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1) Sa se demonstreze ca nu exista \( a,b\in\mathbb{Z^{\ast}} \) astfel incat ecuatia \( x^2-2abx+a^2+b^2=0 \) sa aiba radacini intregi.

I. Cucurezeanu

2) Sa se determine numerele \( a,b\in\mathbb{Z} \), astfel incat ecuatia \( x^2-(a^2+b^2)+ab=0 \) sa aiba radacini intregi.

L. Pananitopol

Indicatie :

a) Se arata ca discriminantul ecuatiei nu poate fi patrat perfect par.

Marius Mainea wrote:1) Sa se demonstreze ca nu exista \( a,b\in\mathbb{Z^{\ast}} \) astfel incat ecuatia \( x^2-2abx+a^2+b^2=0 \) sa aiba radacini intregi.

I. Cucurezeanu

Dem. Daca \( \triangle \) e discriminantul ecuatiei, cum \( x_{1,\ 2}=\frac{2ab\pm\sqrt{\triangle}}{2}\in\mathbb{Z}\Longrightarrow \) discriminantul e patrat perfect. Daca \( \triangle =n^2 \), atunci inlocuind observam ca \( n \) trebuie sa fie par, deci \( n=2k,\ k\in\mathbb{N} \) de unde \( 4a^{2}b^{2}-4\left(a^{2}+b^{2}\right)=4k^{2}\Longrightarrow a^{2}b^{2}-a^{2}-b^{2}=k^{2} \). Daca \( d=\left(a,\ b\right)\Longrightarrow a=da_{1},\ b=db_{1},\ \left(a_{1},\ b_{1}\right)=1 \) atunci trebuie ca \( a_{1}^{2}b_{1}^{2}-a_{1}^{2}-b_{1}^{2}=k_{1}^{2}, \ k_1\in \mathbb{N} \). Dar cum \( \left(a_{1}^{2}-1\right)\left(b_{1}^{2}-1\right)=k_{1}^{2}+1 \), analizam urmatoarele situatii:

\( 1.\ a_{1}\equiv b_{1}\equiv0\left(mod\ 2\right)\Longrightarrow\left(a_{1},\ b_{1}\right)\neq1 \), absurd.
\( 2.\ a_{1}\equiv b_{1}\equiv1\left(mod\ 2\right)\Longrightarrow\left(a_{1}^{2}-1\right)\left(b_{1}^{2}-1\right)\equiv0\left(mod\ 4\right)\Longrightarrow k_{1}^{2}\equiv -1\left(mod\ 4\right) \), absurd.
\( 3.\ a_{1}\equiv0\left(mod\ 2\right),\ b_{1}\equiv1\left(mod\ 2\right)\Longrightarrow a_{1}=2a_{2},\ b_{1}=2b_{2}+1,\ a_{1},\ b_{1}\in\mathbb{N} \). Dar si in acest caz cum \( \left(a_{1}^{2}-1\right)\equiv1\left(mod\ 4\right),\ \left(b_{1}^{2}-1\right)\equiv0\left(mod\ 4\right)\Longrightarrow k_{1}^{2}\equiv-1\left(mod\ 4\right) \).
Prin urmare ecuatia nu are radacini intregi.