Let 'p' be a prime number such that the number p²+11 has exactly 6 positive divisors. Find p .
Answers
Answer:
if the prime factors of p²+1, excluding 1 and itself, are of the form {a²,b | a≠b}, then the possible factors are:
{1,a,b,a², ab, a²b} total exactly 6 factors.
The first few such sums are:
1²+11=12=2²*3 [but 1 is not a prime]
3²+11=20=2²*5 [cannot be the sum of primes]
8²+11=75=3*5² factors are {1,3,5,15,25,75}, and 8=3+5, both primes.
9²+11=92=2²*23; factors are {1,2,4,23,46,92}, and 9=2+2+5
14²+11=207=3²*23; factors are {1,3,9,23,69,207}, and 14=2+5+7
15²+11=236=2²*59; factors are {1,2,4,59,118,236}, and 15=3+5+7
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Given : 'p' be a prime number such that the number p²+11 has exactly 6 positive divisors
To Find : value of p
Solution:
p = 2 => p² + 11 = 15 ahs 4 divisors ( 1 , 3 , 5 , 15)
p =3 => p² + 11 = 20 has 6 exactly divisors ( 1 , 2 , 4 , 5 , 10 , 20 )
Hence p = 3
after that all prime can be written in form :
p = 6q - 1 , 6q + 1
p² + 11 = (6q - 1)² + 11 = 36q² + 1 - 12q + 11 = 36q² - 12q + 12
= 12( 3q² - q + 1)
12 it self has 6 divisors Hence more than 6 divisors
p² + 11 = (6q+ 1)² + 11 = 36q² + 1 + 12q + 11 = 36q² + 12q + 12
= 12( 3q² + q + 1)
12 it self has 6 divisors Hence more than 6 divisors
so only possible p = 3
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