Maths Olympiad Prep

Library / /284 of 520

Number theory Difficulty 6.1 National olympiad Prove it

Lemma 11 Let kk be an integer ⩾3\geqslant 3 and ll be an integer >k>k, aka_{k} be a positive integer, and pp be a prime, then when (p,akk)=1\left(p, a_{k} k\right)=1 we have
∑x=0pl−1e2πiakxkpl=pk−1∑y=0pl−k−1e2πiakykpl−k.\sum_{x=0}^{p^{l}-1} e^{2 \pi i \frac{a_{k} x^{k}}{p^{l}}}=p^{k-1} \sum_{y=0}^{p^{l-k}-1} e^{2 \pi i \frac{a_{k} y^{k}}{p^{l-k}}} .

Solution

The multiples of pp in 0,1,2,⋯ ,pl−1−10,1,2, \cdots, p^{l-1}-1 are
0,p,2p,⋯ ,(pi−2−1)p0, p, 2 p, \cdots,\left(p^{i-2}-1\right) p

Since l>k⩾3l>k \geqslant 3, by (66), (67), and (70), we have
Thus, the lemma is proved.
∑x=0pl−1e2πiakxkpl=p∑y=0pi−2−1e2πiak(py)kpl=p∑y=0pi−2−1e2πiakykpl−k=p∑m=0pk−2−1∑m=0pl−k−1e2xiak(mpl−k+n)kpl−k=p∑m=0pk−2−1∑n=0pl−k−1e2πiaknkpl−k\begin{array}{l} \sum_{x=0}^{p^{l}-1} e^{2 \pi i \frac{a_{k} x^{k}}{p^{l}}}=p \sum_{y=0}^{p^{i-2}-1} e^{2 \pi i \frac{a_{k}(p y)^{k}}{p^{l}}}=p \sum_{y=0}^{p^{i-2}-1} e^{2 \pi i \frac{a_{k} y^{k}}{p^{l-k}}} \\ =p \sum_{m=0}^{p^{k-2}-1} \sum_{m=0}^{p^{l-k}-1} e^{2 x i \frac{a_{k}\left(m p^{l-k}+n\right)^{k}}{p^{l-k}}}=p \sum_{m=0}^{p^{k-2}-1} \sum_{n=0}^{p^{l-k}-1} e^{2 \pi i \frac{a_{k} n^{k}}{p^{l-k}}} \end{array}

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.