To prove that the product of two integers, one of the form $3k_1 + 2$ and the other of the form $3k_2 + 2$, where $k_1$ and $k_2$ are integers, is of the form $3k_3 + 1$ for some integer $k_3$, we can proceed as follows:

Let: $a = 3k_1 + 2$ $b = 3k_2 + 2$

We want to show that: $ab = (3k_1 + 2)(3k_2 + 2)$ is of the form $3k_3 + 1$ for some integer $k_3$.

First, compute the product $ab$: $ab = (3k_1 + 2)(3k_2 + 2)$

Expanding this product using the distributive property: $ab = 3k_1 \cdot 3k_2 + 3k_1 \cdot 2 + 2 \cdot 3k_2 + 2 \cdot 2$

Simplify each term: $ab = 9k_1k_2 + 6k_1 + 6k_2 + 4$

Next, factor out the common factor of 3 from the first three terms: $ab = 3(3k_1k_2 + 2k_1 + 2k_2) + 4$

Now, let $k_3 = 3k_1k_2 + 2k_1 + 2k_2$. Notice that $k_3$ is an integer since $k_1$ and $k_2$ are integers.

Thus, we can write: $ab = 3k_3 + 4$

We need to show that $3k_3 + 4$ is of the form $3k_3 + 1$. To do this, we recognize that 4 can be written as: $4 = 3 + 1$

Therefore: $ab = 3k_3 + 3 + 1$

Combine like terms: $ab = 3(k_3 + 1) + 1$

Let $k_3' = k_3 + 1$. Notice that $k_3'$ is also an integer since $k_3$ is an integer. Hence, we have: $ab = 3k_3' + 1$

Thus, the product $ab$ is indeed of the form $3k_3' + 1$ for some integer $k_3'$.

This completes the proof.

Would you like more details or have any questions?

Here are some related questions for further exploration:

1. What is the product of $3k_1 + 1$ and $3k_2 + 2$?
2. How do the properties of modular arithmetic apply to this problem?
3. Can you prove similar results for numbers of the form $nk + m$?
4. How does this result relate to polynomial congruences?
5. What other properties of numbers of the form $3k + 2$ can be derived?
6. How does the distributive property facilitate this proof?
7. Can you prove the result using induction?
8. How does this relate to quadratic residues in number theory?

Tip: When proving statements involving integers, consider using properties of modular arithmetic to simplify and solve the problem efficiently.