To determine the integer \( k \) for which \( 11^k + 4^{k+1} \) is prime, let's analyze the expression for different values of \( k \).

First, consider \( k = 0 \):
\[ 11^0 + 4^{0+1} = 1 + 4 = 5 \]
Since 5 is a prime number, \( k = 0 \) works.

Next, let's consider \( k = 1 \):
\[ 11^1 + 4^{1+1} = 11 + 4^2 = 11 + 16 = 27 \]
27 is not a prime number.

Now consider \( k = 2 \):
\[ 11^2 + 4^{2+1} = 121 + 4^3 = 121 + 64 = 185 \]
185 is not a prime number.

Let's continue for \( k = 3 \):
\[ 11^3 + 4^{3+1} = 1331 + 4^4 = 1331 + 256 = 1587 \]
1587 is not a prime number.

Since higher values of \( k \) will result in increasingly large numbers, we observe that the expression grows quickly, making it increasingly unlikely to be prime as \( k \) increases.

From our calculations, we can see that the only value of \( k \) that makes \( 11^k + 4^{k+1} \) a prime number is \( k = 0 \).

Thus, the integer \( k \) for which \( 11^k + 4^{k+1} \) is prime is:
\[ k = 0 \]

Would you like more details or have any questions? Here are some questions you might find interesting:

1. How do we determine if a large number is prime?
2. Can we generalize this method to other forms of exponential expressions?
3. What is the significance of prime numbers in number theory?
4. How does the growth rate of exponential functions affect primality?
5. What are other famous sequences or formulas involving prime numbers?

**Tip:** When checking for primality in expressions, evaluating small values first can often provide quick insights before delving into more complex analysis.

Explore how to determine the integer k for which the expression 11^k + 4^(k+1) is prime. Includes step-by-step calculations and insights into exponential functions and primality testing. Suitable for middle school students interested in advanced arithmetic.

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