As in the title. My first thought would be something using Fermat's little theorem, but I'm not sure where to go from there.
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1It's not when $a=0$. – Angina Seng Oct 20 '18 at 16:44
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What about for non-zero $a$? – Oct 20 '18 at 16:48
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2It's not when $a=1$. – Angina Seng Oct 20 '18 at 16:50
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I screwed up the question, it should have been $+a$, my apologies – Oct 20 '18 at 16:52
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2It's still not when $a=1$. – Angina Seng Oct 20 '18 at 16:53
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3We always have $x^p+a\equiv (x+a)^p\pmod p$ so your polynomial is never irreducible. – lulu Oct 20 '18 at 17:01
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@lulu, that's answer. – lhf Oct 20 '18 at 17:15