Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

For any prime $p>3$ show that $$p\sum_{j=0}^{p-1}\dfrac{(-3)^j}{2j+1}\equiv \left(\dfrac{p}{3}\right)\pmod{p^2}$$ where $\left(\dfrac{p}{3}\right)$ denotes the Legendre symbol.

This is proof: we have $$(1\pm i\sqrt{3})^p=2^pe^{\pm i\pi p/3}=2^p(\cos{(\pi p/3)}\pm i\sin{(\pi p/3)}=2^{p-1}\left(1\pm i\left(\dfrac{p}{3}\right)\sqrt{3}\right)$$ On the other hand we have \begin{align*} (1\pm i\sqrt{3})^p=\sum_{k=0}^{p}\binom{p}{k}(\pm i\sqrt{3})^k&\equiv 1\pm (i\sqrt{3})^p-p\sum_{k=1}^{p-1}\dfrac{(\mp i\sqrt{3})^k}{k}\\ &\equiv 1\pm i\sqrt{3}(-3)^{(p-1)/2}-S_{0}\pm i\sqrt{3}S_{1}\pmod{p^2} \end{align*} where $$S_{0}=p\sum_{j=1}^{\frac{p-1}{2}}\dfrac{(-3)^j}{2j},S_{1}=p\sum_{j=0}^{\frac{p-3}{2}}\dfrac{(-3)^j}{2j+1}$$ then $$S_{0}\equiv 1-2^{p-1}\pmod{p^2}, S_{1}\equiv 2^{p-1}\left(\dfrac{p}{3}\right)-(-3)^{(p-1)/2}\pmod{p^2}$$ so \begin{align*} p\sum_{j=0}^{p-1}\dfrac{(-3)^j}{2j+1}&=p\sum_{j=0}^{(p-3)/2}\dfrac{(-3)^j}{2j+1}+(-3)^{(p-1)/2}+(-3)^{(p-1)/2}p\sum_{j=1}^{(p-1)/2}\dfrac{(-3)^j}{p+2j}\\ &\equiv S_{1}+(-3)^{(p-1)/2}+(-3)^{(p-1)/2}S_{0}\\ &\equiv (2^{p-1}-1)\left(\left(\dfrac{p}{3}\right)-(-3)^{(p-1)/2}\right)+\left(\dfrac{p}{3}\right)\\ &\equiv \left(\dfrac{p}{3}\right)\pmod{p^2} \end{align*} where use $$p|(2^{p-1}-1),2|\left(\left(\dfrac{p}{3}\right)-(-3)^{(p-1)/2}\right)$$


My question:

(1):why $$\sum_{k=0}^{p}\binom{p}{k}(\pm i\sqrt{3})^k\equiv 1\pm (i\sqrt{3})^p-p\sum_{k=1}^{p-1}\dfrac{(\mp i\sqrt{3})^k}{k}\pmod{p^2}?$$ (2): why $$S_{0}\equiv 1-2^{p-1}\pmod{p^2}, S_{1}\equiv 2^{p-1}\left(\dfrac{p}{3}\right)-(-3)^{(p-1)/2}\pmod{p^2}?$$

Thank you someone can solve my two problem,Thank you very much,and this $$p\sum_{j=0}^{p-1}\dfrac{(-3)^j}{2j+1}\equiv \left(\dfrac{p}{3}\right)\pmod{p^2}$$ have other methods? Thank you

share|cite|improve this question

Now,I have konw why $$S_{0}\equiv 1-2^{p-1}\pmod {p^2},S_{1}\equiv 2^{p-1}\left(\frac{p}{3}\right)-(-3)^{(p-1)/2}\pmod {p^2}$$

Because use $$(1\pm i\sqrt{3})^p=2^{p-1}(1\pm i\left(\frac{p}{3}\right)\sqrt{3})$$ and other hand we have $$(1\pm i\sqrt{3})^p\equiv 1\pm i\sqrt{3}(-3)^{(p-1)/2}-S_{0}\pm i\sqrt{3}S_{1}\pmod{p^2}$$ so we have done

share|cite|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.