Can there exist a natural number $n$ such that it has a factor $\sqrt{n}+1$?

Question

I'm able to understand that this can never be true for when $n$ is odd but can't prove it for when $n$ is an even number.

Answer 1

You can have $n=0,$ when $ \sqrt n+1=1$ divides evenly into $0$. If you require that $n$ be a square positive integer so that you can take $\sqrt n$ it cannot happen. Let $n=k^2$ then $\sqrt n+1=k+1$ and $n-1=(k+1)(k-1)$ As $n$ is coprime to $n-1$ it cannot have a factor $k+1$

Answer 2

Let $n$ be a perfect square such that $(\sqrt n +1)\mid n$.

Since $(\sqrt n +1)\mid(\sqrt n + 1)(\sqrt n - 1)=n-1$, we must have $(\sqrt n + 1)\mid 1$, hence $n=0$.

Conversely $n=0$ is a solution because $0=(\sqrt0 + 1)\times 0$.