Quote:
Originally Posted by baih
my solution are :qp =c
lets n is integer
e= (c+1)/4
s= n^e Mod c
q= gcd(sn,c)
for your example :

Yes, more detail:
If n=p*q, and (n+1)/4==1 mod (p1) then
(p1)  e=(n3)/4 with Fermat (for every b)
b^e1 is divisible by p so pgcd(b^e1,n) and in general it is in fact p (and not n).
So (partial) factorization is easy when we know pn and n mod (p1).
Say we know that n==r/s mod (p1) for tiny r,s (in solution just try all r,s pair).