, 2010

Let us suppose that we can find such numbers $a, b$, that $p = a + b + 2\sqrt{ab+1}$ is prime. Then $ab+1$ is complete square, $a$ and $b$ are distinct. By Cauchy-Schwartz inequality $a + b \ge 2\sqrt{ab}$. Besides this, we have $2\sqrt{ab} + 1 > 2\sqrt{ab+1} \Rightarrow a + b \ge 2\sqrt{ab} + 1$. $p$ is an odd number, because it is greater than $2$, thus $a + b - 2\sqrt{ab} + 1$ is natural. From the equality $$(a + b + 2\sqrt{ab} + 1)(a + b - 2\sqrt{ab} + 1) = (a - b + 2)(a - b - 2) \ne 0$$ it follows, that either $a - b - 2$ or $a - b + 2$ is divisible by $p$, but absolute value of each of these two numbers does not exceed $a + b + 2$, which is less than $p$. This contradiction finishes the proof.

From the condition of the problem, it follows that $2\sqrt{ab+1}$ is a natural number, which immediately gives us that $z = \sqrt{ab+1}$ is also a natural number. Therefore, $b = \frac{z^2 - 1}{a} = \frac{(z - 1)(z + 1)}{a}$. This implies that $a$ can be factored $a = a_1 a_2$ in such a way that $z - 1$ is divisible by $a_1$, and $z + 1$ is divisible by $a_2$.

$m = a + \frac{(z - 1)(z + 1)}{a} + 2z = \frac{(a + z)^2 - 1}{a} = \frac{(a + z - 1)(a + z + 1)}{a} = \frac{a + z - 1}{a_1} + \frac{a + z + 1}{a_2}$. Both factors $\frac{a + z - 1}{a_1}$ and $\frac{a + z + 1}{a_2}$ are natural and greater than one, thus, $m$ is composite.