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I know that for a quadratic equation we can look at if $ b^2-4ac $ is a quadratic residue and conclude that there are two solutions but I am not sure how to do this with a quartic equation. Is there a way I could apply the method above or is there another way?

GNUSupporter 8964民主女神 地下教會
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chloe loughan
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    The approach I might take would be to look for solutions modulo $2,4,8,16 \dots$ and $5, 25, 125 \dots$ you can use Hensel's lemma to lift solutions to higher powers of the same prime and the chinese remainder theorem to combine the outcomes from the two primes considered separately. – Mark Bennet Apr 29 '18 at 21:45
  • maybe it's helpful: https://math.stackexchange.com/questions/2435913/solutions-of-x2-6x-13-equiv-0-pmod127?rq=1 – Darío A. Gutiérrez Apr 29 '18 at 22:52
  • This graph has only one minimum so it only has 2 solutions per cycle except none at zero, – Phil H Apr 30 '18 at 01:33
  • The proof of my previous comment is that the second derivative is always increasing $(y = 24x^2 + 6x)$ – Phil H Apr 30 '18 at 01:53

1 Answers1

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It is easy to check by hand that $1$ is the unique solution modulo $2$ and modulo $5$. You could then use the nice method proposed by Mark Bennet to solve the problem, but is not so easy to find the solutions predicted by Hensel's lemma.

It is actually a bit depressing to find out that a brute force approach using the Sage software instantaneously gives the unique solution, x = 717071 .

for x in [0..99999]:
  x = 10*x + 1;
  if (((2*x^4 + x^3 - 4*x + 11) % 1000000) == 0):
    print "x = " + str(x)

x = 717071
J.-E. Pin
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