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Fie \( n\in \mathbb{N} \). Exista \( a,b \in \mathbb{N} \) astfel incat \( \sqrt{n+1}=\frac{a\sqrt{n}+b\sqrt{n+2}}{a+b} \)?

\( \sqrt{n+1}(a+b)=a\sqrt{n}+b\sqrt{n+2} \)
ridicam totul la puterea a 2-a si vom avea:
\( (n+1)(a+b)^2=(a\sqrt{n}+b\sqrt{n+2})^2 \)
\( (n+1)(a^2+2ab+b^2)=a^2n+2ab\sqrt{n(n+2)}+b^2(n+2) \)
\( na^2+2abn+b^2n+a^2+2ab+b^2=a^2n+2ab\sqrt{n(n+2)}+b^2+2b^2 \)
\( a^2+2ab+2abn=b^2+2ab\sqrt{n(n+2)}\Longrightarrow 2ab\sqrt{n(n+2)}\in \mathbb N\Longrightarrow n(n+2) \) patrat perfect
dar, pentru \( n\neq 0, n^2<n^2+2n<(n+1)^2=n^2+2n+1\Longrightarrow n=0 \)
inlocuind in fractie,\( b\sqrt2=a+b \), dar \( b\sqrt 2 \)este irational si \( a+b \)este rational\( \Longrightarrow \)nu exista numere sa satisfaca cerinta.