Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.