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.
Can there exist a natural number $n$ such that it has a factor $\sqrt{n}+1$?
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number-theory
elementary-number-theory
factoring
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1Do you mean that the condition is that $n$ is actually divisible by $\sqrt{n}+1$? Also, so that $\sqrt{n}+1$ is an integer itself (because otherwise it doesn't quite make sense to call it a factor in the context of integers), this forces $n$ to be a perfect square - it that what you mean? – 2017-02-15
2 Answers
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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 $0=(\sqrt0 + 1)\times 0$.
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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 $k+1$