How to show that... $$x^4-10x^2+1$$ is reducible modulo $p$ for every prime $p$. I can show it, for say $2$,$3$....but should one proceed if he has to show for all primes.
Just don't seem to get it. Can somebody help me?
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2A solution can be found here: http://math.stackexchange.com/questions/160847/polynomials-irreducible-over-mathbbq-but-reducible-over-mathbbf-p-for – lulu Mar 01 '16 at 19:13
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Your polynomial is equal to each of $$ (x^2-1)^2 - 8x^2 = (x^2 - 1)^2 - 2\cdot 4x^2\\ (x^2 +1)^2 - 12x^2 = (x^2 + 1) - 3\cdot 4x^2\\ (x^2 - 5)^2-24 = (x^2 - 5)^2 - 6\cdot 4 $$ And they factor if one of $2$, $3$ or $6$ is a quadratic residue modulo $p$ (because in that case one of them is the difference of two squares).
$2$ is a quadratic residue when $p \equiv 1$ or $7 \mod8$ and $3$ is a residue when $p \equiv 1$ or $11\mod{12}$. $6$ is a quadratic residue if $p \equiv 1, 5, 19$ or $23\mod{24}$
Any prime $p>3$ is congruent to one of $1, 5, 7, 11, 13, 17, 19$ or $23$ modulo $24$, and we see that we've covered all of them with the congruence classes in the previous paragraph. So, for any prime $p > 3$, at least one of the rewritings above may be used to factor the polynomial. Therefore, once you've checked $p = 2$ and $p = 3$ separately, you're done.
Arthur
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Our polynomial can be represented as $$ (x^2+1)^2-3\cdot(2x)^2, $$ $$ (x^2-1)^2-2\cdot(2x)^2, $$ $$ (x^2-5)^2-6\cdot(2)^2, $$ but for every prime $p>3$, at least one element among $2,3,6$ must be a quadratic residue, since the Legendre symbol is multiplicative: assuming $\left(\frac{2}{p}\right)=\left(\frac{3}{p}\right)=-1$, it follows that $\left(\frac{6}{p}\right)=1$.
That gives that for every prime $p>3$ the original polynomial is the difference of two squares, hence it cannot be irreducible.
Jack D'Aurizio
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