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Prove that the prime divisors of $5a^4-5a^2+1$, for any integer $a > 0$, are congruent $\pm 1 \pmod{20}$.

We have $f(a) = 5a^4-5a^2+1 = 5a^2(a^2-1)+1 \equiv 0 \pmod{p}$. Then since one of $a^2$ and $a^2-1$ is divisible by $4$ we know that $f(a) \equiv 1 \pmod{20}$. Thus the prime divisors of $a$ are of the form $1,3,7,9,11,13,17,19$ modulo $20$. How do we continue from here?

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    $\begingroup$ Multiply with $20$ and complete the square. $\endgroup$ – Daniel Fischer Jan 9 '17 at 22:34
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    $\begingroup$ @DanielFischer Can you explain? We get $100(a^2-1/2)^2-5$. $\endgroup$ – Puzzled417 Jan 9 '17 at 22:53
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    $\begingroup$ $(10a^2-5)^2 - 5 \equiv 0 \pmod{p}$. That gives you $p \equiv \pm 1 \pmod{5}$. Haven't yet had a good idea to eliminate $p \equiv 9,11\pmod{20}$. $\endgroup$ – Daniel Fischer Jan 9 '17 at 22:55
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    $\begingroup$ Puzz, where did you get the problem? $\endgroup$ – Will Jagy Jan 10 '17 at 2:05
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    $\begingroup$ The minimal polynomial of $2\cos(\pi/10)$ is the reciprocal, $x^4-5x^2+5$. It follows that the splitting field of $5x^4-5x^2+1$ is the real subfield of the twentieth cyclotomic field (hence cyclic of degree four). Therefore the claim follows from basic results on splitting of primes in cyclotomic fields (the Frobenius automorphism must be the identity on this intermediate field so $p\equiv\pm1$). But I do think that there is a simpler way :-) $\endgroup$ – Jyrki Lahtonen Jan 10 '17 at 10:12
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The following is an attempt to reverse engineer my comment under the OP to a solution requiring no algebraic number theory. To an extent I replace that with basic facts about finite fields. I think a solution with even more elementary tools is out there.


Assume that $p$ is a prime factor of $f(a)=5a^4-5a^2+1$ for some integer $a$. Clearly $f(a)$ is odd and congruent to $1$ modulo five, so $p\neq2,5$. This implies that any eventual zeros of the twentieth cyclotomic polynomial $$ \Phi_{20}(x)=x^8-x^6+x^4-x^2+1 $$ in some field $K$ of characteristic $p$ have multiplicative order twenty (the zeros of $h(x)=x^{20}-1$ in its splitting field over $K$ are all simple, because $h'(x)=20x^{19}$ is coprime to $h(x)$).

Consider the equation $$ \frac1a=z+\frac1z\Longleftrightarrow z^2-a^{-1}z+1=0\qquad(*) $$ over the field $\Bbb{F}_p$. This is quadratic in $z$, so it has a solution either in the field $\Bbb{F}_p$ itself or in its quadratic extension $\Bbb{F}_{p^2}$. We shall treat these two cases separately, but we first make the observations that $$ (z+\frac1z)^4-5(z+\frac1z)^2+5=\frac{\Phi_{20}(z)}{z^4}, $$ and also that (as a consequence of $(*)$) $$ (z+\frac1z)^4-5(z+\frac1z)^2+5=\frac{1-5a^2+5a^4}{a^4}=\frac{f(a)}{a^4}=0. $$ Therefore we can conclude that $z$ has multiplicative order twenty.

  1. If $z\in\Bbb{F}_p^*$ this immediately implies that $20\mid p-1$. This is because the group $\Bbb{F}_p^*$ is cyclic of order $p-1$. In this case we must have $p\equiv1\pmod{20}$.
  2. If $z$ is an element of the quadratic extension of $\Bbb{F}_p$ instead, then the polynomial $g(x)=x^2-a^{-1}x+1$ is irreducible in $\Bbb{F}_p[x]$. But, from $(*)$ we see that the zeros of $g(x)$ are $z$ and $1/z$. By Galois theory they are also $\Bbb{F}_p$-conjugates, so we can deduce that $1/z=z^p$. Therefore we also have that $z^{p+1}=1$. Consequently $20\mid p+1$, and in this case we have $p\equiv-1\pmod{20}.$
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  • $\begingroup$ I hope it's clear to all that $1/a$ acts in the role of $2\cos(\pi/10)$ in $\Bbb{F}_p$. $\endgroup$ – Jyrki Lahtonen Jan 10 '17 at 13:34
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    $\begingroup$ It seems that (*) would be $\frac{1}{a}=z+\frac{1}{z}\iff z^2-\frac{1}{a}z+1=0$, and that $(z+\frac{1}{z})^4-5(z+\frac{1}{z})^2+1$ should be replaced by $(z+\frac{1}{z})^4-5(z+\frac{1}{z})^2+5$. $\endgroup$ – MathChat Jan 10 '17 at 17:23
  • $\begingroup$ @MathChat Correct. Thanks! Editing. $\endgroup$ – Jyrki Lahtonen Jan 10 '17 at 20:21

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