Question

let p be a prime number.If p divides a square,prove p divides a,where a is a positive integer.

Answer 1

Let the prime factorisation of 'a 'be as follows:
a=p1p2.....p(n) where P1,P2,.....,p(n )are primes not necessarily distinct.
Therefore,(a)2=(p1p2..p(n))=(P1)2(P2)2...p(n)2.
Now We are given that p divides (a)2.therefore, from the Fundamental Theorem of Arithmetic,it follows that p is one of the factors of (a)2.However, using the uniqueness part of the Fundamental Theorem of Arithmetic,we realise that the only prime factors of (a)2 are p1,p2,...,p(n).so p is one of P1,P2,....p(n).
Now,since a=p1p2...p(n),p divides 'a'.
Here in p1,p2,and p(n),1,2,&n are subscripts of p

Answer 2

Answer : it will be positive integer