Which statement is not true for at least one prime $p$?

$ \textbf{(A)}\ \text{If } x^2+x+3 \equiv 0 \pmod p \text{ has a solution, then } \\ \qquad x^2+x+25 \equiv 0 \pmod p \text{ has a solution.}
\\ \\ \textbf{(B)}\ \text{If } x^2+x+3 \equiv 0 \pmod p \text{ does not have a solution, then} \\ \qquad x^2+x+25 \equiv 0 \pmod p \text{ has no solution}
\\ \\ \qquad\textbf{(C)}\ \text{If } x^2+x+25 \equiv 0 \pmod p \text{ has a solution, then} \\ \qquad x^2+x+3 \equiv 0 \pmod p \text{ has a solution}.
\\ \\ \qquad\textbf{(D)}\ \text{If } x^2+x+25 \equiv 0 \pmod p \text{ does not have a solution, then} \\ \qquad x^2+x+3 \equiv 0 \pmod p \text{ has no solution. }
\\ \\ \qquad\textbf{(E)}\ \text{None}
$
To determine which statement is not true for at least one prime \( p \), we need to analyze the given quadratic congruences modulo \( p \).

1. **Analyzing the Quadratic Congruences:**
   - For \( x^2 + x + 3 \equiv 0 \pmod{p} \):
     \[
     x_{1,2} \equiv \frac{-1 \pm \sqrt{1 - 4 \cdot 1 \cdot 3}}{2 \cdot 1} \equiv \frac{-1 \pm \sqrt{-11}}{2} \pmod{p}
     \]
     The discriminant is \( \Delta_1 = -11 \).

   - For \( x^2 + x + 25 \equiv 0 \pmod{p} \):
     \[
     x_{1,2} \equiv \frac{-1 \pm \sqrt{1 - 4 \cdot 1 \cdot 25}}{2 \cdot 1} \equiv \frac{-1 \pm \sqrt{-99}}{2} \pmod{p}
     \]
     The discriminant is \( \Delta_2 = -99 \).

2. **Existence of Solutions:**
   - \( x^2 + x + 3 \equiv 0 \pmod{p} \) has a solution if and only if \( -11 \) is a quadratic residue modulo \( p \).
   - \( x^2 + x + 25 \equiv 0 \pmod{p} \) has a solution if and only if \( -99 \) is a quadratic residue modulo \( p \).

3. **Relationship Between Discriminants:**
   - Notice that \( -99 = -11 \cdot 9 \). Since \( 9 \equiv 3^2 \pmod{p} \), we have:
     \[
     -99 \equiv -11 \cdot 3^2 \pmod{p}
     \]
   - Therefore, if \( -11 \) is a quadratic residue modulo \( p \), then \( -99 \) is also a quadratic residue modulo \( p \), and vice versa.

4. **Checking Specific Primes:**
   - For \( p = 2 \):
     - \( x^2 + x + 3 \equiv 0 \pmod{2} \) simplifies to \( x^2 + x + 1 \equiv 0 \pmod{2} \), which has no solutions.
     - \( x^2 + x + 25 \equiv 0 \pmod{2} \) simplifies to \( x^2 + x + 1 \equiv 0 \pmod{2} \), which also has no solutions.
   - For \( p = 3 \):
     - \( x^2 + x + 3 \equiv 0 \pmod{3} \) simplifies to \( x^2 + x \equiv 0 \pmod{3} \), which has solutions \( x \equiv 0 \) or \( x \equiv 2 \pmod{3} \).
     - \( x^2 + x + 25 \equiv 0 \pmod{3} \) simplifies to \( x^2 + x + 1 \equiv 0 \pmod{3} \), which has solutions \( x \equiv 1 \) or \( x \equiv 2 \pmod{3} \).

5. **Conclusion:**
   - For all primes \( p \), the statements are true. Therefore, the answer is \( \boxed{E} \).