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Sa se rezolve in numere intregi ecuatia:

\( x^{2}+7=2^{n} \).

Ecuatia a fost propusa de Ramanujan in 1913 si rezolvata de Nagell in 1948.
Urmatoarea solutie este elementara (poate cea mai elementara), ea fiind insa destul de lunga.
Presupunem cunoscut faptul ca \( \mathbb{Z}[\frac{1+i\sqrt{7}}{2}] \) este inel euclidian si ca \( U(\mathbb{Z}[\frac{1+i\sqrt{7}}{2}])=\{\pm1\}. \)
Cazul I: n par, \( n=2m\Rightarrow 7=2^{2m}-x^2=(2^m-x)(2^m+x) \).
Dar \( 2^m-x,\ 2^m+x\in\mathbb{N},\ 2^m-x\leq2^m+x\Rightarrow \left\{ \begin{array}{ll}
2^m-x=1\\
2^m+x=7
\end{array} \right. \Rightarrow 2^{m+1}=8=2^3\Rightarrow \)
\( \Rightarrow m+1=3\Rightarrow m=2\Rightarrow n=4\Rightarrow x=\pm3. \)
Deci, \( n=4,\ x=\pm3. \)
Cazul al II-lea: n impar, x impar
\( 2^n\geq 1+7=8\Rightarrow n\geq 3 \)
\( \frac{x+i\sqrt{7}}{2}\cdot\frac{x-i\sqrt{7}}{2}=2^{n-2}=2^y \), unde \( y=n-2,\ y \) impar, \( y\geq1 \).
Daca \( n=3\Rightarrow x=\pm1. \) Presupunem in continuare ca \( y\geq3. \)
\( \frac{x+i\sqrt{7}}{2}=\frac{x-1}{2}+\frac{1+i\sqrt{7}}{2}\in\mathbb{Z}[\frac{1+i\sqrt{7}}{2}] \) si \( \frac{x-i\sqrt{7}}{2}=\frac{x+1}{2}-\frac{1+i\sqrt{7}}{2}\in\mathbb{Z}[\frac{1+i\sqrt{7}}{2}] \)
\( \frac{x-1}{2},\ \frac{x+1}{2}\in\mathbb{Z} \) deoarece \( x \) este impar.
\( 2=\frac{1+i\sqrt{7}}{2}\cdot\frac{1-i\sqrt{7}}{2} \)
Arat ca \( \frac{1+i\sqrt{7}}{2},\ \frac{1-i\sqrt{7}}{2} \) sunt prime in \( \mathbb{Z}[\frac{1+i\sqrt{7}}{2}] \) si ca \( \frac{1+i\sqrt{7}}{2},\ \frac{1-i\sqrt{7}}{2} \) nu sunt asociate in divizibilitate.
Fie \( \frac{1+i\sqrt{7}}{2}=\alpha\cdot\beta,\ \alpha,\ \beta\in\mathbb{Z}[\frac{1+i\sqrt{7}}{2}] \)
Inelul \( \mathbb{Z}[\frac{1+i\sqrt{7}}{2}] \) este euclidian cu functia \( \varphi \), unde \( \varphi(z)=z\cdot\overline{z}=|z|^2. \)
\( 2=\varphi(\frac{1+i\sqrt{7}}{2})=\varphi(\alpha)\cdot\varphi(\beta)\Rightarrow\varphi(\alpha)=1 \) sau \( \varphi(\beta)=1 \)
Daca \( \varphi(\alpha)=1\Rightarrow \alpha\in U(\mathbb{Z}[\frac{1+i\sqrt{7}}{2}]). \) Daca \( \varphi(\beta)=1\Rightarrow \beta\in U(\mathbb{Z}[\frac{1+i\sqrt{7}}{2}]). \)
Deci, \( \frac{1+i\sqrt{7}}{2} \) este prim in \( \mathbb{Z}[\frac{1+i\sqrt{7}}{2}] \). Analog se arata ca \( \frac{1-i\sqrt{7}}{2} \) este prim in \( \mathbb{Z}[\frac{1+i\sqrt{7}}{2}] \).
