Find smallest prime such that, has positive divisors ( is also a divisor.)
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If P 3 + 2 P 2 + P = P ( P + 1 ) 2 has 4 2 divisors, then ( P + 1 ) 2 has 2 1 divisors. Let the powers of the prime factors of P + 1 be a , b , c , … . Then we can write;
( P + 1 ) 2 = ( P 1 a ⋅ P 2 b ⋅ P 3 c … ) 2
Therefore, ( 2 a + 1 ) ( 2 b + 1 ) ( 2 c + 1 ) ⋯ = 2 1
This is possible if P + 1 has 2 factors, 2 a + 1 = 3 and 2 b + 1 = 7 .
Which means that, a = 1 , b = 3
Hence P + 1 is of the form ( P 1 ⋅ P 2 3 )
Since 2 4 = 2 3 ⋅ 3 is the smallest possible case, we get P = 2 3