Prove that for any natural number $n>1$ and $k, n\nmid(kn+1)$
Proof.
(Note: $\mathbb{N}=\mathbb{Z^+}$)
$$\text{WTS }\forall n\in(1,\infty)\cap \mathbb{N}, k\in\mathbb{N}, \forall c \in \mathbb{N}, kn+1 \neq cn$$
Prove this by contradiction,if not then we have the following:
$$\exists\space n\in(1,\infty)\cap\mathbb{N},k\in\mathbb{N},\exists \space c\in \mathbb{N},s.t.kn+1=cn$$
Since the following, we consider two cases: $$\forall n\in(1,\infty)\cap\mathbb{N},$$$$ n\in\{t\in\mathbb{N}| \exists\space a\in\mathbb{N},s.t.t=2a\} \dots\text{(n is even)}$$$$\vee n\in \{t\in\mathbb{N}|\exists\space a\in\mathbb{N}\cup\{0\},s.t.t=2a+1\}\dots\text{(n is odd)} $$
Case 1: n is even
That $$\exists\space a\in\mathbb{N},s.t.n=2a$$
Have $$2ak+1=2ac$$
But left side is odd and right side is even, contradiction!
Case 2: n is odd
That $$\exists\space a\in\mathbb{N}\cup\{0\}, s.t. n=2a+1$$
Have $$k(2a+1)+1=c(2a+1)$$$$2ak+k+1=2ac+c$$$$2a(k-c)=-(k-c+1)$$
Since $k-c=0\rightarrow 0=1$ That $0\neq1\rightarrow k-c\neq0$
Then divide both side by $k-c$ have the following:
$$2a=-\frac{k-c+1}{k-c}$$$$(-\frac{k-c+1}{k-c}=2a\wedge 2a\in\mathbb{Z})\rightarrow k-c=\pm 1$$
But $a\in\mathbb{N}\rightarrow 2a>1$ that:
$$k-c=1\rightarrow-2>1,k-c=-1\rightarrow0>1$$
Both cases implies contradictions, Case 2 also not hold.
Since the negation of the statement can never hold, that the statement hold. $\square$
$1.$Is my proof correct
$2.$Is there better ways to prove this
Any help or hint or suggestion would be appreciated.
Update
Alternative solution
Proof.
$$\text{WTS }\forall n\in(1,\infty)\cap \mathbb{N}, k\in\mathbb{N}, \forall c \in \mathbb{N}, kn+1 \neq cn$$
Prove this by contradiction,if not then we have the following:
$$\exists\space n\in(1,\infty)\cap\mathbb{N},k\in\mathbb{N},\exists \space c\in \mathbb{N},s.t.kn+1=cn$$
Have $$kn-cn=-1$$$$n(c-k)=1\text{ denote as }*$$
Consider 2 cases to show $0<c-k<1$ by contradiction
If not then we have:
$$c-k\ge1\vee c-k\le0$$
Case1: $c-k\ge1$
With $n>1$ and $*$ that $n(c-k)>1$, contradiction!
Case2.1: $c-k=0$
With $n(c-k)=1$ and $*$ That $0=1$, contradiction!
Case2.2: $c-k<0$
With $n>1$ Have $1=n(c-k)<0 $, contradiction!
By contradiction have $0<c-k<1$, implies $c-k\not\in\mathbb{Z}$ contradict with $c,k\in\mathbb{N}.\square$
Best Answer
Your proof is far too long... If $n|kn+1$ then $n|1$. Since $n\geq0$ we have $n=1$, which is a contradiction.