This problem is literally made in China.

Adding the equations together gives us that $(m + 3)x = 10$ .

Next, note that $m \ne 0$ , as this would yield solutions $x = \frac{10}{3}, y = 5$ . So multiply the first equation by $3$ and the second by $m$ and then subtract to find that

$(3mx + 6y) - (3mx - 2my) = 30 - 0 \Rightarrow (6 + 2m)y = 30 \Rightarrow (m + 3)y = 15$ .

So since both $(m + 3)x = 10$ and $(m + 3)y = 15$ , and since we require integer solutions, we can conclude that $(m + 3)$ must be a common factor of $10$ and $15$ , the options being $-5, -1, 1$ and $5$ . This gives us possible values for m of $-8, -4, -2$ and $2$ .

Now the reader can check that each of these values of $m$ , when plugged back into the original system of equations, yields integer solutions for $x$ and $y$ . Thus $m^{2}$ could be any of $64, 16$ or $4$ . To make the solution unique, the question should specify that the system has positive integer solutions, which is only the case for $m$ being either $-2$ or $2$ . This would then make $m^{2} = 4$ the unique solution.

3x - 2y = 0 => y = 3x/2 ------------- (1)

mx + 2y = 10 putting 1 in this.

(m + 3)x = 10

The only possible values of m that would give integral solutions are,

m = 7, 2, -8, -13 for which the values of x become 1, 2, -2, -1

For the solution to have integral solutions we want the y co-ordinate also to be integral. Hence putting 1, 2, -2, -1 in 3x - 2y = 0 and checking the cases where we get integral solutions for y.

1 => 3/2 2 => 3 -2 => -3 -1 => -3/2

Of these cases 2 and -2 are the only values of x for which we get integral values for y. Hence m could be -8 or 2 hence m could be 4 or 64. Given the fact that you had three chances you have to try your luck with both values and see if you hit the correct one.

put m=2 and add first equation to the other we get ,5x=10 then x=2,we back to the equation ,3x-2y=0 -> 2y=6 y=3 and that is integer numbers so m^2 =4 ,thanks