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En utilisant l’algorithme d’Euclide étendu, montrez que $169$ et $45$ sont premiers entre eux et trouvez deux entiers relatifs $u$ et $v$ tels que $169u+ 45v = 1$ + +$169=45\times3+34$ +$45=34\times1+11$ +$34=11\times3+1$ +$11=1\times11+0$ + +Le $PGDC$ est égal à $1$, donc $169$ et $45$ sont premiers entre eux. + +Trouvons $u$ et $v$. Formule: $\boxed{u_{k}=u_{k-2}-u_{k-1}\times q_{k}}$ et $\boxed{v_{k}=v_{k-2}-v_{k-1}\times q_{k}}$ + +| | $u_k$ | $v_k$ | +| ------------------ | ---------- | ---------- | +| | 1 | 0 | +| | 0 | 1 | +| $169=45\times3+34$ | 1 | -3 | +| $45=34\times1+11$ | -1 | 4 | +| $34=11\times3+1$ | 4 | -15 | +| $11=1\times11+0$ | ////////// | ////////// | + +Bilan: $\boxed{au+bv=PGDC(a,b)}$ +Donc, $169\times4+45\times(-15)=1$ + +#### 2. Retrouvez le fait que $169$ et $45$ sont premiers entre eux en utilisant les décompositions en produit de facteurs premiers de $169$ et $45$. + +$169=1\times13^2$ +$45=1\times3^2\times5$ +Le seul diviseur commun est $1$. + +#### 3. Déterminez l’ensemble des couples $(x;y)$ d’entiers relatifs vérifiant les deux conditions suivantes: $(x,y)\in\mathbb{Z}^2,\cases{169x+45y=1 \\ x\in [100,200]\Leftrightarrow x\in\mathbb{Z} \text{ et } 100≤x≤200}$ +Solution particulière: $x_{p}=4$ et $y_{p}=-15$ +$(H)=169x+45y=0$ + +Si $x=45$ et $y=-169$, ça marche. + +Solution homogène: $\cases{x=45k \\ y=-169k}\forall k\in\mathbb{Z}$ +Solution générale: $\cases{x=45k+4 \\ y=-169k-15}\forall k\in\mathbb{Z}$ +Contrainte: $100≤45k+4≤200 \rightarrow96≤45k≤196\rightarrow\frac{96}{45}≤k≤\frac{196}{45}\rightarrow\approx2.≤k≤\approx4.$ + +Avec $k=3$: $x=45\times3+4=139$ et $y-522$ +Avec $k=4$: $x=184$ et $y-691$ + +#### 4. Inverse de $\cases{\overline{34}=\overline{169} \text{ dans }\mathbb{Z}/45\mathbb{Z} \\ \overline{45} \text{ dans }\mathbb{Z}/169\mathbb{Z}}$ + +$a\in(\mathbb{E},+_{\mathbb{E}},\times_{\mathbb{E}})$, $\exists a^{-1}\in\mathbb{E}$ tel que $a\times a^{-1}=1_{\mathbb{E}}$ + +On cherche $u\in\mathbb{Z}/45\mathbb{Z}$ tel que $\overline{34}\times u=\overline{1}\Leftrightarrow\overline{169}\times u=\overline{1}$ +On a vue que $169\times4+45\times(-15)=1$, mais dans $\mathbb{Z}/45\mathbb{Z}$, $45\times(-15)=0$ +Donc $\overline{169}\times\overline{4}=\overline{1}$. +L'inverse de $\overline{34}$, c'est-à-dire l'inverse de $\overline{169}$ dans $\mathbb{Z}/45\mathbb{Z}$ est $4$. $\overline{34}^{-1}=\overline{4}$ + +On cherche $v\in\mathbb{Z}/169\mathbb{Z}$ tel que $\overline{45}\times v=\overline{1}$. +On a vu que $169\times4+45\times(-15)=1$, mais dans $\mathbb{Z}/169\mathbb{Z}$, $169\times4=0$ +Donc $\overline{45}\times(\overline{-15})=\overline{1}$. +L'inverse de $\overline{45}$ dans $\mathbb{Z}/169\mathbb{Z}$ est $\overline{-15}$ soit $\overline{154}$. + +#### 5. $\cases{\mathbb{A}=\mathbb{M}_2(\mathbb{Z}/45\mathbb{Z}) \\ U=\pmatrix{\overline{9}&\overline{2} \\ \overline{1}&\overline{4}}}$ + +$\det{U}=\overline{9}\times\overline{4}-\overline{1}\times\overline{2}=\overline{34}$ + +$\boxed{\forall A\in M_n(\mathbb{R}): \exists A^{-1}\text{ si }\det{A}≠0}$ +$\boxed{\forall A\in M_n(\mathbb{Z}/n\mathbb{Z}): \exists A^{-1}\text{ si }\exists\det{A}^{-1}}$ + +On a $\overline{34}^{-1}=\overline{4}$ donc $\exists U^{-1}$ +$U^{-1}=\overline{4}\pmatrix{\overline{4}&\overline{-2} \\ \overline{-1}&\overline{9}}=\pmatrix{\overline{16}&\overline{-8} \\ \overline{-4}&\overline{36}}=\pmatrix{\overline{16}&\overline{37} \\ \overline{41}&\overline{36}}$ + +#### 6. Trouver $(x,y)\in(\mathbb{Z}/45\mathbb{Z})^2$ tel que $\cases{\overline{9}x+\overline{2}y=\overline{1} \\ x + \overline{4}y=\overline{2}}$ +$\Leftrightarrow U\times\pmatrix{x\\y}=\pmatrix{\overline{1}\\\overline{2}}$ +$\Leftrightarrow\pmatrix{x\\y}=U^{-1}\times\pmatrix{\overline{1}\\\overline{2}}=\pmatrix{\overline{0}\\\overline{68}}=\pmatrix{\overline{0}\\\overline{23}}$ +car $\exists U^{-1}$ + +$x=\overline{0}$ et $y=\overline{23}$. + + +# Exercice 3: + +#### 1. $+$ et $\times$ de $\mathbb{Z}/5\mathbb{Z}$ + +| $+$ | $\overline{0}$ | $\overline{1}$ | $\overline{2}$ | $\overline{3}$ | $\overline{4}$ | +| -------------- | -------------- | -------------- | -------------- | -------------- | -------------- | +| $\overline{0}$ | $\overline{0}$ | $\overline{1}$ | $\overline{2}$ | $\overline{3}$ | $\overline{4}$ | +| $\overline{1}$ | $\overline{1}$ | $\overline{2}$ | $\overline{3}$ | $\overline{4}$ | $\overline{0}$ | +| $\overline{2}$ | $\overline{2}$ | $\overline{3}$ | $\overline{4}$ | $\overline{0}$ | $\overline{1}$ | +| $\overline{3}$ | $\overline{3}$ | $\overline{4}$ | $\overline{0}$ | $\overline{1}$ | $\overline{2}$ | +| $\overline{4}$ | $\overline{4}$ | $\overline{0}$ | $\overline{1}$ | $\overline{2}$ | $\overline{3}$ | + +| $\times$ | $\overline{0}$ | $\overline{1}$ | $\overline{2}$ | $\overline{3}$ | $\overline{4}$ | +| -------------- | -------------- | -------------- | -------------- | -------------- | -------------- | +| $\overline{0}$ | $\overline{0}$ | $\overline{0}$ | $\overline{0}$ | $\overline{0}$ | $\overline{0}$ | +| $\overline{1}$ | $\overline{0}$ | $\overline{1}$ | $\overline{2}$ | $\overline{3}$ | $\overline{4}$ | +| $\overline{2}$ | $\overline{0}$ | $\overline{2}$ | $\overline{4}$ | $\overline{1}$ | $\overline{3}$ | +| $\overline{3}$ | $\overline{0}$ | $\overline{3}$ | $\overline{1}$ | $\overline{4}$ | $\overline{2}$ | +| $\overline{4}$ | $\overline{0}$ | $\overline{4}$ | $\overline{3}$ | $\overline{2}$ | $\overline{1}$ | + + +#### 2. $x\in\cases{\mathbb{Z}/2\mathbb{Z} \\ \mathbb{Z}/3\mathbb{Z} \\ \mathbb{Z}/5\mathbb{Z}}$ tel que $x^2+\overline{1}=\overline{0}$ + +Pour $\mathbb{Z}/2\mathbb{Z}$: + +| $x$ | $\overline{0}$ | $\overline{1}$ | +| ------------------ | -------------- | -------------- | +| $x^2+\overline{1}$ | $\overline{1}$ | $\overline{0}$ | +$S_2=\{\overline{1}\}$ + +Pour $\mathbb{Z}/3\mathbb{Z}$: + +| $x$ | $\overline{0}$ | $\overline{1}$ | $\overline{2}$ | +| ------------------ | -------------- | -------------- | -------------- | +| $x^2+\overline{1}$ | $\overline{1}$ | $\overline{2}$ | $\overline{2}$ | +$S_3=\varnothing$ + +Pour $\mathbb{Z}/5\mathbb{Z}$: + +| $x$ | $\overline{0}$ | $\overline{1}$ | $\overline{2}$ | $\overline{3}$ | $\overline{4}$ | +| ------------------ | -------------- | -------------- | -------------- | -------------- | -------------- | +| $x^2+\overline{1}$ | $\overline{1}$ | $\overline{2}$ | $\overline{0}$ | $\overline{0}$ | $\overline{2}$ | +$S_5=\{\overline{2},\overline{3}\}$ + + +# Exercice 2: +On cherche $x\in[0,100]$ tel que $\cases{x\equiv1[3] \\ x\equiv2[5] \\ x\equiv-1[7]}$ +$3$, $5$ et $7$ sont premiers et différents, donc premiers entre eux, donc d'après le théorème Chinois, il existe des solutions. + +En prenant $\cases{x\equiv1[3] \\ x\equiv2[5]}$, +$\exists k_1\in\mathbb{Z}$ tel que $x=1+3k_1$ +$\exists k_2\in\mathbb{Z}$ tel que $x=2+5k_2$ +$1+3k_1=2+5k_2$ +$3k_1-5k_2=1$ +Solution évidente: on peut prendre $k_1=2$ et $k_2=1$. +$x_p=1+2\times3=2+5=7$. +Solution générale: $x=7+15k\,\forall k\in\mathbb{Z}$. \ No newline at end of file