Find all prime numbers p for which number p^2
+11 has exactly six distinct divisors
(counting 1 and itself).
If the exponents of the prime factorization of an integer N are a,b,c,..., the number of factors of N is the product (a+1)(b+1)(c+1)... If N has exactly 6 factors then N will be the fifth power of a prime or it will be the product of a prime and the square of a prime.
By inspection p=2 is not a solution so p is odd. Setting p=2q+1 it's clear LHS will be divisible by 4 and not 8. So N won't be a fifth power, making N = 4r, r=prime.
Odd primes > 3 are = 1 or 1 mod 6. Then squaring and adding 11 gives a sum divisible by 3, so r is divisible by 3. Thus r=3. But p^2 + 11 = 4*3 makes p=1, a nonprime.
We've excluded p=2 and p>3. Trying p=3 gives the solution 20 which is (2^2)*5 and has the six factors 1,2,4,5,10,20.
p=3 is the unique solution.
Edited on September 16, 2015, 7:26 pm

Posted by xdog
on 20150916 19:23:40 