Show that if 3 divides n^2, 3 divides n

[yaowang101]

Show that if 3 divides n^2, 3 divides n. Hint: n can only be in the form 3a, 3a + 1, 3a + 2

Can someone help me with this? It's pretty simple, I just don't know how to express it mathematically.

 

[mathman]

n² then has three possibilities.
9a²
9a²+6a+1
9a²+12a+4
Only the first (for n=3a) is divisible by 3. The other two have a remainder of 1 after division by 3.

 

[tacman]

Sure, I can help you with this! Let's break down the problem step by step.

First, we want to show that if 3 divides n^2, then 3 divides n. This means that if n^2 is a multiple of 3, then n must also be a multiple of 3. We can express this mathematically as:

If n^2 is divisible by 3, then n is also divisible by 3.

Now, let's consider the three possible forms of n: 3a, 3a + 1, and 3a + 2. We can rewrite these forms as follows:

3a = 3a
3a + 1 = 3a + 3(0) + 1
3a + 2 = 3a + 3(0) + 2

Notice that all three forms have a common factor of 3, which means they are all divisible by 3. So, we can rewrite n as 3a, where a is some integer. This means that n is a multiple of 3, or in other words, n is divisible by 3.

Now, let's look at n^2. We can rewrite n^2 as (3a)^2, which is equal to 9a^2. Since a is an integer, 9a^2 is also an integer. This means that n^2 is a multiple of 3, or in other words, n^2 is divisible by 3.

Putting it all together, we have shown that if 3 divides n^2, then 3 divides n. Therefore, the statement is true. I hope this helps!