Problem 2. a) Determine $(x ; y) \in \mathbb{N} \times \mathbb{N}$ such that $\sqrt{2^{x}}+\sqrt{2^{y}}=33$.

b) Given the number $x=\sqrt{\underbrace{44 . .44}_{\text {2n digits }}-\underbrace{88 \ldots 88}_{\text {n digits }}}$ where $n \in \mathbb{N}^{*}$.

Show that $x$ is a natural number.
Solution:

a) If at least one of the numbers $x$ and $y$ is odd, the expression $\sqrt{2^{x}}+\sqrt{2^{y}}$ is an irrational number, so $\sqrt{2^{x}}+\sqrt{2^{y}} \neq 33$ ..... $.0,5 p$
We deduce that $x$ and $y$ are even numbers. Let $x=2 n, y=2 m$ with $m, n \in \mathbb{N}$. ..... $\mathbf{0 , 5 p}$
The given equality becomes $2^{n}+2^{m}=33$ ..... $.0,5 p$
33 is an odd number, so $2^{n}$ and $2^{m}$ have different parities. ..... $\mathbf{0 , 5 p}$
If $2^{n}=$ even and $2^{m}=$ odd $\Rightarrow m=0$, and $2^{n}+1=33 \Rightarrow 2^{n}=32 \Rightarrow n=5$
Thus, $x=10$ and $y=0$. . . ..... $\mathbf{0 , 5 p}$
If $2^{n}=$ odd and $2^{m}=$ even, the solution is $x=0$ and $y=10 \ldots$ ..... $\mathbf{0 , 2 5 p}$
The problem has two solutions: $(10 ; 0)$ and $(0 ; 10)$ ..... 0,25p
b) Let $a=\underset{\text { ncifre }}{11 \ldots 11}$.
We get $x=\sqrt{4 \cdot\left(a \cdot 10^{n}+a\right)-8 a}$ ..... $1 p$
$x=\sqrt{4 \cdot\left(a \cdot 10^{n}+a-2 a\right)}=2 \cdot \sqrt{a \cdot\left(10^{n}-1\right)}$ ..... $1 p$
It follows that $x=2 \cdot \sqrt{a \cdot \underbrace{99 \ldots 99}_{\text {ncifre }}}=2 \sqrt{a \cdot 9 a}=2 \sqrt{9 a^{2}}=2 \cdot 3 a=6 a$ ..... $1,5 p$
Thus, $x=\underbrace{66 \ldots 66}_{\text {ncifre }}$. ..... $.0,5 p$