Question 1062388
<pre><b><font size=4>
x+y = x×y = x²-y²

Obviously x = 0 = y is a solution.

if x+y = x²-y² then

{{{x+y}}}{{{""=""}}}{{{x^2-y^2}}}

{{{x+y}}}{{{""=""}}}{{{(x-y)(x+y)}}}

{{{(x+y)-(x-y)(x+y)}}}{{{""=""}}}{{{0}}}

{{{(x+y)(1-(x-y)^"")}}}{{{""=""}}}{{{0}}}

{{{(x+y)(1-x+y)}}}{{{""=""}}}{{{0}}}

{{{x+y=0}}}; {{{1-x+y=0}}}

{{{y=-x}}};   {{{y=x-1}}}

But we have to make sure that

{{{x+y=xy}}}

For the first case {{{y=-x}}}

{{{x-x=x(-x)}}}
{{{0=x^2}}}
{{{0=x}}}
{{{y=-x}}}
{{{y=0}}}

So one solution is x = y = 0, the obvious one.

------------------

For the second case {{{y=x-1}}}

{{{y=x-1}}}

{{{x+y=xy}}}

{{{x+(x-1)=x(x-1)}}}

{{{x+x-1=x^2-x}}}

{{{2x-1=x^2-x}}}

{{{0=x^2-3x+1}}}

{{{x^2-3x+1=0}}}

{{{x = (-(-3) +- sqrt((-3)^2-4*1*1 ))/(2*1) }}}

{{{x = (3 +- sqrt(9-4))/2 }}}

{{{x = (3 +- sqrt(5))/2 }}}

{{{y=x-1}}}

For {{{x = (3 + sqrt(5))/2 }}}

{{{y= (3 + sqrt(5))/2-1}}}

{{{y= (3 + sqrt(5))/2-2/2}}}

{{{y= (1 + sqrt(5))/2}}}

So a second solution is:

{{{x = (3 + sqrt(5))/2 }}}, {{{y= (1 + sqrt(5))/2}}} 

----------------

For {{{x = (3 - sqrt(5))/2 }}}

{{{y= (3 - sqrt(5))/2-1}}}

{{{y= (3 - sqrt(5))/2-2/2}}}

{{{y= (1 - sqrt(5))/2}}}

So a third solution is:

{{{x = (3 - sqrt(5))/2 }}}, {{{y= (1 - sqrt(5))/2}}} 

Edwin</pre></b></font>