Quote:
Originally Posted by garambois
1) I am totally unable to do it myself, but wouldn't it be possible to prove conjecture (10) by a method similar to the one used for conjecture (2) ? (see henryzz's post #476 and warachwe's post #480).
Indeed, 2^121 = 3^2*5*7*13 and we do have the 13 which preserves the 7 for the second iteration, except perhaps for some values of k that I did not examine, but the whole problem is precisely there, isn't it ?

The problem is that when the factor 13 are with even power, it does not preserve the 7 for the second iteration.
This only happen when k is multiple of 13, for example 2^(12*13)1 =3^2*5*7*13^2*53*79*...
This is why conjecture 34 ( 3^(18*37) ), 35 ( 3^(36*37) ), and 106 ( 11^(6*37) ) are false.
But this doesn't mean all similar conjecture are false, as there maybe others prime(s) that preserve p.
When we try to 'get rid of' those primes, there maybe yet another that will preserve p instead. Since the size of first iteration grow very quickly, it is hard to find other contradiction this way.
If some of those are true, I imagine the proof might be similar to the proof of conjecture (2).