Your proof is well-organized and generally correct, but let's break it down step-by-step to ensure clarity and correctness. We want to express $n$ in terms of its prime factors, with the goal of rewriting the exponents $k_i$ to leverage their divisibility properties and represent them in a particular form. Here's a closer look:

Step 1: Factorization of $n$

You begin by expressing $n$ as a product of prime factors: $n = p_1^{k_1} p_2^{k_2} \cdots p_m^{k_m}$ where each $p_i$ is a prime and each $k_i \geq 2$. This is correct by the Fundamental Theorem of Arithmetic, which guarantees a unique prime factorization.

Step 2: Expressing $k_i$ in terms of $s_i$ and $t_i$

You want to express each $k_i$ as a sum of multiples of 2 and 3, i.e., in the form $k_i = 2s_i + 3t_i$ for non-negative integers $s_i$ and $t_i$. You do this by considering two cases based on the parity (odd or even nature) of $k_i$:

• If $k_i$ is even, you propose $s_i = \frac{k_i}{2}$ and $t_i = 0$. This works since for even $k_i$, we can write $k_i = 2s_i$, which is of the desired form.

• If $k_i$ is odd, you propose $s_i = \frac{k_i - 3}{2}$ and $t_i = 1$. This works since we can rewrite $k_i$ as $k_i = 2s_i + 3$, which fits the form $2s_i + 3t_i$ with $t_i = 1$.

Thus, in both cases, you have successfully expressed $k_i$ as $k_i = 2s_i + 3t_i$.

Step 3: Rewriting $n$

Now that you have expressed each $k_i$ in the form $k_i = 2s_i + 3t_i$, you substitute this into the original expression for $n$: $n = p_1^{k_1} p_2^{k_2} \cdots p_m^{k_m}$ which becomes: $n = p_1^{2s_1 + 3t_1} p_2^{2s_2 + 3t_2} \cdots p_m^{2s_m + 3t_m}$

Using properties of exponents, you can now rewrite this as: $n = (p_1^{s_1} p_2^{s_2} \cdots p_m^{s_m})^2 (p_1^{t_1} p_2^{t_2} \cdots p_m^{t_m})^3$

This shows that $n$ can be written as a product of a square and a cube: $n = (q_1)^2 (r_1)^3$ where $q_1 = p_1^{s_1} p_2^{s_2} \cdots p_m^{s_m}$ and $r_1 = p_1^{t_1} p_2^{t_2} \cdots p_m^{t_m}$.

Conclusion

Your proof is correct, and the reasoning is sound. You’ve successfully demonstrated that $n$ can be written as the product of a square and a cube using the fact that each $k_i \geq 2$ can be expressed as $2s_i + 3t_i$.

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Do you want more details or have any questions?

Here are 5 related questions for practice:

1. How would this proof change if $k_i$ could be 1 for some $i$?
2. Can every integer $n \geq 2$ be expressed as the product of a square and a cube?
3. How does the Fundamental Theorem of Arithmetic guarantee unique prime factorization?
4. Could this method of splitting exponents be used for different powers, such as representing an exponent as $2s_i + 5t_i$?
5. How would the proof change if some of the $k_i$'s were less than 2?

Tip: Whenever dealing with exponents and factorizations, pay close attention to the cases where values are even or odd; this can significantly simplify or complicate the problem.