Let us call an integer $n>0$ pure if for all integers $d>0$ we have that $n^d + (n+1)^d$ is not prime. Is the set of pure integers non-empty? Is it computable?
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asked Sep 9, 2020 at 22:02
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7$\begingroup$ $n^d+(n+1)^d$ is surely composite if $d$ is not a power of 2. So, one can restrict attention to the case $d=2^k$. That is, $n$ is pure iff the corresponding generalized Fermat numbers are all composite. $\endgroup$– Max AlekseyevCommented Sep 9, 2020 at 22:25
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1$\begingroup$ You may also check the factorisation of fermat numbers $\endgroup$– zeraoulia rafikCommented Sep 9, 2020 at 22:47
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2$\begingroup$ The OP is allowing d to be 1, so the smallest candidate for purity, according to OEIS, is 28 $\endgroup$– paul MonskyCommented Sep 10, 2020 at 4:27
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$\begingroup$ @paulMonsky: Good point! Then a bettert OEIS reference is A057856, which by the way states a conjecture implying that there are no pure numbers. $\endgroup$– Max AlekseyevCommented Sep 10, 2020 at 12:35
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