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The problem is finding an irreducible polynomial in $\mathbb{Z}[x]$ such that it is reducible modulo 2,3 and 5. I can't find anything, any help is appreciated. (Is there some general strategy for doing this?)

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    $\begingroup$ Chinese remainder theorem? $\endgroup$
    – Angina Seng
    Apr 20, 2019 at 11:56
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    $\begingroup$ Just look for a simple quadratic. Try the form $x^2+a$ . $\endgroup$
    – lulu
    Apr 20, 2019 at 11:59
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    $\begingroup$ Have you tried some simple quadratics? $\endgroup$
    – rogerl
    Apr 20, 2019 at 11:59

1 Answer 1

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Hint $:$ Take $f(x)=x^4+1.$ Then it is reducible modulo every prime $p.$ But it is irreducible in $\Bbb Z[x].$

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