$p=3k+1$ is a prime number. For each $m \in \mathbb Z_p$, define function $L$ as follow: $L(m) = \sum_{x \in \mathbb{Z}_p}^{ } \left ( \frac{x(x^3 + m)}{p} \right )$ a) For every $m \in \mathbb Z_p$ and $t \in {\mathbb Z_p}^{*}$ prove that $L(m) = L(mt^3)$. (5 points) b) Prove that there is a partition of ${\mathbb Z_p}^{*} = A \cup B \cup C$ such that $|A| = |B| = |C| = \frac{p-1}{3}$ and $L$ on each set is constant. Equivalently there are $a,b,c$ for which $L(x) = \left\{\begin{matrix} a & & &x \in A \\ b& & &x \in B \\ c& & & x \in C \end{matrix}\right.$ . (7 points) c) Prove that $a+b+c = -3$. (4 points) d) Prove that $a^2 + b^2 + c^2 = 6p+3$. (12 points) e) Let $X= \frac{2a+b+3}{3},Y= \frac{b-a}{3}$, show that $X,Y \in \mathbb Z$ and also show that :$p= X^2 + XY +Y^2$. (2 points) (${\mathbb Z_p}^{*} = \mathbb Z_p \setminus \{0\}$)
Problem
Source: Iran 3rd round 2013 - Number Theory Exam - Problem 5
Tags: function, linear algebra, matrix, number theory proposed, number theory, Quadratic Residues
17.09.2013 05:51
Jacobstal theorem
17.09.2013 16:50
qua96 wrote: Jacobstal theorem More details please?
18.09.2013 06:10
a) $\left ( \frac{t^4}{p} \right )=1$ so we have: $ L(m) =\sum_{x\in\mathbb{Z}_p}^{ }\left (\frac{x(x^3+m)}{p}\right ) =\sum_{x\in\mathbb{Z}_p}^{ }\left (\frac{x(x^3+m)}{p}\right )\left ( \frac{t^4}{p} \right ) = \sum_{x\in\mathbb{Z}_p}^{ }\left (\frac{xt((xt)^3+mt^3)}{p}\right )$ then $p=3k+1$ so $L(mt^3)= \sum_{x\in\mathbb{Z}_p}^{ }\left (\frac{xt((xt)^3+mt^3)}{p}\right )$
29.09.2014 18:50
Any ideas for parts b) , c) , d) and e) ??
12.02.2017 23:54
I think part (b) can use results from part (a) because when $p=3k+1$, the equation $x^3\equiv 1(mod p)$ has 3 answer. So, set $X=\{d \mid \exists x\in \mathbb{Z} , x^3\equiv d (modp)\}$ has exactly $\frac{p-1}{3}$ elements. We can choose $e,f \in \mathbb{Z}$ such that $Y=\{ed \mid \exists x\in \mathbb{Z} , x^3\equiv d (modp)\} \cap X = {\o}$, $Y=\{fd \mid \exists x\in \mathbb{Z} , x^3\equiv d (modp)\} \cap X = {\o}$ and $Y \cap Z= {\o}$ And from part (a), we can partition $\mathbb{Z}/p\mathbb{Z}=X \cup Y \cup Z$
15.01.2018 21:11
15.01.2018 21:14