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Find all incongruent roots modulo $13$ of $x^4+x^2+x+1$

So by Lagrange theorem, there should be $4$, because the highest degree is $4$ and the coefficient in front of it is $1$, which divides $13$.

The solution says that $7$ is the only solution but doesn't say why or how they found it

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Edit Below was the theorem cited in earlier drafts (discussed in comments)

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Bill Dubuque
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user8714896
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    according to my brute force calculations, $\color{red}6$ is the only zero of $x^4+x^2+x+1$ in ${0,1,2,3,4,5,6,7,8,9,10,11,12}$ modulo $13$ – J. W. Tanner Dec 04 '19 at 03:53
  • yea I dunno book says 7, but is that how you can check? Just brute force all relative primes less than x? – user8714896 Dec 04 '19 at 04:04
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    Difficult to answer, for the source of your confusion is in the dark. I am not aware of any result telling that $1\mid 13$ should imply that this polynomial has 4 zeros in $\Bbb{Z}{13}$. For the better, I suppose, for this polynomial has only one such zero. J.W.Tanner got it right, modulo $13$ the polynomial factors as $$x^4+x^2+x+1=(x+7)(x^3+6x^2+11x+2)$$ with the cubic factor irreducible. Which is equivalent to $x=-7=6$ being the only zero in $\Bbb{Z}{13}$. As a quartic the polynomial has four zeros, but the other three are in an extension field $\Bbb{F}_{13^3}$. – Jyrki Lahtonen Dec 04 '19 at 04:07
  • The last phenomenon being somewhat analogous to two of the zeros of $(x-1)(x^2+1)\in\Bbb{R}[x]$ being complex numbers as opposed to real. – Jyrki Lahtonen Dec 04 '19 at 04:09
  • I'll add the theorem to the question – user8714896 Dec 04 '19 at 04:09
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    The theorem you listed requires that $4\mid 13-1=12$, so $1\mid 13$ is rather irrelevant. Also that theorem is only about the specific polynomial $x^4-1$. It does not apply to all polynomials of degree four. – Jyrki Lahtonen Dec 04 '19 at 04:11
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    I agree with @JyrkiLahtonen: Theorem 9.7 refers to polynomial $x^d-1$, not $x^4+x^2+x+1$ – J. W. Tanner Dec 04 '19 at 04:11
  • Sorry I posted the wrong one the right one is up now – user8714896 Dec 04 '19 at 04:12
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    Theorem 9.6 says there are at most $n$ roots; there could be fewer – J. W. Tanner Dec 04 '19 at 04:13
  • Right, but still doesn't really explain how they found the solution besides brute force. – user8714896 Dec 04 '19 at 04:13
  • By changing the theorem(image) you make it much harder to understand the above discussion based on the prior theorem. Please don't do that. Better to update with the new one and mark it as an edit. – Bill Dubuque Dec 04 '19 at 04:18
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    Actually finding those zeros is more taxing. For a small prime such as $p=13$ brute forcing like J.W.Tanner did it may be the fastest way. There are more efficient methods for finding them (for larger primes), see this old thread, but they require more advanced algorithms. I wouldn't bother with those at this point in your studies. – Jyrki Lahtonen Dec 04 '19 at 04:19

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By Fermat $\,x^{12}\!-\!1\,$ has all elements $\not\equiv 0$ as roots so taking its gcd with $\,f(x)\,$ will yield all its roots. The Euclidean algorithm has the monic remainder sequence $\, x^3\!-\!x^2\!-\!6x\!-\!1,\,$ $\,x^2\!+\!x\!-\!3,\,$ $\,x\!+\!7,\,$ $0,\,$ so the gcd $= x\!+\!7,\,$ thus $\,f(x)\,$ has unique root $\,x \equiv -7\equiv 6$.

Remark $ $ We can also use gcds this way to find higher degree factors and do irreducibility tests, e.g. see here and its links.

Note that Lagrange's theorem yields only an upper bound on the number of roots of a polynomial, e.g. $\bmod 13\!:\ f(x) = x^{13}-x+1\,$ has no roots since $\,f(x) \equiv 1\,$ for all $x,\,$ by little Fermat.

Bill Dubuque
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