--- In

primenumbers@yahoogroups.com,

"djbroadhurst" <d.broadhurst@...> wrote:

> This is the best that I could do, by way of an heuristic argument

> that the probability of another (n,b) pair is less than 1/10^6.

There was an error in Part (6), giving an upper bound

for the probability of a solution with n > 100, sorry.

Here I correct it, with no change to the conclusion:

> 6) For n > 6 we know from

> http://mathworld.wolfram.com/TotientFunction.html

> that eulerphi(n) > sqrt(n).

> Hence I estimate the probability of a solution

> with n > 100 as less than

[here I should have computed:]

{print(precision(sum(n=10^2,10^4,1./4^(eulerphi(n)/2))

+suminf(n=10^4,1/2^(sqrt(n)/2)),1));}

0.0000000005135754773441919258

[with the same conclusion:]

> which is far less than the estimate 2/10^7, in Part (4).

>

> 7) In conclusion, I believe that there is a probability

> of less than 1/10^6 for finding a second integer pair (n,b),

> with n>2 and b>1, such that sqrt(Phi(n,b)) is an integer.

David