Avem ca \( \frac{x+i\sqrt{7}}{2}\cdot\frac{x-i\sqrt{7}}{2}=(\frac{1+i\sqrt{7}}{2})^y\cdot(\frac{1-\sqrt{7}}{2})^y. \)
Fie \( d=(\frac{x+i\sqrt{7}}{2},\ \frac{x-i\sqrt{7}}{2}). \) Aratam ca \( d=1. \)
Rezulta ca \( d\ |\ \frac{x+i\sqrt{7}}{2}-\frac{x-i\sqrt{7}}{2}=i\sqrt{7}\Rightarrow d\ |\ i\sqrt{7}. \) Se arata ca mai sus ca \( i\sqrt{7} \) este prim (ireductibil) in \( \mathbb{Z}[\frac{1+i\sqrt{7}}{2}]\Rightarrow d=1 \) sau \( i\sqrt{7} \).
Deoarece \( d\ |\ 2^y\Rightarrow d\neq i\sqrt{7}\Rightarrow d=1 \) si avem urmatoarele egalitati:
\( (1)\ \ \ \left\{ \begin{array}{ll}
\frac{x+i\sqrt{7}}{2}=\pm(\frac{1+i\sqrt{7}}{2})^y\\
\frac{x-i\sqrt{7}}{2}=\pm(\frac{1-i\sqrt{7}}{2})^y
\end{array} \right. \) sau \( (2)\ \ \ \left\{ \begin{array}{ll}
\frac{x+i\sqrt{7}}{2}=\pm(\frac{1-i\sqrt{7}}{2})^y\\
\frac{x-i\sqrt{7}}{2}=\pm(\frac{1+i\sqrt{7}}{2})^y
\end{array} \right. \)
Scadem relatiile din (1) si obtinem: \( \pm i\sqrt{7}=(\frac{1+i\sqrt{7}}{2})^y-(\frac{1-i\sqrt{7}}{2})^y \). Scadem relatiile din (2) si obtinem: \( \pm i\sqrt{7}=(\frac{1-i\sqrt{7}}{2})^y-(\frac{1+i\sqrt{7}}{2})^y \).
Notam \( a=\frac{1+i\sqrt{7}}{2} \) si \( b=\frac{1-i\sqrt{7}}{2} \). Rezulta ca \( a-b=i\sqrt{7},\ a+b=1,\ a\cdot b=2,\ a^y-b^y=\pm i\sqrt{7}=\pm (a-b). \) Aratam ca semnul este \( "-" \).
Presupunem ca \( a^y-b^y=a-b. \) Avem ca \( a^2=(1-b)^2\equiv 1-2b=1-ab^2\equiv 1\ (mod\ b^2). \)
Deoarece \( y \) este impar, \( y\geq 3\Rightarrow a^y=a\cdot (a^2)^{\frac{y-1}{2}}\equiv a\ (mod\ b^2). \)
Avem ca \( a-b=a^y-b^y\equiv a+0=a\ (mod\ b^2)\Rightarrow a\equiv a-b\ (mod\ b^2)\Rightarrow b\equiv 0\ (mod\ b^2)\Rightarrow b^2\ |\ b \) ceea ce este in contradictie cu faptul ca \( b \) este prim. Deci, \( a^y-b^y=-(a-b)=b-a. \)
\( -i\sqrt{7}=(\frac{1+i\sqrt{7}}{2})^y-(\frac{1-i\sqrt{7}}{2})^y=\frac{2[C_y^1 i\sqrt{7}+C_y^3(i\sqrt{7})^3+C_y^5(i\sqrt{7})^5+...]}{2^y}\Rightarrow \)
\( \Rightarrow -2^{y-1}=C_y^1-7C_y^3+7^2C_y^5-7^3C_y^7+...\Rightarrow -2^{y-1}\equiv y\ (mod\ 7). \)
Puterile lui 2 modulo 7 sunt: 1,2,4 (se repeta din 3 in 3). Tinand cont si de faptul ca \( y\equiv 1\ (mod\ 2) \) obtinem:
\( y=3k\Rightarrow y\equiv 3\ (mod 7)\Rightarrow y\equiv 3\ (mod\ 21)\Rightarrow y\equiv 3\ (mod\ 42) \)
\( y=3k+1\Rightarrow y\equiv -1\ (mod 7)\Rightarrow y\equiv 13\ (mod\ 21)\Rightarrow y\equiv 13\ (mod\ 42) \)
\( y=3k+2\Rightarrow y\equiv 5\ (mod 7)\Rightarrow y\equiv 5\ (mod\ 21)\Rightarrow y\equiv 5\ (mod\ 42) \)
Deci, \( y\equiv 3,5,13\ (mod\ 42). \)
Ultimul pas este urmatorul: daca \( y,\ y_1 \) sunt solutii ale ecuatiei si \( y\equiv y_1\ (mod\ 42)\Rightarrow y=y_1. \)
Avem \( \frac{x^2+7}{4}=2^y,\ \frac{x_1^2+7}{4}=2^{y_1},\ y,y_1\geq 3 \) si impare.
