To prove the statement that there exists an integer $n$ such that $2n^2 - 5n + 2$ is prime, we need to find a specific value of $n$ for which $2n^2 - 5n + 2$ evaluates to a prime number.

Let's check a few small integer values for $n$:

1. $n = 1$:

$2(1)^2 - 5(1) + 2 = 2(1) - 5 + 2 = 2 - 5 + 2 = -1 \quad \text{(not prime)}$

2. $n = 2$:

$2(2)^2 - 5(2) + 2 = 2(4) - 10 + 2 = 8 - 10 + 2 = 0 \quad \text{(not prime)}$

3. $n = 3$:

$2(3)^2 - 5(3) + 2 = 2(9) - 15 + 2 = 18 - 15 + 2 = 5 \quad \text{(prime)}$

Here, when $n = 3$, $2n^2 - 5n + 2 = 5$, and 5 is indeed a prime number.

Conclusion:

The integer $n = 3$ satisfies the condition that $2n^2 - 5n + 2$ is prime. Therefore, this proves the statement.

Would you like more details or have any other questions?

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Here are 5 related questions you might consider:

1. What happens if you try other values of $n$?
2. Can $2n^2 - 5n + 2$ be prime for negative values of $n$?
3. How does changing the coefficients in the quadratic expression affect the result?
4. Are there other integer values of $n$ for which $2n^2 - 5n + 2$ is prime?
5. What other types of expressions can yield prime numbers?

Tip: Always check small values first when trying to prove a statement with an example, as they often reveal patterns or solutions quickly.