Let $p \neq 13$ be a prime number of the form $8k+5$ such that $39$ is a quadratic non-residue modulo $p$. Prove that the equation $$x_1^4+x_2^4+x_3^4+x_4^4 \equiv 0 \pmod p$$has a solution in integers such that $p\nmid x_1x_2x_3x_4$.
Problem
Source: TST 3,day 1,P 3
Tags: number theory, prime numbers, modular arithmetic, Quadratic Residues
sqing
16.07.2016 11:31
MRF2017 wrote: $p \neq 13$ is a prime number in form $8k+5$ and we know that $39$ is nonresidue modulo $p$.Prove that the equation $x_1^4+x_2^4+x_3^4+x_4^4 \overset{p}{\equiv} 0$ has a solution in integers such that $p\nmid x_1x_2x_3x_4$.
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thuanz123
10.01.2017 15:20
for63434 wrote: I'm sorry... But, WHY? It's just expanding
mathcool2009
29.08.2018 08:55
Obviously $p = 5$ or $p \ge 29$. If $p = 5$, notice that $39$ is a quadratic residue modulo $p$, contradiction. Hence $p \ge 29$.
Let $g$ be a primitive root modulo $p$ and let $\chi : (\mathbb{Z}/p\mathbb{Z})^* \rightarrow \{ 0,1,2,3\}$ be the unique function such that $\chi(g^x) \equiv x \pmod{4}$ for all $x$.
Notice that $\chi (-1) = 2$ and $\chi (2) \in \{1,3\}$. Furthermore, notice that if $a$ is a quadratic non-residue, we have $\chi (a) \in \{1,3\}$, so $\chi (-a^2) = 0$; i.e. $-a^2$ is a fourth power. In particular, $-2^2$ is a fourth power.
We know that $39 = 3 \cdot 13$ is a quadratic non-residue, so we have the following cases:
Case I. $3$ is a quadratic residue and $13$ is a quadratic non-residue.
In this case we have \[7^4 - 39^2 + (3^2)(2^4) + (-2^2)(4^4) = 0.\]Each term of this equation is a nonzero fourth power modulo $p$.
Case II. $13$ is a quadratic residue such that $\chi (13) = 2$ and $3$ is a quadratic non-residue.
In this case we have \[(-2^2)(2^4) + 3^4 + (-2^2) + (-13) = 0.\]
Case III. $13$ is a quadratic residue such that $\chi (13) = 0$, and $3$ is a quadratic non-residue, and $\chi (15) = 0$.
In this case we have \[(-2^2)(2^4) + (-2^2)(-3^2) + 13 + 15 = 0.\]
Case IV. $13$ is a quadratic residue such that $\chi (13) = 0$, and $3$ is a quadratic non-residue, and $\chi (15) = 2$.
In this case we have \[(-3^2)(3^4) + 5^4 + (-2^2)(13)(13) + (-2^2)(13)(-15) = 0.\]
Case V. $13$ is a quadratic residue such that $\chi (13) = 0$, and $3$ is a quadratic non-residue, and $\chi (30) = 0$.
In this case we have \[(-3^2)(30) + 4^4 + 13 + 1 = 0.\]
Case VI. $13$ is a quadratic residue such that $\chi (13) = 0$, and $3$ is a quadratic non-residue, and $\chi (30) = 2$.
In this case we have \[(-2^2)(2^4) + 3^4 + 13 + (-30) = 0.\]$\blacksquare$
ihategeo_1969
04.01.2025 16:27
Okay I am going to overkill this by proving as long $p \geq 11$, we are fine. Let $T_k$ be the number of solutions to \[x_1^4+\dots+x_k^4=0\]in $\mathbb{F}_p$ where some $x_j$ may or may not be $0$. By PIE shenanigans we want to prove \[T_4-4T_3+6T_2-4T_1+T_0>0\]Figuring out $T_4$ Let $\chi_1$, $\dots$, $\chi_4$ be any Dirichlet characters for now. So see that \[T_4=\sum_{\chi_i^4=\varepsilon} J_0(\chi_1,\chi_2,\chi_3,\chi_4)\]where $J_0$ is the ``alternative" Jacobi Sum of those characters. See that the only relevant characters are $\varepsilon$, $L$, $\chi$ (which has orders $1$, $2$, $4$ respectively). And so by Orthogonality relations we have\begin{align*} T_4&=J_0(\varepsilon,\varepsilon,\varepsilon,\varepsilon)+J_0(L,L,L,L)+4J_0(\chi,L,L,L)+ 6J_0(\chi,\chi,L,L)+4J_0(\chi,\chi,\chi,L)+J_0(\chi,\chi,\chi,\chi) \end{align*}Noting that the only times the product of these characters is $\varepsilon$ is when all $4$ are either $L$ or $\chi$, we get that due to Orthogonality relations, the table in Page $7$ of this handout combined with Triangle inequality we get that \[T_4 \geq p^3-2(p-1)p\]Figuring out $T_3$ Doing the same thing as before we get that \begin{align*} T_3= J_0(\varepsilon,\varepsilon,\varepsilon)+J_0(L,L,L)+3J_0(\chi,L,L)+ 3J_0(\chi,\chi,L)+J_0(\chi,\chi,\chi) \end{align*}And again noting that the only time the product of their Dirichlet characters is $\varepsilon$ is when it is $(L,\chi,\chi)$ we get that due to triangle inequality and that same table that \[T_3 \leq p^2+3(p-1)\sqrt{p}\]Finishing And hence we get that \[T_4-4T_3+6T_2-4T_1+T_0 \geq p^3-2(p-1)p-4(p^2+3(p-1)\sqrt{p})-3>0 \iff p \geq 11\]since $p$ prime and we done.