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

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This discussion proves that if 3 divides \( n^2 \), then 3 must also divide \( n \). The argument is based on the three possible forms of \( n \): \( 3a \), \( 3a + 1 \), and \( 3a + 2 \). The only form that results in \( n^2 \) being divisible by 3 is \( 3a \), as the other two forms yield a remainder of 1 when divided by 3. This establishes a clear mathematical relationship between \( n \) and its divisibility by 3.

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yaowang101
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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.
 
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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.
 

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