Answer:
4 and 5
Step-by-step explanation:
[tex]n,\ n+1-\text{two consecutive numbers}\\\\\sqrt{n}=n+1-3\\\sqrt{n}=n-2\qquad\text{square both sides}\ (domain\ D:\ n\geq2)\\(\sqrt{n})^2=(n-2)^2\qquad\text{use}\ (a-b)^2=a^2-2ab+b^2\\n=n^2-2(n)(2)+2^2\\n=n^2-4n+2\qquad\text{subtract}\ n\ \text{from both sides}\\0=n^2-5n+4\\n^2-5n+4=0\\n^2-n-4n+4=0\\n(n-1)-4(n-1)=0\\(n-1)(n-4)=0\iff n-1=0\ \vee\ n-4=0\\\\n=1\notin D\ \vee\ n=4\in D\\{\qquad\qquad\qquad n+1=5}[/tex]
