Consider the linear equation:
$$x(1 - 1000m) + y (10 - 100m) + z (100-10m) + w(1000-m) = 0 \qquad (1)$$
where $x,y,z,w \in \mathbb N_9, \; xw \neq 0, m \in \{ 2,3,\dots, 9 \}$
Having put this to a computer it outputs the following solutions:
$$(x,y,z,w,m) = (2,1,7,8,4) \\ (x,y,z,w,m) = (1,0,8,9,9)$$
How can I prove that the number of solutions is exactly (or not) 2? (Generally find the number $N$ of solutions to $(1)$ )
(Since the gcd of the coefficients of $x,y,z,w$ does divide 0 , then we are sure that there exist integer solutions.)
Your problem can be enunciated in an equivalent and more clear way: How many integers of four digits, noted $wzyx$, and non-divisible by 10 are there such that $$\frac{wzyx}{xyzw}\in\{2,3,4,5,6,7,8,9\}?$$ $$1000(w-mx)+100(z-my)+10(y-mz)+(x-mw)=0\Rightarrow x-mw=10N$$ ►${x\ne w}$
$x=w$ would be absurd because $$\frac{wzyw}{wyzw}=m>1$$
►${m=2}$
$x-2w=10M\Rightarrow x=2,4,6,8\Rightarrow 1-w, 2-w,3-w,4-w$ are each multiple of $5$. Hence the four possibilities $(x,w)=(2,6),(4,7),(6,8),(8,9)$ which give no solutions because $wzyx=2(xyzw)$ and both $2x$ and $2x+1$ are distinct of $w$; for example for the first case $(2,6) $ we need in $6zy2=2(2yz6)$ that $2\cdot2=6$ or $2\cdot2+1=6$, absurd.
►${m=3}$
$x-3w=10M\Rightarrow (x,w)=(1,7),(2,4),(3,1),(4,8),(5,5),(6,2),(7,9),(8,6),(9,3)$ and because of $w>x$, $(x,w)=(1,7),(2,4),(4,8),(7,9)$ only. These give no solution for $wzyx=3(xyzw)$ by the same argument for $m=2$.
►${m=4}$
$x-4w=10M\Rightarrow (x,w)=(2,3),(2,8),(4,1),(4,6),(6,4),(6,9),(8,2),(8,7)$ from which we must consider only $(x,w)=(2,3),(2,8),(4,6),(6,9)$. For $(2,8)$ there is the solution $(x,y,z,w,m)=(2,1,7,8,4)$ given by the OP and it is clear it is unique. For the other possibilities, $wzyx=4(xyzw)$ have no solution ( $4\cdot2>3$ and $10<4\cdot4<4\cdot6$).
►${m=5}$
$x-5w=10M\Rightarrow (x,w)=(5,1),(5,3),(5,5),(5,7),(5,9)$ Hence $(x,w)=(5,7),(5,9)$ don’t give solution because $5\cdot5>10$.
►${m=6}$
$x-6w=10M\Rightarrow (x,w)=(2,2),(2,7),(4,4),(4,9),(6,1),(6,6),(8,3),(8,8)$. Just $(2,7)$ and $(4,9)$ are allowable and these don’t give solution because $6\cdot2$ and $6\cdot4$ are greater that $10$.
►${m=7}$
$x-7w=10M\Rightarrow (x,w)=(1,3),(2,6),(3,9),(4,2),(5,5),(6,8),(7,1),(8,4),(9,7)$. Just $(1,3),(2,6),(3,9),(6,8)$ must be taken which give no solution because $7$ and $8$ are distinct of $3$ and $7x>10$ for $2,3,6$.
►${m=8}$
$x-8w=10M\Rightarrow (x,w)=(2,4),(2,9),(4,3),(4,8),(6,7),(8,1),(8,6)$ so $(x,w)= (2,4),(2,9),(4,8),(6,7)$. No solution because $8x>10$ for the four values of $x$
►${m=9}$
$x-9w=10M\Rightarrow (x,w)=(1,9),(2,8),(3,7),(4,6),(5,5),(6,4),(7,3),(8,2),(9,1)$ We keep just $(x,w)=(1,9),(2,8),(3,7),(4,6)$. The first couple $(1,9)$ gives the answer $(x,y,z,w,m)=(1,0,8,9,9)$ of the OP which is unique for this couple. The other give no solution because $9x>10$ for $x=2,3,4$
Thus the two given solutions are the only ones.