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.

  • 1
    $\begingroup$ I edited this to put the expression $$\left(\frac{1 \cdot 2}{p}\right) + \left(\frac{2 \cdot 3}{p}\right) + \left(\frac{3 \cdot 4}{p}\right) + \cdots + \left(\frac{(p - 2) (p - 1)}{p}\right) = -1$$ in a single MathJax display. One should not keep alternating in and out of MathJax in such an expression; it should be between a single pair of dollar signs or double dollar signs. Also, one can write $2\cdot3$ or $2\times3$. Use of an asterisk for that is a workaround for situations when one is limited to the characters on the keyboard. $\qquad$ $\endgroup$ – Michael Hardy May 16 '16 at 17:54
  • $\begingroup$ Thank you for fixing my question. $\endgroup$ – Derik May 16 '16 at 17:55
  • $\begingroup$ @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 $n=1, \ldots, p-1$? $\endgroup$ – Erick Wong May 16 '16 at 18:04
  • $\begingroup$ @ErickWong For example, $$(\frac{2 \cdot 3}{p}) = (\frac{\frac{2 + 1}{2}}{p})$$ Is this what you are saying? $\endgroup$ – Derik May 17 '16 at 16:43
  • $\begingroup$ @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. $\endgroup$ – Erick Wong May 17 '16 at 16:45

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