Equation over a finite field

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Equation over a finite field

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#1 h Apr 22, 2025, 6:08 PM • 1 Y

Y by PikaPika999

Find the set of $x\in\mathbb{F}_{5^5}$ such that the equation in the unknown $y\in \mathbb{F}_{5^5}$ :
$x^3y+y^3+x=0$ admits $3$ roots: $a,a,b$ s.t. $a\not=b$ .

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#2 h Apr 22, 2025, 9:30 PM • 2 Y

Y by PikaPika999, loup blanc

Write $K=F_{5^5}$ and we want $f(y)=y^3+x^3y+x$ to have a multiple root $y\in K$ of exact order 2. Hence $x\not=0$ and $y$ is also a root of $f'(y)=3y^2+x^3$ . Then $x^3=2y^2\implies x+3y^3=0\implies x=2y^3$ .
Then $x^9=8y^6=3y^6=7(2y^3)^2=7x^2\implies x^7=2$ .
For $x\in Z$ we have $x^7\equiv 2(5)\iff x^3\equiv 2(5)\iff 1/x\equiv 2(5)\iff x\equiv 3(5)$ hence $3^7=2$ in $K$ . To check if there are other solutions we have to consider $t^7=1$ . It is easy to check that $5^5\not\equiv 1(7)$ since $5^5\equiv 1\iff 5^{-1}\equiv 1\iff 5\equiv 1(7)$ and that is not true. Hence $(7,5^5-1)=1$ and $t=1$ is the only solution to $t^7=1$ in $K$ . Hence $x=3$ is the unique solution in $K$ of $x^7=2$ .
Finally we have to check that $x=3$ works but that is easy since $y=-1=4$ is the double root.
Indeed we have $y^3+2y+3=(y^2+2y+1)(y+3)$ .

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#3 h Apr 23, 2025, 6:16 PM

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Note that in $\mathbb{F}_{5^6}$ , the problem is more complicated. Indeed, $5^6=1 \mod 7$ and
$t^7-1=(t-1)(t^6+\cdots+1)$ (decomposition in irreducibles); then the second factor has roots in $\mathbb{F}_{5^6}$ .
Note that $y^3+x^3y+x$ is the value of the Klein quartic for $z=1$ , cf.
https://artofproblemsolving.com/community/c7t435f7h3530741_how_to_solve_this_problem

This post has been edited 1 time. Last edited by loup blanc, Apr 29, 2025, 6:09 AM

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