deletions | additions       diff --git a/section_Fermat_Like_Problem_Observing__.tex b/section_Fermat_Like_Problem_Observing__.tex  index 3d11d7d..b3920c2 100644  --- a/section_Fermat_Like_Problem_Observing__.tex  +++ b/section_Fermat_Like_Problem_Observing__.tex 
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can be expressed as equivalence of a term in the expansions \begin{equation}  \frac{2}{  x+\frac{\frac{a+b}{2}}{  x+\frac{\frac{(a-b)^2}{a+b}}{ x+\frac{\frac{(a-b)^2}{2(a+b)}}{  x+\frac{\frac{2ab}{a+b}}{  x}}}} \Leftrightarrow \frac{1}{x+\frac{c}{x}}  \end{equation}  which represents the equivalence (as $x\to\infty$) \begin{equation}  \frac{a^0+b^0}{x}-\frac{a^1+b^1}{x^3}+\frac{a^2+b^2}{x^5}-\cdots = \frac{c^0}{x}-\frac{c^1}{x^3}+\frac{c^2}{x^5}-\cdots  \end{equation}  The basic comparison is then \begin{equation}  (n=0)\;\;\;\lim_{x\to\infty} 2=1\;\;\;(\mathrm{no solution})  \end{equation}  then we have \begin{equation}  \frac{2x}{  x+\frac{\frac{a+b}{2}}{  x+\frac{\frac{(a-b)^2}{a+b}}{  x+\frac{\frac{2ab}{a+b}}{  x}}}} -2 \Leftrightarrow \frac{x}{x+\frac{c}{x}} -1  \end{equation}  which represents the equivalence (as $x\to\infty$) \begin{equation}  -\frac{a^1+b^1}{x^2}+\frac{a^2+b^2}{x^4}-\cdots = -\frac{c^1}{x^2}+\frac{c^2}{x^4}-\cdots  \end{equation}  for this we only have solutions $b=-a,c=0$ and $b=a,c=2a$. We perform the next step and solve  \begin{equation}  \lim_{x\to\infty}\left[  \frac{2x^5}{  x+\frac{\frac{a+b}{2}}{  x+\frac{\frac{(a-b)^2}{a+b}}{  x+\frac{\frac{2ab}{a+b}}{  x}}}} -2x^4 + ax^2 + bx^2 -\frac{x^5}{x+\frac{c}{x}} +x^4 -cx^2 \right]=0  \end{equation}  This has solutions \begin{equation}  b=-a,\;c=0\\  \end{equation}