Fermat's Last Theorem related question

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The discussion focuses on proving that the equation x^n + y^n = z^n has a nontrivial integer solution if and only if the equation 1/x^n + 1/y^n = 1/z^n has a nontrivial integer solution. A successful proof in one direction is presented, demonstrating that a solution to the second equation implies a solution to the first. However, the reverse direction remains unresolved, with challenges in applying the same logic backward without additional assumptions. Suggestions include considering the symmetry of the equations to facilitate the proof. The conversation highlights the complexities involved in proving both directions of the equivalence.
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Homework Statement


Show that x^{n}+y^{n}=z^{n} has a nontrivial solution if and only if the equation \frac{1}{x^{n}}+\frac{1}{y^{n}}=\frac{1}{z^{n}} has a nontrivial solution.

Homework Equations


By nontrivial solutions, it is implied that they are integer solutions.

The Attempt at a Solution


I was able to solve in one direction
Given an integer solution to \frac{1}{x^{n}}+\frac{1}{y^{n}}=\frac{1}{z^{n}}
Then it follows that:
\frac{x^{n}+y^{n}}{x^{n}y^{n}}=\frac{1}{z^{n}}
x^{n}y^{n}=z^{n}x^{n}+z^{n}y^{n}
(xy)^{n}=(zx)^{n}+(zy)^{n}
Therefore, since the set of integers is closed under multiplication, then x^{n}+y^{n}=z^{n} has a nontrivial solution.

However, I can't seem to prove the other direction. Working the above proof backwards doesn't work unless I assume that d = gcd(x,y) and somehow prove that z = xy.
 
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Ninty64 said:
Working the above proof backwards doesn't work
Applying exactly the same method in the reverse direction appears to work.
 
I might note that ##\frac{1}{\frac{1}{x^n}}+\frac{1}{\frac{1}{y^n}}=\frac{1}{\frac{1}{z^n}}## is identical to writing ##x^n+y^n=z^n##. Then you can refer back to your earlier proof.
 
alternatively, since the formula you start with is symmetric, try doing something symmetric to it :wink:
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks

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