We are tasked with proving that the square of any integer $a$ is either of the form $3k$ or $3k + 1$ for some integer $k$.

Proof:

We will use modulo 3 to examine the possible forms of $a^2$. Any integer $a$ can be written in one of three forms based on division by 3:

• $a \equiv 0 \pmod{3}$
• $a \equiv 1 \pmod{3}$
• $a \equiv 2 \pmod{3}$

Let's now square each of these possibilities.

Case 1: $a \equiv 0 \pmod{3}$

If $a \equiv 0 \pmod{3}$, then $a = 3n$ for some integer $n$. Squaring both sides: $a^2 = (3n)^2 = 9n^2 \equiv 0 \pmod{3}$ So, $a^2 \equiv 0 \pmod{3}$. This means $a^2$ is of the form $3k$.

Case 2: $a \equiv 1 \pmod{3}$

If $a \equiv 1 \pmod{3}$, then squaring this: $a^2 \equiv 1^2 = 1 \pmod{3}$ So, $a^2 \equiv 1 \pmod{3}$. This means $a^2$ is of the form $3k + 1$.

Case 3: $a \equiv 2 \pmod{3}$

If $a \equiv 2 \pmod{3}$, then squaring this: $a^2 \equiv 2^2 = 4 \equiv 1 \pmod{3}$ So, $a^2 \equiv 1 \pmod{3}$. This means $a^2$ is again of the form $3k + 1$.

Conclusion:

In all cases, the square of $a$ is either congruent to $0 \pmod{3}$ or $1 \pmod{3}$. Therefore, the square of any integer $a$ is either of the form $3k$ or $3k + 1$ for some integer $k$.

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Would you like further details or clarifications? Here are some related questions for further exploration:

1. Can we extend this result to integers modulo 4 or modulo 5?
2. What happens if $a$ is not an integer but a rational number?
3. Can this result help in proving properties of quadratic residues?
4. How would this proof change if we used induction instead of modular arithmetic?
5. What are some real-world applications of knowing that $a^2$ must be in the form $3k$ or $3k+1$?

Tip: Modular arithmetic is often useful in number theory for classifying the behavior of integers based on their remainder when divided by a fixed number.