A problem that I just found

From $11A=9B$ , we have $A=9t$ and $B=11t$ for some integer $t$ .

From $B+C=A$ , we have $C=-2t$ .

From $A+B=C^2$ , $20t=4t^2$ , so $t=0$ or $t=5$ . Both of these give valid solutions - respectively $(A,B,C,X)=(0,0,0,0)$ and $(A,B,C,X)=(45,55,-10,10)$ .

So there are in fact two solutions: $X=\boxed0$ and $X=\boxed{10}$ .

A+B=C^2 from B+C=A we have C=A-B so we put that in the first equation and we get A+B=(A-B)^2 and we expand that to A+B=A^2 + 2AB +B^2 (1) now from the equation: 11A=9B we have B=A11/9 (2) so we can substitute that in (1) and we get, A+A11/9 =A^2 + 2*11(A^2)/9 + 121(A^2)/81 and multiplying with 81 and solving the quadratic we get A=45 and by (2) we have B=55 and that will give us (A+B)/10 = X and we have X=10