2024/014) Solve $x^2-y=111$ and $y^2-x=111$ for $x\ne y$

 We are given 

 $x^2-y=111\cdots(1)$

 $y^2-x=111\cdots(2)$

From (1) and (2)

$x^2-y = y^2 -x$

Or $x^2-y^2 + x -y = 0$

Or $(x-y)(x+y) + (x-y) = 0$

Or  $(x-y)(x+y+1) = 0$

As  $x\ne y$  dividing by x-y we get $x+y+1=0\cdots(3)$

Or $y = -(x+1)$

Putting the value in (1) we get $x^2 + (x+ 1) = 111$ or $x^2 + x - 110 = 0$

Or $(x-10)(x+11) = 0$

Or $x= 10$ or $x = -11$

Putting in (3) if $x= 10$ then $y = -11$

If $x= -11$ then $y = 10$

So Solution set $x=10,y= -11$ or $x=-11,y=10$