Find all postive integer $k$ such $F:R^2\to R^2,F(x,y)=(ax+by,cx+dy)$ and $F(S)=T$ Let $k\in Z_{+}$,and define
$$S=\{(m+\dfrac{1}{k},n)|m,n\in Z\},T=\{(m+\dfrac{2}{k},n)|m,n\in Z\}$$
Find all postive integer $k$ such there exist $a,b,c,d\in R$,and 
$$F:R^2\to R^2,F(x,y)=(ax+by,cx+dy)$$
and $F(S)=T$
Is said to be the American mathematical monthly problem.I think a lot of day, feel this problem is an indefinite equation is obtained
 A: It is necessary and sufficient that $k$ is odd.
First of all, we have
$$F(S)=\left\{\left(am+\frac ak+bn,cm+\frac ck+dn\right)\mid m,n\in\mathbb Z\right\}$$
So, we want to find all positive integers $k$ such that there exist $a,b,c,d\in\mathbb R$ satisfying
$$\left\{\left(am+\frac ak+bn,cm+\frac ck+dn\right)\mid m,n\in\mathbb Z\right\}=\left\{\left(M+\frac 2k,N\right)\mid M,N\in\mathbb Z\right\}$$
In the following, we prove $(1)$ and $(2)$ :
$(1)$ : It is necessary that $k$ is odd. 
$(2)$ : It is sufficient that $k$ is odd.
$(1)$ 
Suppose that $k$ is even.
For $(m,n)=(0,0)$, there has to be a pair $(M_1,N_1)$ such that
$$\frac{a-2}k=M_1\quad\text{and}\quad \frac ck=N_1\implies \frac{a-2}{k},\frac ck\in\mathbb Z\implies a=pk+2,c=qk\quad\text{where}\quad p,q\in\mathbb Z$$
For $(m,n)=(0,1)$, there has to be a pair $(M_2,N_2)$ such that
$$b=M_2-\frac{a-2}k\quad\text{and}\quad d=N_2-\frac ck\implies b,d\in\mathbb Z$$
For $(M,N)=(1,0)$, there has to be a pair $(m_1,n_1)$ such that
$$am_1+\frac ak+bn_1=1+\frac 2k\quad\text{and}\quad cm_1+\frac ck+dn_1=0$$
$$\implies m_1k(bc-ad)=-kd-2d-(bc-ad)\tag3$$
For $(M,N)=(2,0)$, there has to be a pair $(m_2,n_2)$ such that
$$am_2+\frac ak+bn_2=2+\frac 2k\quad\text{and}\quad cm_2+\frac ck+dn_2=0$$
$$\implies m_2k(bc-ad)=-2kd-2d-(bc-ad)\tag4$$
From $(3)-(4)$,
$$(m_1-m_2)k(bc-ad)=kd\implies (m_1-m_2)(bc-ad)=d\tag5$$
Now, for $(M,N)=(0,0)$, there has to be a pair $(m_3,n_3)$ such that
$$am_3+\frac ak+bn_3=0+\frac 2k\quad\text{and}\quad cm_3+\frac ck+dn_3=0$$
$$\implies m_3k(bc-ad)=-2d-(bc-ad)$$
With $(5)$, we get
$$m_3k(bc-ad)=-2(m_1-m_2)(bc-ad)-(bc-ad)\tag6$$
Case 1 : If $bc-ad\not=0$, dividing the both sides of $(6)$ by $bc-ad$ gives
$$m_3k=-2(m_1-m_2)-1$$
This is impossible since the LHS is even and the RHS is odd.
Case 2 : If $bc-ad=0$, from $(5)$, we get $d=0$. So, $bc=0$. 
Case 2-1 : $c=0$, then there is no $(m,n)$ such that $cm+\frac ck+dn=1=N$.
Case 2-2 : $b=0$, then for $M=2$, there has to be an integer $m_4$ such that
$$am_4k+a=2k+2\tag7$$
For $M=1$, there has to be an integer $m_5$ such that
$$am_5k+a=k+2\tag8$$
So, from $(7)-(8)$,
$$ak(m_4-m_5)=k\implies a(m_4-m_5)=1\implies a=\pm 1$$
This is impossible since $a\ (=pk+2\ \text{where}\ p\in\mathbb Z)$ is even.
Hence, it is necessary that $k$ is odd.
$(2)$ 
For every odd $k$,
$$(a,b,c,d)=\left(2k+2,-k-2,k,\frac{-k-1}{2}\right)$$
works :
For any $(m,n)$, $$(M,N)=\left(am+\frac{a-2}k+bn,cm+\frac ck+dn\right)$$
For any $(M,N)$, $$(m,n)=\left(\frac{(k+1)M}{2}-(k+2)N+1,kM-2(k+1)N+2\right)$$
Therefore, from $(1)(2)$, it is necessary and sufficient that $k$ is odd.
