Can $n - 1$ divide $n$, where $n$ is a positive, composite integer?

Question

If $n$ is a positive and composite integer, can I prove that $n - 1$ does not divide $n$ for all $n$?

If not, can you give me a counter example?

• I was trying to prove it by contradiction, but I was just running in a vicious circle.

Any hint or solution please?

Answer 1

$n-(n-1)=1$ so $\operatorname{pgcd}(n,n-1)=1$ by Bezout.

Answer 2

You Can write $$n=n-1+1$$ so $$\frac{n}{n-1}=1+\frac{1}{n-1}$$

Answer 3

No, because the least such integer $k$ such that $n$ could possibly equal $k(n-1)$ is $2$, but $2n-2 > n$ for all $n > 2$.

Answer 4

Hint:

If a prime $p$ divides $n$, and also $n-1$ then $p$ would also divide $n-(n-1)=1$

Answer 5

The hypothesis that $n$ be positive composite implies

$n \ge 4; \tag 1$

thus

$n - 1 \ge 3; \tag 2$

if

$n - 1 \mid n, \tag 3$

there exists $k \in \Bbb Z$ with

$n = k(n - 1); \tag 4$

since

$n, n - 1 \ge 3, \tag 5$

we must have

$k \ge 2; \tag 6$

indeed, $k$ cannot be negative or zero by (4),(5); it cannot be $1$ since $n \ne n - 1$; also,

$k < n, \tag 7$

for if

$k \ge n, \tag 8$

then

$n = k(n - 1) \ge n(n - 1) \ge 3n > n, \tag 9$

impossible.

Then, from (4),

$n = nk - k, \tag{10}$

or

$k = nk - n = (k - 1)n \ge n, \tag 8$

contradicting (7); thus no such $k$ exists, and hence

$n - 1 \not \mid n. \tag 9$