<Question>
What is the smallest positive integer $n$ such that all the roots of $z^4 + z^2 + 1 = 0$ are $n^{\text{th}}$ roots of unity?
</Question>

<think>
<Plan_1>
Okay, so I need to find the smallest positive integer n such that all the roots of the equation z^4 + z^2 + 1 = 0 are nth roots of unity. Hmm, let me start by recalling what roots of unity are. The nth roots of unity are the complex numbers satisfying z^n = 1. So, if all roots of the given quartic equation are nth roots of unity, then each root z must satisfy z^n = 1. My goal is to find the smallest such n.

First, let me try to find the roots of the equation z^4 + z^2 + 1 = 0. Maybe if I can express the equation in a form that relates to roots of unity, I can figure out the necessary n. Let's see. The equation is a quartic, so degree 4. Let me check if it can be factored into quadratics or maybe even linear terms.

<prompt_1.1> Attempt to factor the quartic equation \( z^4 + z^2 + 1 = 0 \) into quadratics. </prompt_1.1>
</Plan_1>

<Execution_1>
<execution_1.1>
Let me try to factor it. Let's see:

z^4 + z^2 + 1. Hmm. If I consider it as a quadratic in z^2, then it's z^4 + z^2 + 1 = (z^2)^2 + z^2 + 1. Maybe I can factor this quadratic. The discriminant would be 1 - 4 = -3, so the roots would be z^2 = [-1 ± i√3]/2. Wait, those are the roots of z^2 + z + 1 = 0, which are the primitive 3rd roots of unity. Interesting. So, z^2 = e^(2πi/3) or z^2 = e^(4πi/3). Therefore, the roots of the original equation would be the square roots of these, right?
</execution_1.1>
</Execution_1>

<Plan_2>
Based on execution_1.1:
So, if z^2 = e^(2πi/3), then z = e^(πi/3 + kπi) for k=0,1. Similarly, for z^2 = e^(4πi/3), z = e^(2πi/3 + kπi) for k=0,1. Let me compute these roots explicitly.

<prompt_2.1> Compute the roots of the equation \( z^4 + z^2 + 1 = 0 \) explicitly as described in the above plan. </prompt_2.1>
-----

Wait, but let me check if there's a smaller n. Let's see. Suppose someone claims n=6.

<prompt_2.2> Check if a smaller \( n \) than 12 is possible, considering the roots found. </prompt_2.2>
</Plan_2>

<Execution_2>
<execution_2.1>
For z^2 = e^(2πi/3) = cos(2π/3) + i sin(2π/3) = (-1/2 + i√3/2). Taking square roots, we get z = ±[cos(π/3) + i sin(π/3)] = ±(1/2 + i√3/2). Similarly, for z^2 = e^(4πi/3) = cos(4π/3) + i sin(4π/3) = (-1/2 - i√3/2), the square roots are ±[cos(2π/3) + i sin(2π/3)] = ±(-1/2 + i√3/2). Wait, let me verify:

Wait, if z^2 = e^(2πi/3), then z = e^(πi/3) or e^(πi/3 + π) = e^(4πi/3). Similarly, z^2 = e^(4πi/3) gives z = e^(2πi/3) or e^(2πi/3 + π) = e^(5πi/3). So, the four roots are e^(πi/3), e^(2πi/3), e^(4πi/3), e^(5πi/3).
</execution_2.1>

<execution_2.2>
Because 12 is a multiple of 6. Let's check if the roots are 6th roots of unity. The primitive 6th roots of unity are e^(πi/3) and e^(5πi/3), which are indeed roots of the equation. But the other roots are e^(πi/6) and e^(3πi/6) (which is e^(πi/2)), but wait, e^(πi/6) is a 12th root of unity, not a 6th root. Because 6th roots of unity are at angles 0, π/3, 2π/3, π, 4π/3, 5π/3. The roots we have here are at π/3, 5π/3 (which are 6th roots), but also π/6, 3π/6 (which is π/2), 7π/6, etc. Wait, π/6 is 30 degrees, which is not a 6th root of unity. So, e^(πi/6) is a primitive 12th root of unity. Therefore, if n=6, then the roots of the equation include e^(πi/6) which is not a 6th root of unity. Therefore, n=6 is too small.
</execution_2.2>
</Execution_2>

<Plan_3>
Based on execution_2.1:
Let me write them down in exponential form:

e^(πi/3), e^(2πi/3), e^(4πi/3), e^(5πi/3).

