[tex]\displaystyle Deoarece~a^2 \geq 0 ~si~4b^2\geq0,~deducem~ca~a^2 \leq4~si~4b^2 \leq4. \\ \\ Deci~a \in [-2;2]~si~b \in [-1;1]. \\ \\ Prin~urmare~a+b \in [-3;3],~iar~cum~a+b \in \mathbb{Z},~deducem~ca \\ \\ a+b \in \{-3;-2;-1;0;1;2;3 \}. \\ \\ Daca~a+b=-3~(respectiv~a+b=3)~am~avea~a=-2~si~b=-1 \\ \\ (respectiv~a=2~si~b=1),~insa~aceste~perechi~nu~convin. \\ \\ Pentru~a+b=-2,~obtinem~sistemul: [/tex]
[tex]\displaystyle \left \{ {{a+b=-2} \atop {a^2+4b^2=4}} \right. .~Sistemul~se~rezolva~rapid~(eventual)~prin~\\ \\ metoda ~ substitutiei,~si~rezulta~(a,b) \in \Big \{ (-2,0); \Big(- \frac{6}{5}, - \frac{4}{5} \Big) \Big\}. \\ \\ Restul~cazurilor~se~trateaza~in~mod~analog...[/tex]
