Find all polynomials $P(x)$ with real coefficients such that for all pairwise distinct positive integers $x, y, z, t$ with $x^2 + y^2 + z^2 = 2t^2$ and $\gcd(x, y, z, t ) = 1,$ the following equality holds
\[2P^2(t ) + 2P(xy + yz + zx) = P^2(x + y + z) .\]

[b]Note.[/b] $P^2(k)=\left( P(k) \right)^2.$
1. **Given Condition and Initial Setup:**
   We are given the polynomial equation:
   \[
   2P^2(t) + 2P(xy + yz + zx) = P^2(x + y + z)
   \]
   for all pairwise distinct positive integers \(x, y, z, t\) such that \(x^2 + y^2 + z^2 = 2t^2\) and \(\gcd(x, y, z, t) = 1\).

2. **Simplifying the Problem:**
   To simplify, let's consider the case where \(x = y = z = t\). This simplifies the equation to:
   \[
   2P^2(t) + 2P(t^2) = P^2(3t)
   \]
   However, since \(x, y, z, t\) are distinct, we need to consider another approach. Let's assume \(x = y = 1\) and \(z = t\). This gives:
   \[
   1^2 + 1^2 + z^2 = 2t^2 \implies 2 + z^2 = 2t^2 \implies z^2 = 2t^2 - 2
   \]
   This is not always possible for integer \(z\) and \(t\). Therefore, we need to consider the general form.

3. **General Polynomial Form:**
   Assume \(P(x) = a_n x^n + a_{n-1} x^{n-1} + \ldots + a_1 x + a_0\). We need to compare the coefficients of \(x^{2n}\) on both sides of the equation:
   \[
   2P^2(t) + 2P(x^2) = P^2(2x)
   \]
   This implies:
   \[
   2(a_n t^n)^2 + 2a_n (x^2)^n = (a_n (2x)^n)^2
   \]
   Simplifying, we get:
   \[
   2a_n^2 t^{2n} + 2a_n x^{2n} = a_n^2 2^{2n} x^{2n}
   \]

4. **Coefficient Comparison:**
   Comparing the coefficients of \(x^{2n}\) on both sides, we get:
   \[
   2a_n^2 + 2a_n = 2^{2n} a_n^2
   \]
   Solving for \(a_n\), we get:
   \[
   2a_n^2 + 2a_n = 2^{2n} a_n^2 \implies a_n (2a_n + 2) = 2^{2n} a_n^2
   \]
   \[
   2a_n + 2 = 2^{2n} a_n \implies a_n (2^{2n} - 2) = 2 \implies a_n = \frac{2}{2^{2n} - 2}
   \]

5. **Induction for Lower Coefficients:**
   We now use induction to show that all lower coefficients \(a_{n-i} = 0\) for \(1 \leq i \leq n\).

   **Base Case:**
   For \(i = 1\), compare the coefficients of \(x^{2n-1}\):
   \[
   2a_n a_{n-1} = 2 \cdot 2^n a_n \cdot 2^{n-1} a_{n-1}
   \]
   Since \(a_n \neq 0\), we get \(a_{n-1} = 0\).

   **Inductive Step:**
   Assume \(a_{n-i} = 0\) for all \(1 \leq i \leq k\). We need to show \(a_{n-k-1} = 0\). Compare the coefficients of \(x^{2n-k-1}\):
   \[
   2^{2n-k} a_n a_{n-k-1} = 4a_n a_{n-k-1}
   \]
   Since \(2n-k > n \geq 2\) and \(a_n \neq 0\), we must have \(a_{n-k-1} = 0\).

6. **Conclusion:**
   By induction, all coefficients \(a_{n-i} = 0\) for \(1 \leq i \leq n\). Therefore, the polynomial \(P(x)\) must be of the form:
   \[
   P(x) = \frac{x^n}{2^{2n-1} - 1}
   \]

The final answer is \( \boxed{ P(x) = \frac{x^n}{2^{2n-1} - 1} } \).