Để \(\sqrt{x^2+x+3}\) nguyên thì
\(\Rightarrow x^2+x+3=a^2\left(a\in Z\right)\)
\(\Leftrightarrow4x^2+4x+12=4a^2\)
\(\Leftrightarrow4a^2-\left(2x+1\right)^2=11\)
\(\Leftrightarrow\left(2a+2x+1\right)\left(2a-2x-1\right)=11\)
\(\Leftrightarrow\left(2a+2x+1,2a-2x-1\right)=\left(1,11;11,1;-1,-11;-11,-1\right)\)
\(\Leftrightarrow\left(a,x\right)=\left(3,-3;3,2;-3,-3;-3,2\right)\)
Vậy ....