Problem 1. Solve in $Z$ the equation: $\sqrt{x^{2}-2000}+\sqrt{2074-x^{2}}=|44 x+1992|$.
## Solution. 1p by default

Condition for the existence of the root:

$$
\begin{array}{r}
x^{2}-2000 \geq 0 \Rightarrow x^{2} \geq 2000 \\
2074-x^{2} \geq 0 \Rightarrow x^{2} \leq 2074, \text { so } \Rightarrow 2000 \leq x^{2} \leq 2074
\end{array}
$$

(1p)

Since $x \in \mathrm{Z}$ and $2000 \leq x^{2} \leq 2074 \Rightarrow x \in\{-45 ; 45\}$ $\qquad$
(1p)

For:

$x= \pm 45 \Rightarrow \sqrt{x^{2}-2000}=\sqrt{25}=5$ and $\sqrt{2074-x^{2}}=\sqrt{49}=7$, so $12=|44 x+1992|$ $\qquad$ (2p)

If $x=45$ we get $12=|44 \cdot 45+1992|(F)$

If $x=-45$ we get $12=|44 \cdot(-45)+1992|(T)$, and thus $x=-45$

Therefore $S=\{-45\}$