1. We consider the equation (E) to solve in $\mathbb { Z }$ : $$7 x - 5 y = 1$$ a. Verify that the pair (3; 4) is a solution of (E). b. Show that the pair of integers $( x ; y )$ is a solution of (E) if and only if $7 ( x - 3 ) = 5 ( y - 4 )$. c. Show that the integer solutions of the equation (E) are exactly the pairs ( $x ; y$ ) of relative integers such that: $$\left\{ \begin{array} { l }
x = 5 k + 3 \\
y = 7 k + 4
\end{array} \text { where } k \in \mathbb { Z } . \right.$$
2. A box contains 25 tokens, some red, some green and some white. Out of the 25 tokens there are $x$ red tokens and $y$ green tokens. Knowing that $7 x - 5 y = 1$, what can be the numbers of red, green and white tokens?
In the rest, we will assume that there are 3 red tokens and 4 green tokens.
3. We consider the following random walk of a pawn on a triangle $A B C$. At each step, we randomly draw one of the tokens from the 25, then put it back in the box.
- When at A: If the token drawn is red, the pawn goes to B. If the token drawn is green, the pawn goes to C. If the token drawn is white, the pawn stays at A.
- When at B: If the token drawn is red, the pawn goes to A. If the token drawn is green, the pawn goes to C. If the token drawn is white, the pawn stays at B.
- When at C: If the token drawn is red, the pawn goes to A. If the token drawn is green, the pawn goes to B. If the token drawn is white, the pawn stays at C.
Initially, the pawn is on vertex A. For any natural integer $n$, we denote $a _ { n } , b _ { n }$ and $c _ { n }$ the probabilities that the pawn is respectively on vertices $\mathrm { A } , \mathrm { B }$ and C at step $n$. We denote $X _ { n }$ the row matrix $\left( a _ { n } \quad b _ { n } \quad c _ { n } \right)$ and $T$ the matrix $\left( \begin{array} { l l l } 0,72 & 0,12 & 0,16 \\ 0,12 & 0,72 & 0,16 \\ 0,12 & 0,16 & 0,72 \end{array} \right)$. Give the row matrix $X _ { 0 }$ and show that, for any natural integer $n$, $X _ { n + 1 } = X _ { n } T$.
4. We admit that $T = P D P ^ { - 1 }$ where $P ^ { - 1 } = \left( \begin{array} { c c c } \frac { 3 } { 10 } & \frac { 37 } { 110 } & \frac { 4 } { 11 } \\ \frac { 1 } { 10 } & - \frac { 1 } { 10 } & 0 \\ 0 & \frac { 1 } { 11 } & - \frac { 1 } { 11 } \end{array} \right)$ and $D = \left( \begin{array} { c c c } 1 & 0 & 0 \\ 0 & 0,6 & 0 \\ 0 & 0 & 0,56 \end{array} \right)$.