##n^2## is divided by 3, then n is also divided by 3.

[Dustinsfl]

Homework Statement

Prove that if $n^2$ is divided by 3, then also n can also be divided by 3.

Homework Equations

The Attempt at a Solution

Is there anything wrong with this argument?

By contradiction, suppose $3|n$ and 3 doesn't divide $n^2$.
Then $3m = n$. Multiple both sides by $n$
$$
3(mn) = n^2
$$
Thus, $3|n^2$ and we have reached a contradiction. Therefore, If $3|n^2$, then $3|n$.

 

[Nathanael]

Right. If n=3m then n²=9m² which is always divisible by 3.

 

[willem2]

By contradiction, suppose 3|n3|n and 3 doesn't divide n2n^2.

This is not right. Your prove that the conjecture 3|n Λ not 3|n^2 produces a contradiction, but that proves that 3|n ⇒ 3|n^2.
You need to prove that 3|n^2 ⇒ 3|n

 

[HallsofIvy]

What you have proved is that "if n is divisible by 3 then $n^2$ is divisible by 3. That's the "wrong way around"!
To give a proof by contradiction, you need to start "suppose n is NOT divisible by 3", not that is divisible by 3.

Do you see that if n is not divisible by 3, it must be of the form "3k+ 1" or "3k+ 2" for some integer k?