Limit and factorization

Note that $x^{m+1}-mx+(m-1) = \left( x-1 \right) \left( x^m + x^{m-1} + \ldots + x^2 + x -\left(m-1\right) \right)$ .

Then

$\begin{aligned} & \lim_{x\to 1} \left( \frac{1}{x^{m+1}-mx+(m-1)} - \frac{1}{x^{n+1}-nx+(n-1)}\right) \\ &=\lim_{x\to 1} \frac{\left( x^n + x^{n-1} + \ldots + x^2 + x -\left(n-1\right) \right) -\left( x^m + x^{m-1} + \ldots + x^2 + x -\left(m-1\right) \right) }{\left( x-1 \right) \left( x^m + x^{m-1} + \ldots + x^2 + x -\left(m-1\right) \right) \left( x^n + x^{n-1} + \ldots + x^2 + x -\left(n-1\right) \right) }. \end{aligned}$

Let $f(x)=\left( x^n + x^{n-1} + \ldots + x^2 + x -\left(n-1\right) \right) -\left( x^m + x^{m-1} + \ldots + x^2 + x -\left(m-1\right) \right)$ and $g(x)=\left( x-1 \right) \left( x^m + x^{m-1} + \ldots + x^2 + x -\left(m-1\right) \right) \left( x^n + x^{n-1} + \ldots + x^2 + x -\left(n-1\right) \right)$ .

As $f(1)=g(1)=0$ , we may use L'Hôpital's Rule.

It is not difficult to see that $f^\prime(1)= \frac{n(n+1)}{2}-\frac{m(m+1)}{2}$ .

For $g(x)$ , let $g(x)=(x-1) h(x)$ where $h(x)=\left( x^m + x^{m-1} + \ldots + x^2 + x -\left(m-1\right) \right) \left( x^n + x^{n-1} + \ldots + x^2 + x -\left(n-1\right) \right)$ . Now $g ^\prime (x)=(x-1)h^\prime(x) + h(x)$ and so $g ^\prime (1)=h(1)=1$ as we know that $h^\prime(1)$ exists.

Now,

$\begin{aligned} & \lim_{x\to 1} \left( \frac{1}{x^{m+1}-mx+(m-1)} - \frac{1}{x^{n+1}-nx+(n-1)}\right) \\ &=\lim_{x\to 1} \frac{\left( x^n + x^{n-1} + \ldots + x^2 + x -\left(n-1\right) \right) -\left( x^m + x^{m-1} + \ldots + x^2 + x -\left(m-1\right) \right) }{\left( x-1 \right) \left( x^m + x^{m-1} + \ldots + x^2 + x -\left(m-1\right) \right) \left( x^n + x^{n-1} + \ldots + x^2 + x -\left(n-1\right) \right) } \\ &=\lim_{x\to 1} \frac{f(x)}{g(x)}\\ &=\lim_{x\to 1} \frac{f^\prime(x)}{g^\prime(x)}\\ &= \frac{f^\prime(1)}{g^\prime(1)}\\ &=\frac{n(n+1)}{2}-\frac{m(m+1)}{2} \end{aligned}$

This means that $\frac{n(n+1)}{2}-\frac{m(m+1)}{2} = 521$ . As 521 is a prime number, there are only two pairs of positive integer solutions $(m,n)$ to this Diophantine Equation: $(n-m)(n+m+1)=2\times 521 = 1\times 1042$

Namely: $n-m=2, n+m+1=521$ or $n-m=1, n+m+1=1042$ . Solve the equations and we obtain $(m,n)=(259, 261)$ or $(520, 521)$ . Hence the minimum value of $n$ is $\boxed{261}$ .

Let $f_L (x) = x^{L+1} - Lx + (L-1)$ . Then, $f_L ' (x) = (L+1)x^L - L \Rightarrow f_L ''(x) = L(L+1)x^{L-1}$ .

And $f_L (1) = 0 , f_L '(1) = 1 , f_L ''(1) = L(L+1)$ .

Given, $\begin{array} { r l l } \displaystyle 521 =& \displaystyle \lim_{x\to 1} \left( \frac{1}{x^{m+1}-mx+(m-1)} - \frac{1}{x^{n+1}-nx+(n-1)}\right) \\ \displaystyle =& \displaystyle \lim_{x\to 1} \left( \dfrac1{f_m(x)} - \dfrac1{f_n(x)} \right) \\ \displaystyle =& \displaystyle \lim_{x\to 1} \left( \dfrac{f_n(x) - f_m (x) }{f_m(x) \cdot f_n (x) } \right) & \qquad \text{Apply L'hopital rule because }0/0 \\ \displaystyle =& \displaystyle \lim_{x\to 1} \left( \dfrac{f_n'(x) - f_m '(x) }{f_m'(x) \cdot f_n (x) + f_m(x) \cdot f_n' (x) } \right) & \qquad \text{Apply L'hopital rule again because }0/0 \\ \displaystyle =& \displaystyle \lim_{x\to 1} \left( \dfrac{f_n''(x) - f_m ''(x) }{f_m''(x) \cdot f_n (x) + f_m'(x) \cdot f_n' (x) + f_m'(x) \cdot f_n' (x) + f_m(x) \cdot f_n'' (x) } \right) \\ =& \dfrac{ n(n+1) - m(m+1) }{0 + 2\times1^2 + 0} \\ 521 \times8 =& 4n(n+1) - 4m(m+1) = (2n + 1)^2 - (2m+1)^2 = (2n - 2m)(2n + 2m + 2) \\ 521 \times2 =& (n-m)(n + m + 1) \\ \end{array}$

Then, $(n-m, n+m+ 1) = (2,521) , (1,521\times1) \Rightarrow (n,m) =(521,520), (261,259)$ . The answer is $\boxed{261}$ .