Presupunem ca \( y-y_1\neq 0\Rightarrow (\exists)\ l\in\mathbb{N}^* \) a.i. \( 7^l\ ||\ y-y_1. \) Existenta lui \( l\in\mathbb{N}^* \) reiese din \( y\equiv y_1\ (mod\ 42)\Rightarrow 7\ |\ 42\ |\ y-y_1. \)
De asemenea, \( a^y-b^y=a^{y_1}-b^{y_1}=b-a,\ a=\frac{1+i\sqrt{7}}{2}. \)
Aratam ca \( a^{y-y_1}\equiv (1+i\sqrt{7})^{y-y_1}\ (mod\ 7^{l+1}). \)
Mai intai aratam prin inductie dupa \( l \) ca \( (1+7\beta)^{7^l}\equiv 1\ (mod\ 7^{l+1}). \)
Pentru \( l=0 \)este evident. Presupunem adevarat pentru \( l \) (pasul inductiv), i.e. \( (1+7\beta)^{7^l}=1+7^{l+1}\gamma. \)
\( (1+7\beta)^{7^{l+1}}=(1+7^{l+1}\gamma)^7=1+C_7^1 7^{l+1}\gamma+...\equiv 1+C_7^1 7^{l+1}\gamma=1+7^{l+2}\gamma\equiv 1\ (mod\ 7^{l+2}). \)
Dar, \( 7^{2(l+1)}\ \vdots\ 7^{l+2},\ (2(l+1)\geq l+2),\ \frac{y-y_1}{6}=7^l\cdot t. \)
Rezulta ca \( 2^{y-y_1}=(2^6)^{\frac{y-y_1}{6}}=(1+7\cdot 9)^{7^l \cdot t}=[(1+7 \cdot 9)^{7^l}]^t\equiv 1^t=1\ (mod\ 7^{l+1}). \)
Deci, \( a^{y-y_1}\equiv (1+i\sqrt{7})^{y-y_1}\ (mod\ 7^{l+1}). \)
In continuare, aratam prin inductie dupa \( l \) ca \( (1+i\sqrt{7})^{7^l}\equiv 1+7^l i\sqrt{7}\ (mod\ 7^{l+1}). \)
Pentru \( l=0 \), avem ca \( 1+i\sqrt{7}\equiv 1+i\sqrt{7}\ (mod\ 7). \) Presupunem adevarata relatia pentru \( l \) si o demonstram pentru \( l+1. \)
\( (1+i\sqrt{7})^{7^l}\equiv 1+7^l i\sqrt{7}\Leftrightarrow (1+i\sqrt{7})^{7^l}=1+7^l i\sqrt{7}+\alpha 7^{l+1}. \)
\( (1+i\sqrt{7})^{7^{l+1}}=(1+7^l i\sqrt{7}+\alpha 7^{l+1})^7\equiv (1+7^l i\sqrt{7})^7+C_7^1(1+7^l i\sqrt{7})^6\cdot \alpha 7^{l+1}\equiv \)
\( \equiv(1+7^l i\sqrt{7})^7\equiv 1+C_7^1 7^l i\sqrt{7}+C_7^2 (7^l i\sqrt{7})^2+...\equiv 1+C_7^1 7^l i\sqrt{7}=1+7^{l+1}i\sqrt{7}\ (mod\ 7^{l+2}). \)
Considerand \( y_1-y=7^l\cdot t \) si ridicand la puterea \( t \) membrii din congruenta \( (1+i\sqrt{7})^{7^l}\equiv 1+7^l i\sqrt{7}\ (mod\ 7^{l+1}) \) obtinem:
\( (1+i\sqrt{7})^{y_1-y}=(1+i\sqrt{7})^{7^l\cdot t}\equiv (1+7^l i\sqrt{7})^t=1+C_t^1 7^l i\sqrt{7}+C_t^2 (7^l i\sqrt{7})^2+...= \)
\( =1+C_t^1 7^l i\sqrt{7}+7^{2l+1}(...)\equiv 1+C_t^1 7^l i\sqrt{7}=1+t7^l i\sqrt{7}=1+(y_1-y)i\sqrt{7}\ (mod\ 7^{l+1}). \)
