If we are on coordinate $(m,n)$ then we can either go to coordinate $(m + n, n)$ or $(m, m + n)$. If we start at $(2,5)$ then what all coordinates we can reach in finitely many moves?
In our case we have: From coordinate $(2,5)$ we can go to $(7,5)$ or $(2,7)$, now if you choose to go to $(7,5)$ then from $(7, 5)$ you can go to $(12,5)$ or $(7,12)$ and so on we keep on iterating this process and we need to find all such reachable coordinates. Any help would be highly appreciated. Thanks.
On
First, note that the map $(m, n) \mapsto (m+n, n)$ corresponds to the matrix $E_1 = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$ and the map $(m, n) \mapsto (m, m+n)$ corresponds to the matrix $E_2 = \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}$. Therefore, it will be useful to study what matrices can be formed as products of these two matrices.
Lemma: A matrix in $M_{2\times 2}(\mathbb{Z})$ can be expressed as a product of $E_1$ and $E_2$ (for example $E_1^5 E_2^2 E_1^7 E_2^3 E_1$) if and only if it has determinant 1 and every entry is a nonnegative integer.
Proof: The forward direction is easy to see. For the reverse direction, we will prove that any such matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ with $ad - bc = 1$ and $a,b,c,d \in \mathbb{N}$ is a product of $E_1$ and $E_2$ by strong induction on $a + c$.
First, if any of $a,b,c$ or $d$ is zero, it certainly cannot be either $a$ or $d$, and $a = d = 1$. Then if $b = 0$, we have $$\begin{bmatrix} 1 & 0 \\ c & 1 \end{bmatrix} = E_2^c.$$ Similarly, if $c = 0$, we have $$\begin{bmatrix} 1 & b \\ 0 & 1 \end{bmatrix} = E_1^b.$$
It now remains to show the case where all of $a,b,c,d > 0$. Then if $a = c$, we must have the common value is $a = c = 1$, and the matrix is of the form $$\begin{bmatrix} 1 & b \\ 1 & b+1 \end{bmatrix} = E_2 \begin{bmatrix} 1 & b \\ 0 & 1 \end{bmatrix} = E_2 E_1^b.$$
If $a > c$, then $ad - bc = 1$ implies that $\frac{a}{c} - \frac{b}{d} = \frac{1}{cd}$. Therefore, $\frac{b}{d} = \frac{a}{c} - \frac{1}{cd} \ge \frac{a}{c} - \frac{1}{c} \ge \frac{c}{c} = 1$, so $b \ge d$. We now see that $$\begin{bmatrix} a & b \\ c & d \end{bmatrix} = E_1 \begin{bmatrix} a - c & b - d \\ c & d \end{bmatrix}.$$ However, by inductive hypothesis since $(a-c) + c < a+c$, we have $\begin{bmatrix} a-c & b-d \\ c & d \end{bmatrix}$ is a product of $E_1$ and $E_2$, so $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ is also a product of $E_1$ and $E_2$.
If $a < c$, then a similar argument shows that $b \le d$, and $\begin{bmatrix} a & b \\ c & d \end{bmatrix} = E_2 \begin{bmatrix} a & b \\ c - a & d - b \end{bmatrix}$ where the second factor can be expressed as a product of $E_1$ and $E_2$ by inductive hypothesis. $\square$
Therefore, the points reachable from $(2, 5)$ via the operations $E_1$ and $E_2$ are precisely the points which can be expressed as $(2a + 5b, 2c + 5d)$ where $a,b,c,d \in \mathbb{N}$ and $ad - bc = 1$.
(I'm unsure of exactly what kind of characterization the original problem is asking for. This does provide a straightforward way of generating reachable points - for example by choosing random natural numbers $b,c$, factoring $bc + 1$ and assigning a random factorization to $a,d$. On the other hand, if you're given a pair and you want to determine whether it's reachable, it's probably easier to use the procedure described by Hagen von Eitzen's answer rather than trying to solve for $a,b,c,d$ in this description.)
Given $(x,y)$, can it be reached? Clearly (or formally: by induction), $x,y$ must be integers and $x\ge 2$, $y\ge 5$. Also, we see (again formally: by induction) that $x\ne y$ because the allowed steps can only produce different coordinates $m+1\ne n$ or $m\ne m+n$ from positive $n,m$. Thus $(x,y)\ne(2,5)$ can be reached only via $(x-y,y)$ if $x>y$ and only via $(x,y-x)$ if $y>x$ and not at all if $x=y$.
Thus to check if $(x,y)$ can be reached, we only have to repeatedly go backwards and subtract the smaller from the larger coordinate and have won iff the last position before making $x<2$ or $y<5$ is $(2,5)$. Note that these steps are really just Euclid's algorithm in disguise, except that we stop a few steps before one coordinate becomes $0$ and the other the gcd. In particular, necessarily $\gcd(x,y)=\gcd(2,5)=1$. But this is not sufficient as many such pairs will rather go via points like $(3,7)$ for example.