p1. Determine all pairs of non -negative integers $ (a, b, x, y) $ that satisfy the system of equations: 
$a + b = xy$
$x + y = ab$


p2. Find all pairs of real numbers $(x, y, z)$ that satisfy the system of equations
$x=1+\sqrt{y-z^2}$
$y=1+\sqrt{z-x^2}$
$z=1+\sqrt{x-y^2}$

p3. A man has $6$ friends. One night at a restaurant, he met with each of them $11$ times, every $2$ of them $6$ times, every $3$ of them $4$ times, every $4$ of them $3$ times, every $5$ of them $3$ times, and all of them $10$ times. He eats out $9$ times without meeting them. How many times did he eat at the restaurant in total?

[url=https://artofproblemsolving.com/community/c6h2371607p19389327]p4.[/url] Given an acute triangle $ABC$. Point $H$ denotes the foot of the altitude drawn from $A$. Prove that $$AB + AC \ge  BC cos \angle BAC + 2AH sin  \angle BAC$$

p5. It is known that $p_0 = 1$ and $p_i$ is the i-th prime number, for $i = 1, 2, ... $ namely $p_1 = 2$, $p_2 = 3$,$...$.
The prime number of $p_i$ is said to be [i]moderate [/i]if $$p_i^{(n^2)} >p_{i-1} (n!)^4$$ for all positive integers $n$. Determine all moderate prime numbers.
To solve the system of equations:
\[ x = 1 + \sqrt{y - z^2} \]
\[ y = 1 + \sqrt{z - x^2} \]
\[ z = 1 + \sqrt{x - y^2} \]

we start by noting that since the left-hand sides (LHS) of the equations are real numbers, the right-hand sides (RHS) must also be real. This implies that the expressions under the square roots must be non-negative:
\[ y - z^2 \geq 0 \]
\[ z - x^2 \geq 0 \]
\[ x - y^2 \geq 0 \]

From these inequalities, we get:
\[ y \geq z^2 \]
\[ z \geq x^2 \]
\[ x \geq y^2 \]

Next, we can chain these inequalities together:
\[ y \geq z^2 \geq (x^2)^2 = x^4 \]
\[ z \geq x^2 \geq (y^2)^2 = y^4 \]
\[ x \geq y^2 \geq (z^2)^2 = z^4 \]

Thus, we have:
\[ y \geq x^4 \]
\[ z \geq y^4 \]
\[ x \geq z^4 \]

Combining these, we get:
\[ y \geq y^8 \]
\[ z \geq z^8 \]
\[ x \geq x^8 \]

Since \( y \geq y^8 \), \( z \geq z^8 \), and \( x \geq x^8 \), and knowing that \( 0 \leq y, z, x \leq 1 \), the only values that satisfy these inequalities are \( y = 1 \), \( z = 1 \), and \( x = 1 \).

To verify, we substitute \( x = 1 \), \( y = 1 \), and \( z = 1 \) back into the original equations:
\[ x = 1 + \sqrt{y - z^2} \Rightarrow 1 = 1 + \sqrt{1 - 1^2} \Rightarrow 1 = 1 + \sqrt{0} \Rightarrow 1 = 1 \]
\[ y = 1 + \sqrt{z - x^2} \Rightarrow 1 = 1 + \sqrt{1 - 1^2} \Rightarrow 1 = 1 + \sqrt{0} \Rightarrow 1 = 1 \]
\[ z = 1 + \sqrt{x - y^2} \Rightarrow 1 = 1 + \sqrt{1 - 1^2} \Rightarrow 1 = 1 + \sqrt{0} \Rightarrow 1 = 1 \]

Thus, the solution \( (x, y, z) = (1, 1, 1) \) satisfies all the equations.

The final answer is \( \boxed{ (x, y, z) = (1, 1, 1) } \).