Finding unknown exponent

Question

How do you count the n for this general formula, when you know a, b and c, please?

$a^n + b^n = c^n$ (optional conditions $c<(a+b)<2c$ and $a,b,c ∈ N$ and $n ∉ N$)

Answer 1

You are looking for the zero of function $$f(x)=c^x-a^x - b^x $$ and numerical methods need to be considered.

In fact, it is better to search for the zero of $$g(x)=\log(c^x)-\log(a^x+b^x)=x \log(c)-\log(a^x+b^x)$$ and use Newton method.

To generate a good starting point, you can develop $g(x)$ as a Taylor series built at $x=0$ and get $$g(x)=-\log (2)+x \left(\log (c)-\frac{\log (a)}{2}-\frac{\log (b)}{2}\right)+O\left(x^2\right)$$ Ignoring the higher order terms, this will give, as an estimate, $$x_0=\frac{2 \log (2)}{2 \log (c)-\log (a)-\log (b)}$$

Using, as Ondřej Dvořák did in a comment, $a=5$, $b=7$, $c=10$ and using either $x_0=0$ or this estimate (which corresponds to the first iteration of Newton method), the iterates would be $$\left( \begin{array}{cc} n & x_n \\ 0 & 0.000000000 \\ 1 & 1.320504044 \\ 2 & 1.370646293 \\ 3 & 1.370715807 \end{array} \right)$$

You will not have an overshoot of the solution since $$f(0)=-\log(2) \qquad \text{and} \qquad f''(0)=-\frac{1}{4} (\log (a)-\log (b))^2 \implies f(0) \times f''(0)>0$$ (refer to Darboux-Fourier theorem).