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How to prove that the following equation: $$1+x^8y^4+x^4y^8-x^2y^4-x^6y^6-x^4y^2=0$$

has for solution(in real numbers): $|x|=|y|=1~$ only.

Any hint would be appreciated.

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Note that the proposed equation is equivalent to $$ (1-x^2y^4)^2+(1-x^4y^2)^2+x^4y^4(x^2-y^2)^2=0 $$ So, any solution satisfies $x^2y^4=x^4y^2=1$ and $xy(x^2-y^2)=0$. This implies clearly that $x^2=y^2=1$.$\qquad\square$

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Pick any number $\alpha\neq 0$ and assume there is a solution with $y =\alpha x$, then the equation becomes $$ 1-(\alpha^2+\alpha^4)y^6 + (\alpha^4-\alpha^6+\alpha^8)y^{12}=0.$$ This is a quadratic in $z=y^6$ that can be solved. For $\alpha=\pm 1$ you get the solutions $|x|=|y|=1$, but you should also get real solutions for other values of $\alpha$.

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  • $\begingroup$ Hint: put multicharacter exponents in braces. So y^{12} instead of y^12 gives $y^{12}$ instead of $y^12$ This works generally: fractions, subscripts, etc. $\endgroup$ – Ross Millikan Jun 18 '14 at 19:59
  • $\begingroup$ Yes, indeed. In my case the problem is that I didn't read properly what I wrote. Sometimes one gets blind to ones own errors... Thanks for the correction. $\endgroup$ – Peder Jul 7 '14 at 20:19

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