Find the smallest odd prime $p$, such that there exist coprime positive integers $k$ and $\ell$ which satisfy \[4k-3\ell=12\quad \text{ and }\quad \ell^2+\ell k +k^2\equiv 3\text{ }(\text{mod }p)\]
Problem
Source: 2022 Bulgarian Spring Math Competition, Problem 10.4
Tags: algebra, system of equations, prime
Slave
27.03.2022 12:25
//wrong//
VicKmath7
27.03.2022 12:54
Proceed as below to get $37x^2-44x+13 \equiv 0 (mod p)$ and multiply it by $37$ and transform it into the form $m^2 \equiv 3 (mod p)$ then use quadratic residues and Quadratic Reciprocity. Answer is $p=11$; construction can be done with CRT as well, if you're lazy to find concrete example.
Assassino9931
27.03.2022 13:17
No need to use overcomplicated ideas. (Hopefully I have not made a mistake in these rushed calculations.) Clearly $k=3x$ and then $\ell = 4x-4$, so the congruence becomes $37x^2 - 44x + 13 \equiv 0 \pmod p$. For $p=3$ we get $(x-1)^2 \equiv 0 \pmod 3$ and $k$ and $\ell$ are both divisible by $3$. For $p=5$ we get $2x^2 - 4x + 3 \equiv 0 \pmod 5$, impossible by direct check. For $p=7$ we get $2x^2 - 2x + 6 \equiv 0 \pmod 7$, again impossible by direct check. For $p=11$ we get $4x^2 + 2 \equiv 0 \pmod {11}$, with $x\equiv 4 \pmod {11}$ as a solution. To ensure that $k$ and $\ell$ are coprime just pick $x=59$ (hmm even $x=15$ works actually).