## AW: Fwd: [PrimeNumbers] Re: Small prime divisors of very large numbers

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• I m not sure I get only 5 factors. 2,3,5,11,821 . Note , I have also 821 ! No other found. Here my UBASIC program    10   P=1    20   P=nxtprm(P)
Message 1 of 2 , Jul 1, 2009
I'm not sure I get only 5 factors.

2,3,5,11,821 . Note , I have also 821 ! No other found.

Here my UBASIC program

10   P=1
20   P=nxtprm(P)
30   S1=modpow(137,137,P-1)
40   S2=modpow(137,S1,P-1)
50   S3=modpow(137,S2,P-1)
60   S4=modpow(137,S3,P-1)
70   S5=modpow(137,S4,P)+73
80   if S5@P=0 then print P;
90   goto 20
OK

Hm ?

> > The first 7 primes that divide

> > 137^(137^(137^ (137^137) )) + 73

> > are 2, 3, 5, 29, 821, 23339, 67525153.
• ... In line 50, use the modulus m3 = eulerphi(P-1). In line 40, use the modulus m4 = eulerphi(m3). In line 30, use the modulus m5 = eulerphi(m4). Below is a
Message 2 of 2 , Jul 1, 2009
Norman Luhn <nluhn@...> wrote:

> I get only 5 factors.

These lines were wrong:

>    30   S1=modpow(137,137,P-1)
>    40   S2=modpow(137,S1,P-1)
>    50   S3=modpow(137,S2,P-1)

In line 50, use the modulus m3 = eulerphi(P-1).
In line 40, use the modulus m4 = eulerphi(m3).
In line 30, use the modulus m5 = eulerphi(m4).

Below is a Pari-GP procedure "pmod(a,m)" to compute
a[1]^(a[2]^(a[3]^ ... ^(a[k-1]^a[k]) ... ) modulo m
where the modulus "m" need not be prime.
I shall use it to show that
3^6 * 11 * 13^2 * 277 * 1063 * 7459 * 93408839
divides
137^(137^(137^(137^(137^137)))) + 184

{pmod(a,m)=local(k,q,t);k=#a;q=[m];t=a[k];
for(j=2,k-1,q=concat(eulerphi(q[1]),q));
for(j=1,k-1,t=Mod(a[k-j],q[j])^lift(t));t}

m = 3^6 * 11 * 13^2 * 277 * 1063 * 7459 * 93408839;
print(pmod([137,137,137,137,137,137],m)+184)

Mod(0, 278027998329615967980261)

David
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