Question

, Let k be an arbitrary natural number and p a prime number if k² is a multiple of p then necessarily;

A.     K is an odd number

B.     K is a prime number

C.     K is a multiple of P

D.     None of the above

Answer 1

If $p|k^2,$ then we have that.

$V_p(k^2) = 2m, where\\ m \in \Z^+\\$

Taking the square root of K, then

$V_p(k)=\frac{V_p(k^2)}{2}=m \in \Z^+$

Therefore, k is a multiple of p