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I am struggling with this proof. I really cannot think of where to start:

Let $p$ be an odd prime, prove that:

$$\left(\frac{1 \cdot 2}{p}\right) + \left(\frac{2 \cdot 3}{p}\right) + \left(\frac{3 \cdot 4}{p}\right) + \cdots + \left(\frac{(p - 2) (p - 1)}{p}\right) = -1$$

If someone could just give me a push in the right direction, I am sure I could get it.

Derik
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  • @Derik Hint: for any invertible $n$, $n(n+1)$ has the same quadratic character as $(n+1)/n = 1 + n^{-1}$. What set of values does this range over for $n=1, \ldots, p-1$? – Erick Wong May 16 '16 at 18:04
  • @ErickWong For example, $$(\frac{2 \cdot 3}{p}) = (\frac{\frac{2 + 1}{2}}{p})$$ Is this what you are saying? – Derik May 17 '16 at 16:43
  • @Derik Yes, in the sense that you would interpret $\frac32$ as $3$ times the modular inverse of $2$. That is what I meant by "same quadratic character", that the Legendre symbols mod $p$ will be the same. – Erick Wong May 17 '16 at 16:45
  • @ErickWong Hmm, how do we know that $$(\frac{2 \cdot 3}{p}) = (\frac{1 + 2^{-1}}{p})$$? – Derik May 17 '16 at 17:06
  • @Derik (Assuming you mean $3/2$ instead of $2/3$). Just write it out carefully using $3 = 2+1$. The distributive law holds modulo $p$ (indeed in any ring). Unlike your earlier question on Euler's criterion, this problem is relatively challenging and requires some mastery of fundamentals to actually navigate the proof. – Erick Wong May 17 '16 at 17:08
  • @ErickWong Well maybe I need a little more than a push. I am still confused with how we know that $$(\frac{n(n + 1)}{p}) = (\frac{1 + n^{-1}}{p})$$ – Derik May 18 '16 at 02:54
  • @Derik, Oh, just multiply by $(n^{-1})^2$, which is always a square. – Erick Wong May 18 '16 at 05:21

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