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I am trying to prove that the diophantine equation $$1998^2x^2+1997x+1995-1998x^{1998}=1998y^4+1993y^3-1991y^{1998}-2001y$$ has no solution in integers (given that $1997$ is a prime).

To do so, I tried to make use of the cyclicity of $\mathbb{F}_{1997}^*=\langle g \rangle$. Hence the equation become $g^{2n}+1995-g^{1998n}=g^{4m}-4g^{3m}+6g^{1998m}-4g^m$. But I don't know how to continue.

Alternatively, I want to use the fact that in $\mathbb{F}_{1997}^*$, every number $a$ is a cube. Hence the equation become $a^6+1995-a^{5994}=b^{12}-4b^9+6b^{5994}-4b^3$, but again I have no idea how to continue, could someone gives me some hints, thanks in advance.

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1 Answer 1

up vote 3 down vote accepted

$t^{1998}\equiv t^{1997}\cdot t\equiv t^2\pmod {1997} \\\implies x^2-2-x^2\equiv y^4-4y^3+6y^2-4y\pmod {1997} \\\implies (y-1)^4\equiv-1\pmod {1997}.$

But $1997\equiv 3\pmod 4\implies z^2\equiv -1\pmod {1997}$ has no integer solutions, a contradiction.

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