How many primes $p$ are there such that $2p^4-7p^2+1$ is equal to square of an integer?

$ 
\textbf{(A)}\ 0
\qquad\textbf{(B)}\ 1
\qquad\textbf{(C)}\ 4
\qquad\textbf{(D)}\ \text{Infinitely many}
\qquad\textbf{(E)}\ \text{None of the preceding}
$
1. We need to find the number of prime numbers \( p \) such that \( 2p^4 - 7p^2 + 1 \) is a perfect square. Let's denote this perfect square by \( k^2 \), where \( k \) is an integer. Therefore, we have:
   \[
   2p^4 - 7p^2 + 1 = k^2
   \]

2. First, let's check small prime numbers to see if they satisfy the equation. 

   - For \( p = 2 \):
     \[
     2(2^4) - 7(2^2) + 1 = 2 \cdot 16 - 7 \cdot 4 + 1 = 32 - 28 + 1 = 5
     \]
     Since 5 is not a perfect square, \( p = 2 \) is not a solution.

   - For \( p = 3 \):
     \[
     2(3^4) - 7(3^2) + 1 = 2 \cdot 81 - 7 \cdot 9 + 1 = 162 - 63 + 1 = 100
     \]
     Since 100 is a perfect square (\( 10^2 \)), \( p = 3 \) is a solution.

3. Next, we need to consider all other primes. All primes greater than 3 are of the form \( 6k \pm 1 \). We will analyze the expression \( 2p^4 - 7p^2 + 1 \) modulo 6 to see if it can be a perfect square.

   - If \( p \equiv 1 \pmod{6} \):
     \[
     p^2 \equiv 1 \pmod{6} \quad \text{and} \quad p^4 \equiv 1 \pmod{6}
     \]
     \[
     2p^4 - 7p^2 + 1 \equiv 2 \cdot 1 - 7 \cdot 1 + 1 \equiv 2 - 7 + 1 \equiv -4 \equiv 2 \pmod{6}
     \]
     Since 2 is not a quadratic residue modulo 6, \( 2p^4 - 7p^2 + 1 \) cannot be a perfect square for \( p \equiv 1 \pmod{6} \).

   - If \( p \equiv -1 \pmod{6} \):
     \[
     p^2 \equiv 1 \pmod{6} \quad \text{and} \quad p^4 \equiv 1 \pmod{6}
     \]
     \[
     2p^4 - 7p^2 + 1 \equiv 2 \cdot 1 - 7 \cdot 1 + 1 \equiv 2 - 7 + 1 \equiv -4 \equiv 2 \pmod{6}
     \]
     Similarly, since 2 is not a quadratic residue modulo 6, \( 2p^4 - 7p^2 + 1 \) cannot be a perfect square for \( p \equiv -1 \pmod{6} \).

4. Therefore, the only prime number \( p \) that satisfies the equation \( 2p^4 - 7p^2 + 1 = k^2 \) is \( p = 3 \).

The final answer is \(\boxed{1}\)