Consider the function $f(x)=5x^4-12x^3+30x^2-12x+5$. Let $f(x_1)=p$, wher $x_1$ and $p$ are non-negative integers, and $p$ is prime. Find with proof the largest possible value of $p$.
[i]Proposed by tkhalid[/i]
1. **Rewrite the polynomial using Sophie Germain Identity:**
   The given polynomial is \( f(x) = 5x^4 - 12x^3 + 30x^2 - 12x + 5 \). We can rewrite this polynomial using the Sophie Germain Identity:
   \[
   f(x) = (x+1)^4 + 4(x-1)^4
   \]
   This identity states that \( a^4 + 4b^4 = (a^2 - 2ab + 2b^2)(a^2 + 2ab + 2b^2) \).

2. **Factor the polynomial:**
   Using the Sophie Germain Identity, we can factor \( f(x) \) as follows:
   \[
   f(x) = (x^2 - 2x + 5)(5x^2 - 2x + 1)
   \]

3. **Determine when \( f(x) \) is prime:**
   For \( f(x) \) to be prime, one of the factors must be 1, and the other must be a prime number. We need to solve the equations:
   \[
   x^2 - 2x + 5 = 1 \quad \text{or} \quad 5x^2 - 2x + 1 = 1
   \]

4. **Solve \( x^2 - 2x + 5 = 1 \):**
   \[
   x^2 - 2x + 5 = 1 \implies x^2 - 2x + 4 = 0
   \]
   Solving this quadratic equation:
   \[
   x = \frac{2 \pm \sqrt{4 - 16}}{2} = \frac{2 \pm \sqrt{-12}}{2} = 1 \pm i\sqrt{3}
   \]
   Since \( x \) must be a real number, there are no real solutions for this equation.

5. **Solve \( 5x^2 - 2x + 1 = 1 \):**
   \[
   5x^2 - 2x + 1 = 1 \implies 5x^2 - 2x = 0 \implies x(5x - 2) = 0
   \]
   Solving this equation:
   \[
   x = 0 \quad \text{or} \quad x = \frac{2}{5}
   \]

6. **Evaluate \( f(x) \) at the solutions:**
   - For \( x = 0 \):
     \[
     f(0) = 5(0)^4 - 12(0)^3 + 30(0)^2 - 12(0) + 5 = 5
     \]
   - For \( x = \frac{2}{5} \):
     \[
     f\left(\frac{2}{5}\right) = 5\left(\frac{2}{5}\right)^4 - 12\left(\frac{2}{5}\right)^3 + 30\left(\frac{2}{5}\right)^2 - 12\left(\frac{2}{5}\right) + 5
     \]
     Simplifying this:
     \[
     f\left(\frac{2}{5}\right) = 5 \left(\frac{16}{625}\right) - 12 \left(\frac{8}{125}\right) + 30 \left(\frac{4}{25}\right) - 12 \left(\frac{2}{5}\right) + 5
     \]
     \[
     = \frac{80}{625} - \frac{96}{125} + \frac{120}{25} - \frac{24}{5} + 5
     \]
     \[
     = \frac{80}{625} - \frac{480}{625} + \frac{3000}{625} - \frac{3000}{625} + \frac{3125}{625}
     \]
     \[
     = \frac{80 - 480 + 3000 - 3000 + 3125}{625} = \frac{3725}{625} = 5.96
     \]
     This is not an integer, so it is not a valid solution.

7. **Conclusion:**
   The only valid solution is \( x = 0 \), and \( f(0) = 5 \).

The final answer is \( \boxed{5} \).