Do consecutive integers never divide each other? If not, why?

Question

I was solving some number theory problems and it struck my mind. I am a beginner, so don't go hard on me.

Answer 1

Simply put: The difference between $n$ and $n+1$ is only a single unit. But the difference between different multiples of $n$ must be at least $n$ or a multiple of $n$.

But notice: $1$ divides $2$. And $-1$ divides $0$. (Everything divides $0$).

.... or more formally:

If $n > 0$ and $1*n$ is the first multiple of $n$ then $2*n$ is the second multiple of $n$. And $n < n+1 \le 2n$. So $n +1$ can only be a multiple of $n$ if if $n+1 = 2n$ or $n = 1$.

Method 2:

In general $\frac {n+1}{n} = \frac nn + \frac 1n = 1 + \frac 1n$. This is an integer if and only if $\frac 1n$ is an integer.

So $n =\pm 1$. And $n=1$ divides $n+1 =2$. And $n=-1$ divides $n+1 = 0$.

Those are the only two exceptions.

......

Method 2:

If $n|k$ then $n|k - n$. That is because there is an integer $m$ so that $n*m = k$ so $k - n = (m-1)n$.

So if $n|n+1$ then $n|1$. That can only happen if $n =\pm 1$.

Method 4:

If $n$ is a natural number: $n|k$ then $k = 0 \pmod n$. (We don't want to deal with negative modulus right now.)

So if $n|n+1$ then $n+1 \equiv 0 \pmod{n}$. So $n+1 \equiv 1\equiv 0 \pmod {n}$

The only way for $1 \equiv 0 \pmod n$ is if $n = 1$.

....

All four of those arguments are all basically the same argument.

Answer 2

If $n$ divide $n+1$ , we must have $n+1=kn$ for $k$ an integer larger than $1$ and so we get $$1=n(k-1)\ge n(2-1)=n$$ So the only condition then $n$ divide $n+1$ is $n=1$

Answer 3

On the other hand, $1\mid0$ and $-1\mid 0$. Actually everything divides $0$ except itself.