Attemption to prove that a discrete memoryless distribution is a geometric distribution leading to nonsenses Let's try to prove that any memoryless discrete distribution whose associated random variable's values can only be natural, is a geometric distribution:
For any $k$: $q:=\frac{\operatorname{P}(X=2)}{\operatorname{P}(X=1)}=\frac{\operatorname{P}(X=k+1|X>k-1)}{\operatorname{P}(X=k+1|X>k)}=\frac{\frac{\operatorname{P}(X=k+1)}{\operatorname{P}(X>k-1)}}{\frac{\operatorname{P}(X=k+1)}{\operatorname{P}(X>k)}}=\frac{\operatorname{P}(X>k)}{\operatorname{P}(X>k-1)}=\operatorname{const.}$
Thus for any $k$: $\operatorname{P}(X>k)=q^{k-1}\operatorname{P}(X=1)$
Thus for any $k\geq2$: $\operatorname{P}(X=k)=\operatorname{P}(X>k-1)-\operatorname{P}(X>k)=\operatorname{P}(X=1)(q^{k-2}-q^{k-1})=q^{k-2}\operatorname{P}(X=1)(1-q)$
This already seems to contradict what we had to prove. Since for a geometric distribution we have: $\operatorname{P}(X=k)=(1-p)^{k-1}p=q^{k-1}(1-q)$ for $q:=1-p$.
Clearly $q^{k-2}\operatorname{P}(X=1)(1-q)=q^{k-1}(1-q)$ if and only if $\operatorname{P}(X=1)=q$; and we know that for a geometric distribution $\operatorname{P}(X=1)=p=1-q$. This leaves us the equation $q=1-q$ which clearly has only one solution: $p=q=\frac12$.
Therefore a discrete memoryless distribution is a geometric distribution only for $p=\frac12$.
And this is nonsense. Since, as we know, a geometric distribution can be defined as a memoryless discrete distribution.
It is likely because I'm too sleepy and loopy now. But I can't find my error. And I must've made some error.
Could you point me to my error?
 A: The error is here:

$q:=(...)=\frac{\operatorname{P}(X>k)}{\operatorname{P}(X>k-1)}=\operatorname{const.}$
Thus for any $k$: $\operatorname{P}(X>k)=q^{k-1}\operatorname{P}(X=1)$

Clearly this is not the case, rather, this implies that for any $k$: $\operatorname{P}(X>k)=q^{k-1}\operatorname{P}(X>1)$
Thanks for @Peter who prompted me to write this "proof" more verbosely. Up to the point of error it would go like that:

For any $k$:
Frome the definition of memorylessness:
  $\operatorname{P}(X=2)=\operatorname{P}(X=(k+1)-(k-1))=\operatorname{P}(X=k+1|X>k-1)$
From the definition of conditional probability:
  $\operatorname{P}(X=k+1|X>k-1)=\frac{\operatorname{P}(X=k+1\wedge
> X>k-1)}{\operatorname{P}(X>k-1)}=\frac{\operatorname{P}(X=k+1)}{\operatorname{P}(X>k-1)}$
Simirarily:
  $\operatorname{P}(X=1)=\operatorname{P}(X=k+1|X>k)=\frac{\operatorname{P}(X=k+1)}{\operatorname{P}(X>k)}$
Therefore for any $k$:
  $\frac{\operatorname{P}(X=2)}{\operatorname{P}(X=1)}=\frac{\frac{\operatorname{P}(X=k+1)}{\operatorname{P}(X>k-1)}}{\frac{\operatorname{P}(X=k+1)}{\operatorname{P}(X>k)}}=\frac{\operatorname{P}(X>k)}{\operatorname{P}(X>k-1)}$
Now for a given distribution
  $\frac{\operatorname{P}(X=2)}{\operatorname{P}(X=1)}$ is constant and
  let $q:=\frac{\operatorname{P}(X=2)}{\operatorname{P}(X=1)}$
Therefore $\frac{\operatorname{P}(X>k)}{\operatorname{P}(X>k-1)}$ is
  also constant and equals $q$.
In that case, by induction, for any $k$:
  $\operatorname{P}(X>k)=q^{k-1}\operatorname{P}(X=1)$ <- and the error
  is here.

BTW I found out that this line of thought is not yet sufficient to prove what we have to prove, even having corrected this error we still get a noticeably weaker condition that we had to achieve, but that's a different topic.
