Let $p$ be a prime, and $b$ an integer such that $b \not \equiv 0,1 \pmod{p}$. Prove that $$\sum_{i=0}^{p-1}(i+1)b^{-i} \equiv b(b-1)^{-1} \pmod{p}.$$

I thought about expanding the left-hand side to get \begin{align*}\sum_{i=0}^{p-1}(i+1)b^{-i} &\equiv 1+2b^{-1}+3b^{-2}+\cdots+pb^{-(p-1)}\\&\equiv 1+2b^{-1}+\cdots+(p-1)b^{-(p-2)} \pmod{p}.\end{align*} How can we continue from here to get the right-hand side?

  • $\begingroup$ I wonder what happens when you instead consider LHS $\times (b-1)$. To me it looks like it might be a little bit simpler. $\endgroup$ – mdave16 Apr 23 '17 at 23:08
  • $\begingroup$ Just to clarify, is $b^{-i} = b^i \cdot a$ where $a$ is the multiplicative inverse $\pmod p$ of $b$? $\endgroup$ – enedil Apr 23 '17 at 23:15
  • $\begingroup$ @enedil Yes, $b^{-1}$ denotes the multiplicative inverse taken modulo $p$. $\endgroup$ – user19405892 Apr 23 '17 at 23:16

$$\sum_{i=0}^{p-1}(i+1)x^i = \sum_{i=0}^{p-1} \sum_{j=0}^i x^i = \sum_{j=0}^{p-1} \sum_{i=j}^{p-1} x^i = \sum_{j=0}^{p-1}\frac{x^j-x^p}{1-x}$$

$$=\frac{\frac{1-x^p}{1-x}-px^p}{1-x} = \frac{1-x^p}{(1-x)^2}-p \frac{x^p}{(1-x)^2} \equiv\frac{1-x^p}{(1-x)^2} \equiv \frac{1-x}{(1-x)^2}=\frac{1}{1-x}$$

Taking $x=b^{-1}$ gives the desired result.

Note that the only $\mod p$-specific features used here are simply that

  1. $p\equiv0$
  2. $x^p\equiv x$.
  • 1
    $\begingroup$ Can be obtained too with $\bigg(\sum\limits_{i=1}^{p}x^i\bigg)'$. $\endgroup$ – zwim Apr 24 '17 at 0:30

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