Show that if p is a prime and p divides n^2 then p divides n. Use Fundamental Theorem Of Arithmetic?

Answer 1

n^2 = n*n

n can either be a prime number or the product of many prime numbers according to Fundamental Theory of Arithmetic.

so, if the prime number p divides n^2, then n=p in case 1 discussed above.

If we consider the second case, n itself will be a product of various primes. One of these prime numbers divide n^2.

i.e:- the prime number which divides n^2 is a factor of n

eg:- case1: 169=13*13
here, n=13,p=13

case2: 196=14*14
here n=14,p=2(or)7

Also note that p^2 always divides n^2 since p divides n.

Answer 2

Yes, this is a consequence of the fundamental Theorem of Arithmetic. When you factor n into prime numbers, you get n = p1^a1 * p2^a2 *...p_m^a_m, where p_1, ....p_m are the m >=1 primes that appear in the factorization of n. Then, n^2 = p1^(2a1) * p2^(2a2) *...p_m^(2a_m). So, the primes that appear in the factoring ion of n^2 are exactly the same that appear in the factoring of n It follows that, if p is a prime factor of n^2, then p must be a prime factor of n, proving the assertion.