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Algebra Level 4

How many integers $n > 1$ are there such that $x^n-x+a$ is reducible over the rationals for some positive integer $a$ ?

1 240 8 Infinitely many 60 120 None 2

2 solutions

Christopher Criscitiello
Dec 2, 2017

Hint: Prove that

$(-2)^{2^{n+1}}-1=\prod _{j=0}^n \left(2^{2^j}+1\right).$

Hence, $x^{1+2^k}-x+2*\prod _{j=0}^n \left(2^{2^j}+1\right)$ has the root $x=-2$ , and this polynomial is reducible.

Actually, this is a very easy problem. Let $a = 2^n - 2$ , where $n$ is an odd positive integer. Then $x^n - x + a$ has $x + 2$ as a factor, and is thus reducible.

Jon Haussmann - 3 years, 6 months ago

@Jon Haussmann - totally overlooked that, thanks!!

Christopher Criscitiello - 3 years, 6 months ago

Patrick Corn
Dec 14, 2017

In fact, there are infinitely many integers $n>1$ such that $x^n-x+1$ is reducible over the rationals. If $n \equiv 2 \pmod 6,$ $x^n-x+1$ is divisible by $x^2-x+1$ (proof: plug in a sixth root of unity).

You could avoid Jon's solution by stipulating that the polynomial be reducible but have no rational roots, although that invalidates your solution too.

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