Observing other solutions we can infer a general rule for any \(a^n+b^n\) \[\begin{matrix} 1^n+2^n & 2,3/2,1/6,4/3,0\\ 2^n+3^n & 2,5/2,1/10,12/5,0\\ 3^n+4^n & 2,7/2,1/14,24/7,0 & A074605\\ 3^n+5^n & 2,8/2,4/16,30/8,0 & A074606\\ 2^n+7^n & 2,9/2,25/18,28/9,0 & A074602\\ 117^n+234^n & 2,351/2,39/2,156,0 & ...\ a^n+b^n & \mathrm{size}(\{a,b\}),\frac{a+b}{\mathrm{size}\{a,b\}},\frac{(a-b)^2}{2(a+b)},\frac{2ab}{a+b},0\\ c^n & 1,c,0,0 \\ \end{matrix}\] Then we can write \[\mathcal{I}(2;\frac{a+b}{2},\frac{2ab}{a+b};\frac{(a-b)^2}{2(a+b)},0)=a^n+b^n\]
We then see that Fermat’s last theorem’s problem \[a^n+b^n=c^n\] can be expressed as equivalence of a term in the expansions \[\frac{2}{ x+\frac{\frac{a+b}{2}}{ x+\frac{\frac{(a-b)^2}{2(a+b)}}{ x+\frac{\frac{2ab}{a+b}}{ x}}}} \Leftrightarrow \frac{1}{x+\frac{c}{x}}\] which represents the equivalence (as \(x\to\infty\)) \[\frac{a^0+b^0}{x}-\frac{a^1+b^1}{x^3}+\frac{a^2+b^2}{x^5}-\cdots = \frac{c^0}{x}-\frac{c^1}{x^3}+\frac{c^2}{x^5}-\cdots\]
Therefore to gain the equation that must be solved for a potential solution of each \(n\) we have the set of equations to solve \[\frac{2}{ x+\frac{\frac{a+b}{2}}{ x+\frac{\frac{(a-b)^2}{2(a+b)}}{ x+\frac{\frac{2ab}{a+b}}{ x}}}} = \frac{1}{x+\frac{c}{x}}\] \[\frac{2x}{ x+\frac{\frac{a+b}{2}}{ x+\frac{\frac{(a-b)^2}{2(a+b)}}{ x+\frac{\frac{2ab}{a+b}}{ x}}}} -2 = \frac{x}{x+\frac{c}{x}} -1\] \[\frac{2x^3}{ x+\frac{\frac{a+b}{2}}{ x+\frac{\frac{(a-b)^2}{2(a+b)}}{ x+\frac{\frac{2ab}{a+b}}{ x}}}} -2x^2 +a +b= \frac{x^3}{x+\frac{c}{x}} -x^2 +c\] \[\frac{2x^5}{ x+\frac{\frac{a+b}{2}}{ x+\frac{\frac{(a-b)^2}{2(a+b)}}{ x+\frac{\frac{2ab}{a+b}}{ x}}}} -2x^4 +ax^2 +bx^2 -a^2 -b^2= \frac{x^5}{x+\frac{c}{x}} -x^4 +cx^2 -c^2\]
For the greatest of these here, we can see when \(x\to0\) we reclaim the next equation down \(a^2+b^2=c^2\). The real solution of that equation in the limit \(x\to 0\) is given by \[c=\frac{(3i+\sqrt{3})a^2b^2(a^2+b^2)+(\sqrt{3}-3i)(-a^6b^6(a^2+b^2)^3)^{1/3}}{6ab(-a^6b^6(a^2+b^2)^3)^{1/6}}\]