Exercise (7.4).11

Let $p= 13$ and $a = 16$. Show that

$$\sum_{k=1}^{(p-1)/2} \lfloor \frac{a \times k}{p} \rfloor = 23 $$

but $g= 2$ where $g$ is as defined in Gauss’s Lemma. Explain why Lemma (7.20) fails in this case.

The following table shows the calculations which confirm the sum is 23.

k	floor (16k/13)
1	1
2	2
3	3
4	4
5	6
6	7
	23

The Lemma is not applicable because it requires $a$ to be odd, but here $a=16$ is even.