To prove that $2015^{12} + 2^{10}$ is not a prime number, we need to show that it can be expressed as the product of two or more integers greater than 1. Let's proceed step by step:

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Step 1: Express $2015^{12}$ in terms of its prime factorization.

The number $2015$ can be factored as: $2015 = 5 \times 13 \times 31$ Thus: $2015^{12} = (5 \times 13 \times 31)^{12} = 5^{12} \cdot 13^{12} \cdot 31^{12}$

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Step 2: Rewrite $2015^{12} + 2^{10}$ as a sum.

Let: $N = 2015^{12} + 2^{10}$ Substitute: $N = (5^{12} \cdot 13^{12} \cdot 31^{12}) + 2^{10}$

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Step 3: Check divisibility or factorization.

To prove $N$ is not a prime number, we check if $N$ is divisible by smaller integers or find two factors that are greater than 1.

Observation:

Since $2015^{12}$ is even (all prime factors raised to $12$ are powers of $2$ or higher), and $2^{10} = 1024$, the entire sum $N$ is even. Therefore: $N \text{ is divisible by } 2.$ Since $N$ is divisible by $2$ and greater than $2$, $N$ cannot be prime.

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Conclusion:

$2015^{12} + 2^{10} \text{ is not a prime number because it is even and divisible by 2.}$

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Would you like a deeper explanation of how this divisibility works or additional analysis of $N$? 😊

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5 Related Questions:

1. How can you determine the primality of large numbers in general?
2. What are some techniques to check divisibility quickly?
3. How does prime factorization assist in proving non-primality?
4. Can modular arithmetic simplify this proof further?
5. How would you compute $2015^{12} + 2^{10}$ explicitly?

Tip:

When working with large exponents, factorize the base first and look for patterns or divisibility rules to simplify calculations.