Check my proof: By the Fundamental Theorem of Arithmetic, we say that n = pk1
1 pk2
2 ...pkm
m
where pi’s are primes and, by supposition, ki ≥ 2 for all i. Now, we can
write ki = 2si + 3ti for some non-negative integers si, ti. This can be done
by allowing si = ki
2 and ti = 0 for even ki. If ki is odd, let si = ki−3
2 and
ti = 1. Then, n = (ps1
1 ...psm
m )2(pt1
1 ...ptm
m )3.

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 \( 2s_i + 3t_i \).

---

### 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.

Check my proof: By the Fundamental Theorem of Arithmetic, we say that n = pk1 1 pk2 2 ...pkm m where pi’s are primes and, by supposition, ki ≥ 2 for all i. Now, we can write ki = 2si + 3ti for some non-negative integers si, ti. This can be done by allowing si = ki 2 and ti = 0 for even ki. If ki is odd, let si = ki−3 2 and ti = 1. Then, n = (ps1 1 ...psm m )2(pt1 1 ...ptm m )3.

This proof demonstrates how to express a number n as a product of a square and a cube using prime factorization and the Fundamental Theorem of Arithmetic. The proof covers the concepts of exponents, prime factorization, and number theory, making it suitable for students in Grades 11-12.

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