there is relation like $\frac{1}{2},{\frac{1}{2}}^2 ,{\frac{1}{2}}^3 $) – Arnab Dutta Jun 02 '13 at 06:12