Does this prove I made seem correct to show that if $6$ divides $a$ then $9$ divides $a^2$
If $6\mid a$, then $a = 6k$ (k is some integer).
Then $a^2 = 36k^2 = 9(4k^2)$.
Which means that $9\mid a^2$.
perhaps if not is there any other way?
Does this prove I made seem correct to show that if $6$ divides $a$ then $9$ divides $a^2$
If $6\mid a$, then $a = 6k$ (k is some integer).
Then $a^2 = 36k^2 = 9(4k^2)$.
Which means that $9\mid a^2$.
perhaps if not is there any other way?
No comments:
Post a Comment