Find all polynomials $p(x)\in\mathbb{R}[x]$ such that for all $x\in \mathbb{R}$:
$p(5x)^2-3=p(5x^2+1)$ such that:
$a) p(0)\neq 0$
$b) p(0)=0$
To find all polynomials \( p(x) \in \mathbb{R}[x] \) such that for all \( x \in \mathbb{R} \):

\[ p(5x)^2 - 3 = p(5x^2 + 1) \]

we will consider two cases: \( p(0) \neq 0 \) and \( p(0) = 0 \).

### Case 1: \( p(0) \neq 0 \)

1. **Assume \( p(x) \) is a polynomial of degree \( n \):**
   \[
   p(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0
   \]
   where \( a_n \neq 0 \).

2. **Substitute \( x = 0 \) into the given equation:**
   \[
   p(0)^2 - 3 = p(1)
   \]
   Since \( p(0) \neq 0 \), let \( p(0) = c \). Then:
   \[
   c^2 - 3 = p(1)
   \]

3. **Assume \( p(x) \) is non-constant and compare the degrees:**
   The left-hand side \( p(5x)^2 \) has degree \( 2n \), while the right-hand side \( p(5x^2 + 1) \) has degree \( n \). For these to be equal, \( n \) must be 0, which contradicts the assumption that \( p(x) \) is non-constant.

4. **Thus, \( p(x) \) must be a constant polynomial:**
   Let \( p(x) = c \). Then:
   \[
   c^2 - 3 = c
   \]
   Solving for \( c \):
   \[
   c^2 - c - 3 = 0
   \]
   Using the quadratic formula:
   \[
   c = \frac{1 \pm \sqrt{1 + 12}}{2} = \frac{1 \pm \sqrt{13}}{2}
   \]
   Therefore, the solutions are:
   \[
   c = \frac{1 + \sqrt{13}}{2} \quad \text{or} \quad c = \frac{1 - \sqrt{13}}{2}
   \]

### Case 2: \( p(0) = 0 \)

1. **Assume \( p(x) \) is a polynomial of degree \( n \):**
   \[
   p(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x
   \]
   where \( a_n \neq 0 \).

2. **Substitute \( x = 0 \) into the given equation:**
   \[
   p(0)^2 - 3 = p(1)
   \]
   Since \( p(0) = 0 \):
   \[
   -3 = p(1)
   \]

3. **Assume \( p(x) \) is non-constant and compare the degrees:**
   The left-hand side \( p(5x)^2 \) has degree \( 2n \), while the right-hand side \( p(5x^2 + 1) \) has degree \( n \). For these to be equal, \( n \) must be 0, which contradicts the assumption that \( p(x) \) is non-constant.

4. **Thus, \( p(x) \) must be a constant polynomial:**
   However, this contradicts \( p(0) = 0 \). Therefore, there are no non-constant polynomials satisfying the given conditions with \( p(0) = 0 \).

Conclusion:
The only polynomials \( p(x) \) that satisfy the given equation are the constant polynomials \( p(x) = \frac{1 + \sqrt{13}}{2} \) and \( p(x) = \frac{1 - \sqrt{13}}{2} \).

The final answer is \( \boxed{ p(x) = \frac{1 + \sqrt{13}}{2} } \) or \( p(x) = \frac{1 - \sqrt{13}}{2} \).