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A Number Theorist Who Solves the Hardest Easy Problems
- kevinventullo 6y agoThis left a bit of bad taste in my mouth: “It’s easy to imagine an alternative timeline in which Zhang proved his result six months after Maynard instead of six months before.” I don’t think even Maynard has claimed he would have gotten to bounded gaps without Zhang.
- Alkhwarizmi 6y agoVery good point! Shame on these writers, I hope they edit that at the least.
- alimw 6y agoThe article does call Maynard's approach "completely independent". So while Maynard himself might not want to make any such claim, clearly the impression left with the interviewer is that he would indeed have got there without Zhang.
- amflare 6y agoDoes this mean that for any given prime, there will always be a prime within 600 digits? Or that given a set of infinite numbers, you can also construct an infinite set of pairs of primes 600 digits apart?
- deleted 6y ago[deleted]
- ogogmad 6y ago> Does this mean that for any given prime, there will always be a prime within 600 digits? No. Euclid showed that the gap between consecutive primes can be arbitrarily large. In particular, n!+2, n!+3, ..., n!+n are all composite whenever n>1. > Or that given a set of infinite numbers, you can also construct an infinite set of pairs of primes 600 digits apart? It means that there are infinitely many pairs of primes p and q such that 0 < |p - q| < 600.
- ColinWright 6y agoThere are several ways of saying the same thing, and different things resonate with different people. The one I like is this. Go out as far as you like. Somewhere out there ahead of you there are two primes p and q who differ by no more than 600. So there's a prime p with another prime q less than p+600. And now if you start from there and go still further, then you can do it again. No matter how far you will still find primes p and q with p<q<p+600. So there are infinitely many such pairs, because you can always find another one.
- nullc 6y agoNo, although there is always a prime between x and 2x (for x>3).
- magneticnorth 6y agoNow I'm curious about that last result discussed, for bases less than 10 - "proving that there are infinitely many primes that don’t have any 7s (or any other digit you might choose)". It sounds like Maynard decided to stop at proving the result for all bases 10 and above, but could it be true for the smaller bases too? In the case of binary numbers, proving that there are infinitely many prime binary numbers without a 0 is the same as proving that there are infinitely many Mersenne primes, so the (non-trivial) binary case seems to still be a difficult open problem. It would be fascinating if there's a cutoff in base expansions where this stops being true.
- kevinventullo 6y agoI suspect in that in any base b, and for any collection of base b digits such that the gcd of the whole set is 1, there are infinitely many primes which only use those digits. I also suspect the Mersenne case b=2, set={1}, is the hardest.
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