Wait, but hold on, πi/3 is 60 degrees, 2πi/3 is 120 degrees, 4πi/3 is 240 degrees, and 5πi/3 is 300 degrees. So, these are the 6th roots of unity, but only every other one. Wait, the primitive 6th roots of unity are e^(πi/3) and e^(5πi/3), right? Because 6th roots of unity are at angles 0, π/3, 2π/3, π, 4π/3, 5πi/3. But the roots we have here are π/3, 2π/3, 4π/3, 5π/3. Wait, so these are actually the 6th roots of unity but excluding 0 and π. Wait, but 2π/3 is 120 degrees, which is a primitive 3rd root of unity. Hmm, confusing.
-----

But let me check n=12. Let's see.

<prompt_3.1> Verify if \( n = 12 \) is a possible solution by checking if the roots are 12th roots of unity. </prompt_3.1>
-----

But let me check n=12. Let's see.

<prompt_3.2> Verify if \( n = 12 \) is a possible solution by checking if the roots are 12th roots of unity. </prompt_3.2>
-----

But let me check n=12. Let's see.

<prompt_3.3> Verify if \( n = 12 \) is a possible solution by checking if the roots are 12th roots of unity. </prompt_3.3>
-----

But let me check n=12. Let's see.

<prompt_3.4> Verify if \( n = 12 \) is the minimal solution by checking if any smaller \( n \) can work. </prompt_3.4>
</Plan_3>

<Execution_3>
<execution_3.1>
The 12th roots of unity are at angles kπ/6 for k=0,1,...,11. So, e^(πi/3) = e^(2πi/12) which is a 12th root. Similarly, e^(2πi/3) = e^(8πi/12) = e^(2πi*2/3), which is also a 12th root. Wait, 2π/3 is 8π/12. Similarly, 4π/3 is 16π/12, which is equivalent to 4π/3 - 2π = -8π/12, which is the same as e^(-2πi*2/3). And 5π/3 is 10π/12, which is e^(5πi/6). So all these roots are indeed 12th roots of unity.
</execution_3.1>

<execution_3.2>
The 12th roots of unity are the roots of z^12 = 1, which includes all the roots of z^k = 1 for k dividing 12. So, e^(πi/3) = e^(2πi/12 * 2) = e^(2πi/6), which is a 6th root. Wait, but 6 divides 12, so yes, these roots are 12th roots of unity. Specifically, e^(πi/3) = e^(2πi/6) = e^(4πi/12), which is a primitive 6th root but also a 12th root. Wait, but 12th roots include all 6th roots as well, since 6 divides 12. Therefore, if n=12, then all roots of the equation are 12th roots of unity.
</execution_3.2>

<execution_3.3>
The 12th roots of unity are e^(2πik/12) for k=0,1,...,11. That is, angles 0, π/6, π/3, π/2, 2π/3, 5π/6, etc. So, the roots we found are π/3 (which is 4π/12), 2π/3 (8π/12), 4π/3 (16π/12), 5π/3 (20π/12). So, these correspond to k=4,8,16 (which is equivalent to 4 mod 12), and 20 (equivalent to 8 mod 12). Wait, but 16π/12 is 4π/3, which is 16π/12 = 4π/3, and 20π/12 = 5π/3. So, these are all included in the 12th roots of unity. So, each of these roots is a 12th root of unity. Therefore, n=12 works.
</execution_3.3>

<execution_3.4>
First, the roots are e^(πi/3), e^(2πi/3), e^(4πi/3), e^(5πi/3). Let's note that e^(πi/3) is a primitive 12th root of unity because (e^(πi/3))^12 = e^(4πi) = 1. Similarly, e^(2πi/3) is (e^(2πi/3))^3 = 1, so it's a primitive 3rd root. But 3 divides 12, so the 3rd roots of unity are included in the 12th roots of unity. Similarly, e^(4πi/3) is also a primitive 3rd root, and e^(5πi/3) is a primitive 6th root? Wait, e^(5πi/3) is the same as e^(-πi/3), which is a primitive 6th root. Wait, but 6 divides 12, so all these roots are 12th roots of unity. Therefore, if we take n=12, all roots are 12th roots of unity.
</execution_3.4>
</Execution_3>

<Plan_4>
Based on execution_3.1:
But is 12 the smallest such n? Let's check if maybe n=6 or n=8 or n=12. Let's see.

<prompt_4.1> Verify if \( n = 12 \) is the smallest solution by checking smaller values of \( n \). </prompt_4.1>
</Plan_4>

<Execution_4>
<execution_4.1>
If n=6, as before, the 6th roots of unity are e^(2πik/6) for k=0,1,...,5. The roots we found are e^(πi/3) = e^(2πi*1/6), e^(2πi*2/6), e^(4πi/3) = e^(2πi*4/6), e^(5πi/3) = e^(2πi*5/6). Wait, but hold on, e^(πi/3) is e^(2πi/6), which is a primitive 6th root, but e^(4πi/3) is e^(8πi/6) = e^(2πi*4/6), which is also a primitive 6th root. Similarly, e^(5πi/3) is e^(10πi/6) = e^(2πi*5/6). So, actually, all four roots are 6th roots of unity? Wait, but hold on, the 6th roots of unity are {1, e^(πi/3), e^(2πi/3), e^(πi), e^(4πi/3), e^(5πi/3)}. So, the roots of our equation are exactly the primitive 6th roots of unity, excluding 1 and -1. Wait, but e^(πi/3) and e^(5πi/3) are complex conjugates, as are e^(2πi/3) and e^(4πi/3). Wait, but hold on, the roots of the equation z^4 + z^2 +1 =0 are e^(πi/3), e^(2πi/3), e^(4πi/3), e^(5πi/3). Comparing to 6th roots of unity: 6th roots are at 0, π/3, 2π/3, π, 4π/3, 5π/3. So, the roots of the equation are the 6th roots of unity excluding 0 and π (i.e., 1 and -1). But wait, z^4 + z^2 +1 factors as (z^2 + z +1)(z^2 - z +1). Wait, let me check:

Wait, (z^2 + z +1)(z^2 - z +1) = z^4 - z^2 +1. But our equation is z^4 + z^2 +1. So, that's different. Therefore, my initial factorization was wrong. Wait, so perhaps I need to factor z^4 + z^2 +1 differently.

Wait, perhaps it's (z^4 + z^2 +1) = (z^2 + ω z +1)(z^2 + ω^2 z +1), where ω is a primitive 3rd root of unity. Let me check.
</execution_4.1>
</Execution_4>

<Plan_5>
Based on execution_4.1:
Let's see: Let ω = e^(2πi/3). Then (z^2 + ω z +1)(z^2 + ω^2 z +1) = z^4 + (ω + ω^2) z^3 + (ω ω^2 + 1 +1) z^2 + (ω + ω^2) z +1.

<prompt_5.1> Factor the quartic equation \( z^4 + z^2 + 1 = 0 \) correctly using the above plan. </prompt_5.1>
</Plan_5>

<Execution_5>
<execution_5.1>
Since ω + ω^2 = -3/2? Wait, ω + ω^2 = -1, because 1 + ω + ω^2 =0. So, ω + ω^2 = -1. Then the coefficient of z^3 is -1. The coefficient of z^2 is (ω*ω^2) +1 +1. But ω*ω^2 = ω^3 =1. So, 1 +1 +1=3. Wait, but our original equation has coefficient 1 for z^2. So, that's not matching. Therefore, that factorization is incorrect.

Wait, perhaps (z^4 + z^2 +1) factors as (z^4 + 2z^2 +1) - z^2 = (z^2 +1)^2 - z^2 = (z^2 + z +1)(z^2 - z +1). Wait, but (z^2 + z +1)(z^2 - z +1) = z^4 - z^2 +1, which is different from z^4 + z^2 +1. So, that's not it. Hmm. Maybe adding z^2 instead of subtracting? Let me check.

Alternatively, perhaps z^4 + z^2 +1 = (z^4 - z +1) + z^2 + z. Not sure. Alternatively, maybe use the identity z^6 -1 = (z^2 -1)(z^4 + z^2 +1). Therefore, z^4 + z^2 +1 = (z^6 -1)/(z^2 -1). Therefore, the roots of z^4 + z^2 +1 =0 are the 6th roots of unity except z=1 and z=-1. Wait, but if z^6 =1, then z^6 -1=0, so z^4 + z^2 +1 =0 when z^6=1 and z≠1, -1. Therefore, the roots are the primitive 6th roots of unity. But wait, the primitive 6th roots of unity are e^(πi/3) and e^(5πi/3), which are two roots. But our equation is quartic, so degree 4. That suggests there are four roots. Therefore, perhaps my previous approach is wrong.

Wait, perhaps z^4 + z^2 +1 =0 can be rewritten as z^4 = -z^2 -1. Let me square both sides: z^8 = (z^2 +1)^2. So, z^8 = z^4 + 2z^2 +1. But substituting z^4 = -z^2 -1 into the right-hand side: z^8 = (-z^2 -1) + 2z^2 +1 = z^2. Therefore, z^8 = z^2, so z^8 - z^2 =0, z^2(z^6 -1)=0. Therefore, z^6=1, provided z≠0. Therefore, the roots of z^4 + z^2 +1=0 are the 6th roots of unity except z=0, but also excluding z=1 and z=-1? Wait, let's check.

If z^6=1, then z is a 6th root of unity. Then, substituting into z^4 + z^2 +1: Let's take z = e^(2πik/6) for k=0,1,....,5. Then z^4 + z^2 +1 = e^(8πik/6) + e^(4πik/6) +1. Let's compute for each k:

For k=0: z=1. Then 1 +1 +1=3≠0.

For k=1: z=e^(πi/3). Then z^4 = e^(4πi/3), z^2 = e^(2πi/3). So, e^(4πi/3) + e^(2πi/3) +1. Let's compute this: e^(2πi/3) is -1/2 + i√3/2, e^(4πi/3) is -1/2 - i√3/2. Adding them gives -1, plus 1 is 0. So, z=e^(πi/3) is a root.

For k=2: z=e^(2πi/3). Then z^4 = e^(8πi/3)=e^(2πi/3 + 2πi)=e^(2πi/3). z^2 = e^(4πi/3). So, same as above, sum is e^(2πi/3) + e^(4πi/3) +1=0. So, z=e^(2πi/3) is a root.

For k=3: z=e^(πi) = -1. Then z^4=1, z^2=1. So, 1+1+1=3≠0.

For k=4: z=e^(4πi/3). Similar to k=2, z^4=e^(16πi/3)=e^(4πi/3 + 4πi)=e^(4πi/3). z^2=e^(8πi/3)=e^(2πi/3). So, sum is e^(4πi/3) + e^(2πi/3) +1=0. So, root.

For k=5: z=e^(5πi/3). Then z^4=e^(20πi/3)=e^(2πi/3 + 6πi)=e^(2πi/3). z^2=e^(10πi/3)=e^(4πi/3). So, same as above, sum is 0. So, roots are k=1,2,4,5. Therefore, the roots are the primitive 6th roots of unity, which are e^(πi/3), e^(2πi/3), e^(4πi/3), e^(5πi/3). Therefore, these are the four roots. Therefore, all roots of the equation are 6th roots of unity. But wait, hold on, earlier I thought they were 12th roots. But according to this, they are 6th roots. But wait, the 6th roots of unity are at angles 60°, 120°, 180°, 240°, 300°, 360°, which are k=1,2,3,4,5,0. But the roots of the equation are k=1,2,4,5. So, they are the primitive 6th roots (excluding k=0 and k=3, which are 1 and -1). Therefore, the roots are the primitive 6th roots of unity. Therefore, they are 6th roots of unity. So, does that mean n=6? But wait, hold on, the primitive 6th roots are e^(2πi/6) and e^(10πi/6), which are e^(πi/3) and e^(5πi/3). Wait, but in our case, we have four roots: e^(πi/3), e^(2πi/3), e^(4πi/3), e^(5πi/3). Wait, but e^(2πi/3) is e^(4πi/6), which is a primitive 3rd root of unity. Similarly, e^(4πi/3) is e^(8πi/6) = e^(4πi/3), which is also a primitive 3rd root. So, these roots are actually the primitive 3rd roots and primitive 6th roots. Wait, but 3rd roots are included in 6th roots. So, the roots of the equation are the primitive 3rd roots (e^(2πi/3), e^(4πi/3)) and the primitive 6th roots (e^(πi/3), e^(5πi/3)). Therefore, all roots are 6th roots of unity. Therefore, n=6 would suffice. But earlier I thought n=12. But according to this, the roots are 6th roots.
</execution_5.1>
</Execution_5>

<Plan_6>
Based on execution_5.1:
Wait, but let's check if z^4 + z^2 +1 factors as (z^2 + z +1)(z^2 - z +1). Let's multiply them:

<prompt_6.1> Factor the quartic equation \( z^4 + z^2 + 1 = 0 \) and verify the factorization using the above plan. </prompt_6.1>
</Plan_6>

<Execution_6>
<execution_6.1>
(z^2 + z +1)(z^2 - z +1) = z^4 - z^3 + z^2 + z^3 - z^2 + z + z^2 - z +1 = z^4 + z^2 +1. Wait, that's correct! So, z^4 + z^2 +1 factors into (z^2 + z +1)(z^2 - z +1). Therefore, the roots are the roots of z^2 + z +1 =0 and z^2 - z +1=0.

Solving z^2 + z +1 =0 gives roots e^(2πi/3) and e^(4πi/3), which are the primitive 3rd roots of unity. Solving z^2 - z +1 =0 gives roots [1 ± sqrt(-3)]/2 = e^(πi/3) and e^(5πi/3), which are primitive 6th roots of unity. Therefore, the four roots are the primitive 3rd and 6th roots of unity. Therefore, all roots are 6th roots of unity. Therefore, n=6.
</execution_6.1>
</Execution_6>

<Plan_7>
Based on execution_6.1:
Wait, but hold on. If n=6, then all roots are 6th roots of unity. But the roots are the primitive 3rd and 6th roots. But 3rd roots are also 6th roots. Therefore, all roots are 6th roots of unity. Therefore, n=6. But wait, let's verify if indeed all roots are 6th roots. For example, e^(πi/3) is a primitive 6th root, e^(5πi/3) is also a primitive 6th root. The other two roots are e^(2πi/3) and e^(4πi/3), which are primitive 3rd roots, but 3rd roots are also 6th roots (since 3 divides 6). Therefore, all roots are 6th roots of unity. Therefore, n=6. But this contradicts my initial thought that n=12. So, which is correct?

Wait, but let me check if z= e^(πi/3) is a 6th root of unity. Yes, because (e^(πi/3))^6 = e^(2πi) =1. Similarly, e^(2πi/3)^3 = e^(2πi)=1. Therefore, all roots are 6th roots of unity. Therefore, n=6. But then why did I initially think n=12?

Because when I considered the roots as e^(πi/3), e^(2πi/3), e^(4πi/3), e^(5πi/3), I thought of them as 12th roots, but actually, they are 6th roots. Therefore, the minimal n is 6. But wait, let me check if these roots are primitive 6th roots or not. e^(πi/3) is primitive 6th root, since its order is 6. e^(2πi/3) is a primitive 3rd root, but 3 divides 6, so it's also a 6th root. Similarly, e^(4πi/3) is primitive 3rd root, and e^(5πi/3) is primitive 6th root. Therefore, all roots are 6th roots of unity, but not all primitive 6th roots. However, the question is to find the smallest n such that all roots are n-th roots of unity. So, even if some are primitive k-th roots with k dividing n, as long as they are n-th roots, it's okay. Therefore, n=6 works.
</Plan_7>

<Execution_7>
No prompt to execute.
</Execution_7>

<Final_answer>
Therefore, the smallest positive integer \( n \) such that all the roots of \( z^4 + z^2 + 1 = 0 \) are \( n^{\text{th}} \) roots of unity is $\boxed{6}$.
</Final_answer>
</think>