Deci, \( 1+(y_1-y)i\sqrt{7}\equiv a^{y_1-y}\equiv (1+i\sqrt{7})^{y_1-y}\ (mod\ 7^{l+1}). \)
Folosind \( y_1-y=7^l\cdot t \) si \( a^y\equiv \frac{1+yi\sqrt{7}}{2^y}\ (mod\ 7)\Leftrightarrow a^y=\frac{1+yi\sqrt{7}}{2^y}+7\cdot\alpha \), obtinem:
\( a^{y_1}=a^{y_1-y}\cdot a^y\equiv [1+(y_1-y)i\sqrt{7}]a^y=a^y+a^y(y_1-y)i\sqrt{7}\equiv a^y+\frac{1+yi\sqrt{7}}{2^y}(y_1-y)i\sqrt{7}\ (mod\ 7^{l+1}). \)
Am folosit mai sus \( \frac{1+yi\sqrt{7}}{2^y}(y_1-y)\equiv a^y(y_1-y)=\frac{1+yi\sqrt{7}}{2^y}(y_1-y)+(y_1-y)7\alpha\ (mod\ 7^{l+1}) \) si \( (y_1-y)7\alpha\ \vdots\ 7^{l+1}. \)
Avem ca \( b^{y_1}\equiv [1+(y-y_1)i\sqrt{7}]b^y\equiv b^y+\frac{(y-y_1)i\sqrt{7}(1-yi\sqrt{7})}{2^y}\ (mod\ 7^{l+1}) \), deoarece \( b=\frac{1-i\sqrt{7}}{2},\ b^{y_1-y}\equiv (1-i\sqrt{7})^{y_1-y}\equiv 1+(y-y_1)i\sqrt{7}\ (mod\ 7^{l+1}) \) si \( b^y\equiv \frac{1-yi\sqrt{7}}{2^y}. \)
Asadar, \( \frac{(y-y_1)i\sqrt{7}(1-yi\sqrt{7})}{2^y}\equiv b^{y_1}-b^y=a^{y_1}-a^y\equiv \frac{(y_1-y)i\sqrt{7}(1+yi\sqrt{7})}{2^y}\ (mod\ 7^{l+1})\Rightarrow \)
\( \Rightarrow\frac{1+yi\sqrt{7}}{2^y}(y_1-y)i\sqrt{7}\equiv\frac{-1+yi\sqrt{7}}{2^y}(y_1-y)i\sqrt{7}\ (mod\ 7^{l+1})\Rightarrow (y_1-y)i\sqrt{7}\equiv 0\ (mod\ 7^{l+1})\Rightarrow \)
\( \Rightarrow (y_1-y)i\sqrt{7}=7^{l+1}\cdot\gamma,\ \gamma\in\mathbb{Z}[\frac{1+i\sqrt{7}}{2}]\Rightarrow \)
\( \Rightarrow (y_1-y)^2\cdot 7=\varphi((y_1-y)i\sqrt{7})=7^{2(l+1)}\cdot\varphi(\gamma)\Rightarrow (y_1-y)\ \vdots\ 7^{2l+1}. \)
Dar, \( 7^l\ ||\ y_1-y\Rightarrow 7^{2l}\ ||\ (y_1-y)^2\Rightarrow (y_1-y)^2\neq 0\ (mod\ 7^{2l+1}) \) ceea ce reprezinta o contradictie \( \Rightarrow y_1=y. \)
Deci, pentru \( y \) impar, \( y\geq 3 \) avem \( y\in\{3,5,13\}. \) Dar, \( n=y+2\Rightarrow n\in\{5,7,15\}\Rightarrow x\in\{\pm5,\pm11,\pm181\}. \)
Asadar, solutiile ecuatiei lui Ramanujan in \( \mathbb{Z} \) sunt: \( (x,n)\in\{(\pm1,3),(\pm3,4),(\pm5,5),(\pm11,7),(\pm181,15)\}. \)

Remarca. In solutia data de Nagell se folosea ecuatia Pell:
\( x^2+7=2^{2n+1}\Leftrightarrow x^2-2^{2n+1}=-7\Leftrightarrow x^2-2y^2=-7. \)