1. **Problem Statement:** We need to prove that either $2 \cdot 10^{500} + 15$ or $2 \cdot 10^{500} + 16$ is not a perfect square. 2. **Key Idea:** Two consecutive integers cannot both be perfect squares because the difference between consecutive perfect squares grows as the numbers get larger. 3. **Step 1: Understand the numbers involved.** - Let $N = 2 \cdot 10^{500}$. - The two numbers are $N + 15$ and $N + 16$. 4. **Step 2: Consider the difference between the two numbers.** - The difference is $(N + 16) - (N + 15) = 1$. 5. **Step 3: Use the property of perfect squares.** - If both $N + 15$ and $N + 16$ were perfect squares, say $a^2$ and $b^2$ respectively, then their difference would be: $$b^2 - a^2 = (b - a)(b + a) = 1$$ 6. **Step 4: Analyze the factorization.** - Since $b$ and $a$ are integers, $(b - a)$ and $(b + a)$ are integers. - The product of two integers is 1 only if both are 1 or both are -1. - But $b + a > b - a$ and both are positive (since squares are non-negative), so the only possibility is: $$b - a = 1 \quad \text{and} \quad b + a = 1$$ - Adding these two equations gives: $$2b = 2 \Rightarrow b = 1$$ - Substituting back: $$1 + a = 1 \Rightarrow a = 0$$ 7. **Step 5: Check if this is possible.** - $a = 0$ and $b = 1$ means the two perfect squares are $0^2 = 0$ and $1^2 = 1$. - But our numbers $N + 15$ and $N + 16$ are huge (on the order of $10^{500}$), so this is impossible. 8. **Conclusion:** - Therefore, it is impossible for both $N + 15$ and $N + 16$ to be perfect squares. - Hence, at least one of them is not a perfect square. 9. **Type of proof:** - This is a **nonconstructive proof** because it shows the existence of a non-perfect square number among the two without explicitly identifying which one it is. **Final answer:** Either $2 \cdot 10^{500} + 15$ or $2 \cdot 10^{500} + 16$ is not a perfect square, and the proof is nonconstructive.