Given the equation $$(1+x)^n=x^m$$ where $m$ and $n$ are two different natural numbers, I was trying to find as many solution as possible expressing them without transcendent functions.

WLOG we can suppose $n<m$ and coprime, in this case by the Abel–Ruffini theorem we know we can find a closed form solution for $m<4$. So I found solutions for $(n,m) \in \{(1,2),(1,3),(1,4),(3,4)\}$

Indeed, by playing with the numbers, I noticed that also $(1+x)^4=x^5$ has closed form solutions as the equation is divisible by $x^2+x+1$.

How can I find other solutions violating Abel-Ruffini?

  • $\begingroup$ what you mean by violating Abel-Ruffini? $\endgroup$
    – Masacroso
    May 25 '17 at 14:56
  • 5
    $\begingroup$ A remark: The polynomial $P(x)=(1+x)^n-x^m$ is divisible by $x²+x+1$ if $n$ is even and $3$ divide $2n-m$. Simply show that $P(j)=0$ using $1+j=-j^2$ with $j=\exp(2i\pi/3)$. $\endgroup$
    – Kelenner
    May 25 '17 at 14:57
  • 1
    $\begingroup$ @Masacroso it was just a joke: I just meant peculiar cases still solvable with $m>4$. $\endgroup$
    – N74
    May 25 '17 at 15:57

The difference of powers formula can yield solutions for $m>4$: $0 = (1+x)^{n} - x^{m} = (1+x)^{n} - (x^\frac{m}{n})^{n} = (1+x-x^\frac{m}{n})((1+x)^{n-1} +(1+x)^{n-2}x^{\frac{m}{n}} + \ldots + (x^{\frac{m}{n}})^{n-1})$.

Under the assumption that $m>n$, then the zeros $1+x-x^{\frac{m}{n}}$ will yield all possible solutions. The huge benefit here is that zeros of this polynomial can be quickly found using Newton's Method. This can help you find your desired other solutions.

  • $\begingroup$ I am trying to understand, so the equation $\left(1+x\right)^5-x^4$ has a solution for $x=-0.461$. When I construct this equation: $1+x-x^{\frac{5}{4}}$ I get a root of x=3.08. What is it I did wrong. I would like to ask you to provide an example. Thanks. $\endgroup$
    – NoChance
    Sep 3 '19 at 2:42
  • $\begingroup$ @NoChance, my approach only works in the instances where the power of $x^{m}$ is higher than the power of $(1+x)^{n}$. Otherwise, it will only capture 1 of potentially 2 solutions. In your case, you want to solve roots of $1+x-x^{\frac{4}{5}}$. $\endgroup$
    – Oreomair
    Sep 3 '19 at 3:06
  • $\begingroup$ OK, I tested with $(1+x)^4 - x^5$ with $1+x+x^{4/5}$ and it worked. Root is about 3.08 - Very nice. I will test more....Here is the link: desmos.com/calculator/dzcmy0y3as $\endgroup$
    – NoChance
    Sep 3 '19 at 3:18

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