Answer:
To prove that there exists an integer \( n \) such that \( 2n^2 - 5n + 2 \) is prime, we can consider the expression for different values of \( n \). Let's analyze the quadratic expression:
\[ 2n^2 - 5n + 2 \]
Factorizing this quadratic expression may help us identify potential values of \( n \). However, in this case, it's not easily factorizable into two linear factors with integer coefficients. So, let's explore some values of \( n \) to demonstrate the existence of an integer \( n \) for which \( 2n^2 - 5n + 2 \) is prime.
Consider \( n = 1 \):
\[ 2(1)^2 - 5(1) + 2 = 2 - 5 + 2 = -1 \]
Consider \( n = 2 \):
\[ 2(2)^2 - 5(2) + 2 = 8 - 10 + 2 = 0 \]
Consider \( n = 3 \):
\[ 2(3)^2 - 5(3) + 2 = 18 - 15 + 2 = 5 \]
Now, we've found that for \( n = 3 \), the expression \( 2n^2 - 5n + 2 \) evaluates to a prime number (5). Thus, we have demonstrated the existence of an integer \( n \) (specifically \( n = 3 \)) for which \( 2n^2 - 5n + 2 \) is prime.
