Question

Tìm tất cả STN n để gt bt Q = \(\sqrt{n+2}+\sqrt{n+\sqrt{n+2}}\) là số nguyên

Answer 1

Để Q nguyên \(\Rightarrow\sqrt{n+2}\) nguyên

Đặt \(\sqrt{n+2}=a>1\Rightarrow n=a^2-2\)

\(\Rightarrow Q=a+\sqrt{a^2+a-2}\)

Để Q nguyên \(\Rightarrow a^2+a-2=k^2\Leftrightarrow4a^2+4a-8=4k^2\)\(\left(k\in N\right)\)

\(\Leftrightarrow\left(2a+1-2k\right)\left(2a+1+2k\right)=9\)

\(\Rightarrow\left\{{}\begin{matrix}2a+1+2k=9\\2a+1-2k=1\end{matrix}\right.\)

\(\Rightarrow2a+1=5\Rightarrow a=2\)

\(\Rightarrow\sqrt{n+2}=2\Rightarrow n=2\)