[seqfan] Re: Primes of the form (4^p+1)/5^t

[Jack Brennen]

The form 4^p+1 for odd p is factorable by Aurifeuillean factorization into:

4^p+1 == 4^(2k+1)+1 == 2^(4k+2)+1.

That factors as:

    2^(4k+2)+1 == (2^(2k+1) - 2^(k+1) + 1) * (2^(2k+1) +2^(k+1) +1)

Note that at most one of those two factors is divisible by 5. (Their 
difference is a power of 2.)

So in order for N to be prime, the Aurifeuillean factorization would 
need to have one factor a power of 5, and the other factor a prime.

For p=3, k=1, and 2^(2k+1) - 2^(k+1) + 1 == 2^3 - 2^2 + 1 == 5.
For p=5, k=2, and 2^(2k+1) - 2^(k+1) + 1 == 2^5 - 2^3 + 1 == 25.

In order for 4^n+1 to be divisible by 125, which would be required in 
order for there to be any further solutions, n would have to be 
divisible by 25.  Solutions for 4^n+1 == 0 (mod 125) are n=25,75,125,175,...

So 4^p+1, for p>5, can't be divisible by 125.

So there are no more such numbers.  The three already given (p=2,3,5) 
are the only ones.



On 3/14/2017 11:31 PM, israel at math.ubc.ca wrote:
> That may tell you something about t, but what does it have to do with 
> whether N is prime?
>
> On Mar 14 2017, Don Reble wrote:
>
>>> Let N=(4^p+1)/5^t, where p is prime, 5^t is the most power of 5 
>>> dividing
>>> 4^p+1. For p=2,3,5, N=17,13,41. What is the next prime p for which N is
>>> prime?
>>
>>    To prove there aren't any more prime N's, a first step is to
>>    show that if 5^n divides either of (2^(2a+1) +- 2^(a+1) + 1),
>>    then 5^(n-1) divides (2a+1). Calculations suggest it's true,
>>    but I'm stuck.
>>
>>
>
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