Re: Is 6 the difference of two powers?

In article <bfdf2998-4ecf-4932-8822-b670d13e3c25@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
Pubkeybreaker <pubkeybreaker@xxxxxxx> wrote:

Up to N, the number of powers P(N) is O(N^(1/2 + o(1)). The set of
powers is dominated by squares. This is easy to see.

If we take PAIRS of prime powers, how many different pairs are there?
Clearly there are C(P(N), 2). Their differences are all less than N,
so it would seem that there should be enough different pairs of powers
such that their differences "cover" all of the integers less than N.

All differences of squares are either odd or a multiple of 4. So for
the N/4 differences that are equal to 2 mod 4 we have only the
differences between a square and a higher power or between two higher
powers. So the dominant term of the number of pairs is going to be
N^1/2 . N^1/3 = N^5/6, which is much less than N/4 for large N.

But this doesn't help much, because the differences less then N
may arise from numbers larger than N.

-- Richard
--
:wq
.

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