[tex]\displaystyle\it\\a+1|b-1 \implies \exists m \in\mathbb{N}^*~a.i.~b-1=m(a+1).\\b|a^2+a+2 \implies \exists n \in\mathbb{N}^*~.a.i.~bn=(a^2+a+2).\\din~prima~relatie~deducem~ca~b=m(a+1)+1,~inlocuim~in~a~doua~relatie.\\n\bigg[m\bigg(a+1\bigg)+1\bigg]=a^2+a+2 \Leftrightarrow\\mn(a+1)+n=a^2+a+2,~reducem~modulo~(a+1)~aceasta~relatie.\\n\equiv a^2+a+2 \equiv a(a+1)+2\equiv 2(mod~a+1),~deci~\\n\equiv~2(mod~a+1).\\daca~n\geq a+3 \implies mn(a+1)+n\geq n(a+1)+n\geq (a+3)(a+1)+1>\\a^2+a+2,~contradictie.\\[/tex]
[tex]\displaystyle\it\\si~deci~n<a+3,~si~cum~n\equiv 2(mod~a+1) \implies \boxed{\it n=2}.\\inlocuim,~si~obtinem~ca~a^2+a+2=2m(a+1) +2 \Leftrightarrow\\a(a+1)=2m(a+1) \implies \boxed{\it a=2m}.\\stim~ca~\boxed{\it b=m(2m+1)+1=2m^2+m+1}.\\deci,~solutiile~care~verifica~cele~doua~conditii~sunt~de~forma\\(a,b)=(2m,~2m^2+m+1),~iar~b~este~un~numar~de~trei~cifre\implies\\100\leq b\leq 999 \Leftrightarrow 100\leq 2m^2+m+1\leq 999.\\[/tex]
[tex]\displaystyle\it\\\left \{ {{2m^2+m+1\geq 100} \atop {2m^2+m+1\leq 999}} \right. \Leftrightarrow \left \{ {{m\geq 7 } \atop {m\leq 22}} \right. \\de~unde~se~deduce~ca~m\in\left\{7,8,9,...,23\right\},~iar~in~aceasta~multime\\sunt~16~elemente,~deci~numarul~perechilor~este~de~\boxed{\it16}.[/tex]
