2
$\begingroup$

Here's the context of this problem.

Solve:

$x^2=y^3-3y^2+2y$

$y^2=x^3-3x^2+2x$

We subtract the second equation from the first and obtain $$(x-y)(x^2+y^2+xy-2x-2y+2)=0$$ The first factor yields the solutions $0,2\pm\sqrt2$ and the second one can be rewritten as $$(x+y)^2+(x-2)^2+(y-2)^2=4$$ I have checked Wolfram Alpha and it tells me that there are no more real solution to the above system.This means that the above equation has no solution.Now I have to prove it.

Obviously,none of the expressions $(x+y)^2,(x-2)^2,(y-2)^2$ can be $4$ otherwise the LHS would be larger than $4$. Hence,we have the inequalities $$|x+y|<2$$$$|x-2|<2$$$$|y-2|<2$$ which(unless I am mistaken) yield $$0<x<2$$ $$0<y<2$$ I cannot proceed any further.Perhaps there is some clever way to write the whole expression solely using squares but I cannot see it immediately.

I would appreciate it if the method used to solve this is elementary.Some help will be appreciated.

$\endgroup$
5
  • $\begingroup$ Without any calculus, correct? $\endgroup$
    – JohnD
    Dec 16, 2014 at 13:39
  • $\begingroup$ @JohnD,yes,correct. $\endgroup$
    – rah4927
    Dec 16, 2014 at 13:40
  • $\begingroup$ if you are looking for non trivial solution of course.. (0,0) $\endgroup$
    – Chinny84
    Dec 16, 2014 at 13:42
  • $\begingroup$ $f(x,y)=(x+y)^2+(x-2)^2+(y-2)^2>4$ for all $x,y$. In fact, the minimum of $f$ is $16/3$ and occurs at $(2/3,2/3)$. $\endgroup$
    – JohnD
    Dec 16, 2014 at 13:42
  • $\begingroup$ @Chinny84,oh,the first factor yields that solution.Forgot to mention it. $\endgroup$
    – rah4927
    Dec 16, 2014 at 13:43

1 Answer 1

6
$\begingroup$

Write $$f(x) = (x+y)^2 + (x-2)^2 + (y-2)^2 - 4$$ for a fixed $y$.

The zeros of $f(x)$ are the solutions you want. If you compute the discriminant of $f(x)=0$, you get $-8y^2 + 16y -32$. This is negative for every $y$, hence $f(x)$ has no real solution for every $y$. This suffice!

$\endgroup$
0

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .