<Question>
Let
\[x^8 + 3x^4 - 4 = p_1(x) p_2(x) \dotsm p_k(x),\]where each non-constant polynomial $p_i(x)$ is monic with integer coefficients, and cannot be factored further over the integers.  Compute $p_1(1) + p_2(1) + \dots + p_k(1).$
</Question>

<think>
<Plan_1>
Okay, so I need to factor the polynomial \( x^8 + 3x^4 - 4 \) into irreducible monic polynomials with integer coefficients and then evaluate the sum of each of these polynomials at \( x = 1 \). Hmm, let's start by recalling some factoring techniques. The given polynomial is a quartic in terms of \( x^4 \), so maybe I can substitute \( y = x^4 \) to make it a quadratic equation: \( y^2 + 3y - 4 \).
Let me try factoring that.

Looking for two numbers that multiply to -4 and add to 3. Let's see, 4 and -1: 4 * (-1) = -4 and 4 + (-1) = 3.
So, the quadratic factors as \( (y + 4)(y - 1) \). Substituting back \( y = x^4 \), we get \( (x^4 + 4)(x^4 - 1) \).

Wait, so the original polynomial factors into \( (x^4 + 4)(x^4 - 1) \). But are these factors irreducible over the integers?

Let me check each one.

Starting with \( x^4 - 1 \). That's a difference of squares: \( (x^2)^2 - 1^2 = (x^2 - 1)(x^2 + 1) \). Then, \( x^2 - 1 \) is another difference of squares: \( (x - 1)(x + 1) \). So, \( x^4 - 1 \) factors into \( (x - 1)(x + 1)(x^2 + 1) \).

Now, \( x^4 + 4 \). Hmm, this looks like a sum of squares, but over the integers, sums of squares can sometimes be factored using the Sophie Germain identity. Let me recall: \( a^4 + 4b^4 = (a^2 + 2ab + 2b^2)(a^2 - 2ab + 2b^2) \). If I set \( b = 1 \), then \( x^4 + 4 = (x^2 + 2x + 2)(x^2 - 2x + 2) \). Let me check that:

<prompt_1.1> Verify the factorization of \( x^4 + 4 \) using the Sophie Germain identity. </prompt_1.1>

So putting it all together, the original polynomial factors as:
\( (x^2 + 2x + 2)(x^2 - 2x + 2)(x - 1)(x + 1)(x^2 + 1) \).

Now, we need to check if each of these factors is irreducible over the integers. Let's see:

<prompt_1.2> Check if each factor is irreducible over the integers. </prompt_1.2>
-----

Now, each of these polynomials is monic with integer coefficients and irreducible. So the factorization is complete.

Now, compute \( p_i(1) \) for each factor:

<prompt_1.3> Compute \( p_i(1) \) for each factor and find the sum. </prompt_1.3>
</Plan_1>

<Execution_1>
<execution_1.1>
Multiplying out \( (x^2 + 2x + 2)(x^2 - 2x + 2) \):
First, \( x^2 * x^2 = x^4 \).
Then, \( x^2 * (-2x) = -2x^3 \).
Then, \( x^2 * 2 = 2x^2 \).
Next, \( 2x * x^2 = 2x^3 \).
\( 2x * (-2x) = -4x^2 \).
\( 2x * 2 = 4x \).
Next, \( 2 * x^2 = 2x^2 \).
\( 2 * (-2x) = -4x \).
\( 2 * 2 = 4 \).
Combine like terms:
\( x^4 + (-2x^3 + 2x^3) + (2x^2 - 4x^2 + 2x^2) + (4x - 4x) + 4 \).
Simplifies to \( x^4 + 0x^3 + 0x^2 + 0x + 4 = x^4 + 4 \). Perfect, that works.
</execution_1.1>

<execution_1.2>
- \( x^2 + 2x + 2 \): Let's check its discriminant: \( (2)^2 - 4*1*2 = 4 - 8 = -4 \). Since the discriminant is negative, it has no real roots and is irreducible over the reals, hence over the integers.
- \( x^2 - 2x + 2 \): Discriminant: \( (-2)^2 - 4*1*2 = 4 - 8 = -4 \). Same as above, irreducible.
- \( x - 1 \): Linear polynomial, irreducible.
- \( x + 1 \): Linear, irreducible.
- \( x^2 + 1 \): Discriminant: \( 0^2 - 4*1*1 = -4 \). Also irreducible.
</execution_1.2>

<execution_1.3>
1. For \( x^2 + 2x + 2 \): Substitute x = 1: \( 1 + 2 + 2 = 5 \).
2. For \( x^2 - 2x + 2 \): Substitute x = 1: \( 1 - 2 + 2 = 1 \).
3. For \( x - 1 \): Substitute x = 1: \( 1 - 1 = 0 \).
4. For \( x + 1 \): Substitute x = 1: \( 1 + 1 = 2 \).
5. For \( x^2 + 1 \): Substitute x = 1: \( 1 + 1 = 2 \).
</execution_1.3>
</Execution_1>

<Plan_2>
Based on execution_1.1, execution_1.2:
Therefore, the factorization is complete, and we have 5 factors.
Wait, but the problem states the factorization is \( p_1(x) p_2(x) \dotsm p_k(x) \).
So here, k = 5.

Now, the problem asks for the sum \( p_1(1) + p_2(1) + \dots + p_k(1) \). Let's compute each \( p_i(1) \):

<prompt_2.1> Compute the sum of \( p_i(1) \) for each factor. </prompt_2.1>
-----

Based on execution_1.3:
But let me double-check each step to ensure there are no mistakes.

First, the original polynomial: \( x^8 + 3x^4 - 4 \). Substituting \( y = x^4 \), getting \( y^2 + 3y - 4 \), which factors into \( (y + 4)(y - 1) \).

<prompt_2.2> Verify the entire process step-by-step to ensure accuracy. </prompt_2.2>
</Plan_2>

<Execution_2>
<execution_2.1>
1. \( x^2 + 2x + 2 \) evaluated at 1: \( 1 + 2 + 2 = 5 \).
2. \( x^2 - 2x + 2 \) evaluated at 1: \( 1 - 2 + 2 = 1 \).
3. \( x - 1 \) evaluated at 1: \( 1 - 1 = 0 \).
4. \( x + 1 \) evaluated at 1: \( 1 + 1 = 2 \).
5. \( x^2 + 1 \) evaluated at 1: \( 1 + 1 = 2 \).

Adding these up: 5 + 1 + 0 + 2 + 2 = 10.
</execution_2.1>

<execution_2.2>
That seems correct. Then substituting back, we get the quartic factors. Then factoring each quartic further. The \( x^4 - 1 \) breaks down into linear and quadratic factors, and \( x^4 + 4 \) into two quadratics. All checks out. Each quadratic is irreducible because of negative discriminant. Then evaluating at x=1:

- \( x^2 + 2x + 2 \) at 1: 1 + 2 + 2 = 5.
- \( x^2 - 2x + 2 \) at 1: 1 - 2 + 2 = 1.
- \( x - 1 \) at 1: 0.
- \( x + 1 \) at 1: 2.
- \( x^2 + 1 \) at 1: 2.

Adding them up: 5 + 1 + 0 + 2 + 2 = 10.
</execution_2.2>
</Execution_2>

<Plan_3>
Based on execution_2.1:
Wait, but let me verify if all these factors are indeed irreducible. For polynomials of degree 2, they are irreducible over the integers if their discriminants are not perfect squares.

<prompt_3.1> Verify the irreducibility of each factor using their discriminants. </prompt_3.1>
-----

Based on execution_2.1, execution_2.2:
But let me check if there's another way to factor the original polynomial that might result in different factors, but I don't think so.

Original polynomial: \( x^8 + 3x^4 - 4 \). The way to factor it is by substitution, which seems the only straightforward method.

Alternatively, could there be a different factorization over integers? For example, could \( x^8 + 3x^4 - 4 \) factor into four quadratics instead of the current combination? Let's see.

<prompt_3.2> Explore alternative factorizations and confirm the final answer. </prompt_3.2>
</Plan_3>

<Execution_3>
<execution_3.1>
Let's check each quadratic:

1. \( x^2 + 2x + 2 \): Discriminant \( D = 4 - 8 = -4 \), which is not a perfect square in integers, so irreducible.
2. \( x^2 - 2x + 2 \): Discriminant same as above, \( D = 4 - 8 = -4 \), irreducible.
3. \( x - 1 \): Linear, irreducible.
4. \( x + 1 \): Linear, irreducible.
5. \( x^2 + 1 \): Discriminant \( D = 0 - 4 = -4 \), irreducible.

So all factors are indeed irreducible. Therefore, the factorization is complete, and the sum is 10.
</execution_3.1>

<execution_3.2>
If we suppose that, but given the substitution earlier, it's more natural to factor as quadratic times quadratic. However, given that \( x^4 - 1 \) further factors into linear terms, which are degree 1, and \( x^4 + 4 \) factors into quadratics, the factorization is unique up to ordering and associates (since we're in a UFD). So the factorization is as we found, and hence the sum is 10.

Therefore, the answer should be 10.
</execution_3.2>
</Execution_3>

<Plan_4>
Based on execution_3.2:
**Final Answer**
\boxed{10}
</Plan_4>

<Execution_4>
No prompt to execute.
</Execution_4>

<Final_answer>
Therefore, the sum of the evaluated factors is $\boxed{10}$.
</Final_answer>
</think>