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I have the following nonlinear system of two equations \begin{align} 1+4x-y =\frac{49}{15}\qquad\text{ together with }\qquad \frac{x}{x-1} =\left(\frac{y}{y-1}\right)^4\\ \end{align}

I can solve it by substitution, and find the solution of the resulting cubic equation $34 - 121 y + 144 y^2 - 46 y^3=0$, yielding $y=2$, and thus $x=16/15$. However, the simplicity of the solution, and the form of the nonlinear equation of the two, suggests that a smarter and faster way than my brute force attempt may exist. I would appreciate any suggestion.

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We have that

$$\frac{x}{x-1} =\left(\frac{y}{y-1}\right)^4 \iff \frac{x}{x-1} =\left(\frac{f(x)}{f(x)-1}\right)^4$$

with $f(x)=4x-\frac{34}{15}$.

Since

  • $g(x)=\frac{x}{x-1}:(1,\infty)\to (1,\infty)$ is bijective
  • $f(x)=4x-\frac{34}{15}$ is bijective
  • $h(x)=\sqrt[4]x:[0,\infty)\to [0,\infty)$ is bijective

we expect exactly one solution that is indeed $x=\frac{16}{15}$.

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Eliminating $y$ from your system you will get $$2484000 x^3-6166800 x^2+5004495 x=1336336$$ with only one real solution $$x=\frac{16}{15}$$

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  • $\begingroup$ Yes, I know...you would need to solve a cubic equation, though. What I am asking is: is there a smarter way that avoids cubic equations? $\endgroup$ – Pierpaolo Vivo Dec 2 '19 at 18:05
  • $\begingroup$ I think there isn't. $\endgroup$ – Dr. Sonnhard Graubner Dec 2 '19 at 18:07
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Probably not "faster", but: suppose you guess that there is a rational solution. If $y/(y-1) = \alpha/\beta$ with $\alpha$ and $\beta$ coprime integers, $x/(x-1) = \alpha^4/\beta^4$. Then $y = \alpha/(\alpha - \beta)$ and $x = \alpha^4/(\alpha^4 - \beta^4)$, so your first equation becomes $$ 1 + \frac{4 \alpha^4}{\alpha^4-\beta^4} - \frac{\alpha}{\alpha- \beta} = \frac{49}{15}$$ Now $\alpha^4 - \beta^4 = (\alpha - \beta)(\alpha^3 + \alpha^2 \beta + \alpha \beta^2 + \beta^3)$, where $\alpha^3 + \alpha^2 \beta + \alpha \beta^2 + \beta^3$ is coprime to $\alpha$. Unless $\alpha = 0$ (which certainly does not work), the denominator of the left side will be divisible by $\alpha^3 + \alpha^2 \beta + \alpha \beta^2 + \beta^3$, and the only way to make this equation work is that this factor divides $15$. It's easy to check that $\alpha = 2$, $\beta = 1$ makes $\alpha^3 + \alpha^2 \beta + \alpha \beta^2 + \beta^3 = 15$, and this does happen to satisfy the equation.

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  • $\begingroup$ Would you not need to check all factors of $60$ because the expression could have a common factor of $4$ with the numerator of $4\alpha^4$? $\endgroup$ – S. Dolan Dec 2 '19 at 19:13
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For a rational solution, let $$\frac{y}{y-1}=\frac{u}{v}$$ with $u$ and $v$ coprime. Then $$11u^4-15u^3v-15u^2v^2-15uv^3+34v^4=0.$$ Therefore $u$ is a factor of $34$ and $v$ is a factor of $11$.

The solution is $u=2,v=1$.

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