Summing the reciprocal of the polynomial appearing in Fermat's Last Theorem

Source abstract

Consider the polynomial $f=x^N+y^N-z^N$ where $x,\,y$ and $z$ are positive integers and $N \ge 3$ is an integer. By Fermat's Last Theorem, $f$ is never zero so that its reciprocal, $1/f$, has no singularities. We therefore study the finite sum of the reciprocal: $S(m,N)=\sum_{x=1}^m\sum_{y=1}^m\sum_{z=1}^{m}\frac{1}{f}$. The terms $1/f$ can be positive, negative and their magnitude is less than unity. A key observation is that $S(m,N)$ can be split into two convenient parts: a dominant contribution $D(m,N)$ that has a simple analytical expression and a remainder $R(m,N)$ which is more complicated but negligible compared to $D(m,N)$. Therefore, $S(m,N)$ is almost identical to $D(m,N)$. The analytical expression for $D(m,N)$ is $(2\,m-1)\,H_m^{(N)}$ where $H_m^{(N)}=\sum_{x=1}^m\frac{1}{x^N}$ approaches quickly the Riemann zeta function $ζ(N)$ as $m$ increases. Therefore, the original sum $S(m,N)$ has a simple expression: it is basically linear in $m$ with slope equal to $2\,ζ(N)$. Its linear behavior is not an asymptotic result; plots of $S(m,N)$ vs. $m$ for different $N$ show a straight line starting at $m=1$. $S(m,N)$ deviates slightly from a straight line over a small interval $8\le m\le 12$ for the case $N=3$. This slight deviation is due to Fermat near misses where $x^3+y^3-z^3=\pm 1$ (for $z\ne x$ and $z\ne y$); these create a jump in the remainder $R(m,3)$ at $m=9$. We make a numerical and analytical study of the remainder $R(m,N)$. From the numerical analysis, $R(m,N)$ converges for $N\ge 4$ but it was harder to tell whether $N=3$ converged. An analytical study based on a comparison of $R(m,N)$ to its Cauchy principal value integral, shows that $R(m,3)$ likely diverges logarithmically. It also shows that $R(m,N)$ converges for $N\ge 4$ in agreement with the numerical analysis. We discuss in the conclusion some interesting questions for future investigation.