-(b + c - p - 2q)(b + c + p + 2q)
p2 - 2p + 2 can be factored as (p - 1)(p - 1)which can be written as ( p - 1)2.
p2+10d+7
p2+2pq+q2=1
P2 + l 2 + w
p2 + 3p = p (p + 3)
p2 + 3p - 4What two factors of - 4 add up to 3 ??(p - 1)(p + 4)===============so,p = 1p = - 4
p2 + 2pq + q2 = 1q2 + 2pq + (p2 - 1) = 0q = 1/2 [ -2p plus or minus sqrt( 4p2 - 4p2 + 4 ) ]q = -1 - pq = 1 - p
rutherford
By definition, prime number p has two factors, 1 and p. P2 has three factors, 1, p and p2 Therefore, p2 is composite.
No. Let p1 be a prime number. Let p2 be a multiple of p1 such that p2 = p1 * k. Then the factors of p2 are: 1, p1, k and p2. ==> p2 is not a prime number. Hence, a multiple of a prime number cannot be a prime number.
Even
A number will have 12 factors if it satisfies any of the following prime factor configurations: P11 (P1)5*(P2) (P1)3*(P2)2 (P1)2*(P2)*(P3)