Ecuatie diofantica

[Beniamin Bogosel]

Rezolvati ecuatia \( x^2+2=y^3 \) in numere intregi.

[Cristi Popa]

Pentru a rezolva in numere intregi ecuatia lui Fermat, vom presupune cunoscute:
1) \( \mathbb{Z}[i\sqrt{2}] \) este inel euclidian;
2) \( U(\mathbb{Z}[i\sqrt{2}])=\{\pm1\} \);
3) Fie \( R \) un inel factorial si \( a\cdot b=c^k,\ k\in\mathbb{N},\ k\geq 2, \ (a,b)=1,\ a,b,c\in R \). Atunci \( a\sim a_1^k,\ b\sim b_1^k,\ a_1,b_1\in R \).

\( x^2+2=y^3\Leftrightarrow (x+i\sqrt{2})(x-i\sqrt{2})=y^3 \). Fie \( d=(x+i\sqrt{2},\ x-i\sqrt{2}) \). Vom arata ca \( d=1 \).
\( d\ |\ x+i\sqrt{2}-(x-i\sqrt{2})=2i\sqrt{2}=-(i\sqrt{2})^3 \). Aratam ca \( i\sqrt{2} \) este prim (ireductibil) in \( \mathbb{Z}[i\sqrt{2}] \), folosind functia norma \( \varphi\ :\ \mathbb{Z}[i\sqrt{2}]\rightarrow \mathbb{N},\ \varphi(a+bi\sqrt{2})=(a+bi\sqrt{2})\cdot (a-bi\sqrt{2})=a^2+2b^2 \).
Fie \( i\sqrt{2}=\alpha\cdot\beta,\ \alpha,\beta\in\mathbb{Z}[i\sqrt{2}]\Rightarrow 2=\varphi(i\sqrt{2})=\varphi(\alpha)\cdot\varphi(\beta)\Rightarrow\varphi(\alpha)=1 \) sau \( \varphi(\beta)=1\Rightarrow \alpha\in U(\mathbb{Z}[i\sqrt{2}]) \) sau \( \beta\in U(\mathbb{Z}[i\sqrt{2}])\Rightarrow i\sqrt{2} \) este ireductibil (prim).
Fie \( \alpha=m+ni\sqrt{2}\in\mathbb{Z}[i\sqrt{2}] \) si \( \varphi(\alpha)=1\Rightarrow m^2+2n^2=1\Rightarrow n=0\Rightarrow m=\pm1\Rightarrow\alpha=\pm 1\Rightarrow\alpha\in U(\mathbb{Z}[i\sqrt{2}]) \).
Din \( d\ |\ (i\sqrt{2})^3 \) si \( i\sqrt{2} \) prim \( d=(i\sqrt{2})^j,\ 0\leq j\leq3 \). Presupunem ca \( j\geq1\Rightarrow i\sqrt{2}\ |\ d\ |\ x+i\sqrt{2}\Rightarrow x+i\sqrt{2}=i\sqrt{2}(m+ni\sqrt{2}),\ m,n\in\mathbb{Z}\Rightarrow x=-2n \), deci \( x \) este par.
Dar \( x^2+2=y^3\Rightarrow y \) este par \( \Rightarrow 2=y^3-x^2\ \vdots\ 4 \) - contradictie. Deci, \( j=0\Rightarrow d=1 \).
Folosind propozitia 3) \( \Rightarrow x+i\sqrt{2}=u\cdot a^3,\ a\in\mathbb{Z}[i\sqrt{2}],\ u\in U(\mathbb{Z}[i\sqrt{2}])\Rightarrow x+i\sqrt{2}=\pm a^3=(\pm a)^3=b^3,\ b=s+ti\sqrt{2}\in\mathbb{Z}[i\sqrt{2}] \).
\( x+i\sqrt{2}=(s+ti\sqrt{2})^3=s^3+3s^2ti\sqrt{2}+3s(ti\sqrt{2})^2+(ti\sqrt{2})^3=s^3-6st^2+i\sqrt{2}(3s^2t-2t^3)\Rightarrow\left\{\begin{array}{II} x=s^3-6st^2 \\ 1=3s^2t-2t^3=t(3s^2-2t^2)\Rightarrow t=\pm 1\end{array} \right. \)
Daca \( t=-1\Rightarrow3s^2=1\Rightarrow s^2=\frac{1}{3} \) - contradictie cu \( s\in\mathbb{Z} \).
Daca \( t=1\Rightarrow 3s^2-2t^2=1\Rightarrow 3s^2=3\Rightarrow s=\pm1 \).
Rezulta ca \( x=s^3-6st^2=s(s^2-6t^2)=-5s=\pm 5 \) si \( y^3=x^2+2=27\Rightarrow y=3 \).
Deci, \( x=\pm5 \) si \( y=3 \).