RMO 1994

From $5x=(12-5x)(x^2+y^2),\quad 5y=(4+5y)(x^2+y^2)$ it follows that $\frac{x}{y}=\frac{12-5x}{4+5y}$ , or: $(5x-6)(5y-2)=12.$ Since we are looking for integer solutions, we have to write twelve as a product of two integers, the first one congruent to $4\pmod{5}$ and the second one congruent to $3\pmod{5}$ , so the only possibilities are $4\cdot 3$ and $(-1)\cdot(-12)$ and the only integer solution is $(x,y)=(2,1)$ .

$Let\quad z=x+iy\\ Then\quad z\bar { z } ={ x }^{ 2 }+{ y }^{ 2 }\\ Multiply\quad second\quad equation\quad by\quad i\quad and\quad add\quad it\quad to\quad first\quad equation\quad then\quad we\quad have\\ z+\bar { \frac { z }{ z\bar { z } } } =\frac { 12+4i }{ 5 } \quad or\quad z+\frac { 1 }{ z } =\frac { 12+4i }{ 5 } \quad Solving\quad we\quad get\quad z=\frac { 2-i }{ 5 } \quad or\quad 2+i\quad hence\quad to\quad ensure\\ integral\quad values\quad for\quad x\quad and\quad y\quad we\quad have\quad z=2+i\quad hence\quad x=2\quad and\quad y=1\quad therefore\quad x+y=3\quad is\quad the\quad answer$

From solving:

$5x(1+\frac{1}{x^{2}+y^{2}}) = 12$ $, 5x(1-\frac{1}{x^{2}+y^{2}}) = 4$

We get $x=2$ and $y=1$

So, $a=2$ and $b=1$

Thus, the answer is: $a+b=2+1=\boxed{3}$

set x^2+y^2=a and solve the 2 equations and tind the parralel solutions