[tex]\left.\begin{aligned} \bf 2a-3b=5 \\ \\ \bf \Big(a\; ; b \Big) = 1 \end{aligned} \; \; \right\} \implies \bf 2a=5 + 3b \; \bigg| : 2[/tex]
[tex]\bf a = \dfrac{5 + 3b}{2} \; \; ;\;\; \forall \; \; a\; ;b \in \mathbb{N}[/tex]
[tex]\bf b=0 \implies a = \dfrac{5+3b}{2} = \dfrac{5+3\cdot 0}{2} = \dfrac{5}{2} \notin \mathbb{N}[/tex]
[tex]\bf b=1 \implies a = \dfrac{5+3b}{2} = \dfrac{5+3}{2} = \dfrac{8}{2}=4 \in \mathbb{N}[/tex]
[tex]\bf b=2 \implies a = \dfrac{5+3b}{2} = \dfrac{5+3\cdot 2}{2} = \dfrac{11}{2} \notin \mathbb{N}[/tex]
[tex]\bf b=3 \implies a = \dfrac{5+3b}{2} = \dfrac{5+3\cdot 3 }{2} = \dfrac{14}{2}=7 \in \mathbb{N}[/tex]
[tex]\bf b=4 \implies a = \dfrac{5+3b}{2} = \dfrac{5+3\cdot 4}{2} = \dfrac{17}{2} \notin \mathbb{N}[/tex]
[tex]\bf b=5 \implies a = \dfrac{5+3b}{2} = \dfrac{5+3\cdot 5}{2} = \dfrac{20}{2}=10 \in \mathbb{N}[/tex]
Observam ca daca b este un număr impar vom avea o relație convenabila.
Observam ca a se va repeta din 3 in 3 la fiecare nr impar b.
Nu uitam ca (a ; b) = 1 deci excludem din variante.
Fără o condiție vom avea o infinitate de termeni.
#copaceibrainly
