Morteza wishes to take two real numbers $S$ and $P$, and then to arrange six pairwise distinct real numbers on a circle so that for each three consecutive numbers at least one of the two following conditions holds: 
1) their sum equals $S$
2) their product equals $P$.
 Determine if Morteza’s wish could be fulfilled.
To determine if Morteza’s wish could be fulfilled, we need to analyze the conditions given for the arrangement of six pairwise distinct real numbers on a circle. We will consider two cases: \( P \neq 0 \) and \( P = 0 \).

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

Let \( x_1, x_2, \dots, x_6 \) be the six pairwise distinct real numbers arranged on a circle. According to the problem, for each three consecutive numbers, at least one of the following conditions must hold:
1. Their sum equals \( S \).
2. Their product equals \( P \).

Since the numbers are pairwise distinct, two consecutive triplets cannot follow the same condition. Otherwise, we would have:
\[ x_1 + x_2 + x_3 = x_2 + x_3 + x_4 \quad \text{or} \quad x_1 x_2 x_3 = x_2 x_3 x_4 \]
which implies:
\[ x_1 = x_4 \quad \text{(if sums are equal)} \quad \text{or} \quad x_1 = x_4 \quad \text{(if products are equal)} \]
This leads to a contradiction since the numbers are pairwise distinct.

Assume without loss of generality (WLOG) that:
\[ x_1 + x_2 + x_3 = S \]
\[ x_3 + x_4 + x_5 = S \]
\[ x_5 + x_6 + x_1 = S \]
and
\[ x_2 x_3 x_4 = P \]
\[ x_4 x_5 x_6 = P \]
\[ x_6 x_1 x_2 = P \]

From the sum conditions, we have:
\[ x_1 + x_2 + x_3 = x_3 + x_4 + x_5 \implies x_1 + x_2 = x_4 + x_5 \]
\[ x_3 + x_4 + x_5 = x_5 + x_6 + x_1 \implies x_3 + x_4 = x_6 + x_1 \]

From the product conditions, we have:
\[ x_2 x_3 x_4 = x_4 x_5 x_6 \implies x_2 x_3 = x_5 x_6 \]
\[ x_4 x_5 x_6 = x_6 x_1 x_2 \implies x_4 x_5 = x_1 x_2 \]

Squaring the sum condition \( x_1 + x_2 = x_4 + x_5 \) and using the product condition \( x_1 x_2 = x_4 x_5 \), we get:
\[ (x_1 + x_2)^2 = (x_4 + x_5)^2 \]
\[ x_1^2 + 2x_1 x_2 + x_2^2 = x_4^2 + 2x_4 x_5 + x_5^2 \]
Since \( x_1 x_2 = x_4 x_5 \), we have:
\[ x_1^2 + x_2^2 = x_4^2 + x_5^2 \]
This implies:
\[ (x_1 - x_2)^2 = (x_4 - x_5)^2 \]
Thus:
\[ x_1 - x_2 = \pm (x_4 - x_5) \]
This leads to either \( x_1 = x_4 \) or \( x_1 = x_5 \), which is a contradiction since the numbers are pairwise distinct.

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

If \( P = 0 \), then at least one of the products of three consecutive numbers must be zero. This means at least one of the \( x_i \)'s must be zero. Assume WLOG that \( x_1 = 0 \). Then, at least one of the following must be zero:
\[ x_2 x_3 x_4 = 0 \]
\[ x_3 x_4 x_5 = 0 \]
\[ x_4 x_5 x_6 = 0 \]
This implies that at least one of \( x_2, x_3, x_4, x_5, x_6 \) must be zero. However, this would mean that there are at least two zeros among the six numbers, which contradicts the condition that the numbers are pairwise distinct.

Therefore, no such arrangement of six pairwise distinct real numbers exists that satisfies the given conditions.

The final answer is \(\boxed{\text{False}}\).