Evaluate $\sum_x\big(\frac{x}{p}\big), \sum_x\big(\frac{x+k}{p}\big), \sum_x\big(\frac{ax+b}{p}\big)$ where $p$ is prime and the rests are integers.

Question

Evaluate the following sums:

$$\sum^{p}_{x=1}\left(\frac{x}{p}\right), \sum^{p}_{x=1}\left(\frac{x+k}{p}\right), \sum^{p}_{x=1}\left(\frac{ax+b}{p}\right)$$

Let $p$ be a prime and $k, a,$ and $b$ be arbitrary integers.

If someone could explain this to me and perhaps share a resource where I could learn how to solve these types of questions better, I would be indebted.

Answer 1

$$\sum^{p}_{x=1}(\frac{x}{p})=\frac{1}{p}\sum^{p}_{x=1}x=\frac{1}{p}\frac{p(p-1)}{2}=\frac{p-1}{2}$$

$$\sum^{p}_{x=1}(\frac{x+k}{p})=\frac{1}{p}\sum^{p}_{x=1}x+k=\frac{1}{p}\big (\frac{p(p-1)}{2}+pk \big)=\frac{p-1}{2}+k $$

The last summation follows in the exact same way... you split it up.

$$\sum^{p}_{x=1}(\frac{ax+b}{p})$$

Answer 2

Observe the first sum is equals to sum of p-first natural numbers times the inverse of p, use the formule of sum arithimetic progression, the others are analogous.