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Solve in positive integers: $$(x+1)^n-x^n=p^m$$

$p$ is prime, $n\ge 2$. Seemingly Zsigmondy's Theorem and LTE won't work here.

Though you can tell (as suggested by user barto), using Zsigmondy, that $n=q$ is prime and $p\equiv 1\pmod q$ (since $\text{ord}_p (1+x^{-1})=q\mid p-1$ by FLT).

This is just a problem I've thought of (because I was solving diophantine equations of the form $x^n-y^n=p^m$ and the case $(x+1)^n-x^n=p^m$ is the only one that doesn't let me use Zsigmondy's Theorem directly to finish the problem) and I'm interested to hear a solution.


Zsigmondy's Theorem tells us that $x^n-y^n$
(if $(x,y)=1, x>y+1, (x,y,n)\neq (2,1,6), \lnot(n=2\wedge (x+y=2^l, l\in\mathbb N))$) has at least $d(n)$ different prime divisors ($d(n)$ -- number of $n$ divisors), which is impossible ($d(n)\ge 2$).

If $n=2\wedge (x+y=2^l, l\in\mathbb N)$, then we're left with $x-y=2^{m-l}$, which has infinitely many solutions, fully characterized by $(x,y)=(2^{m-l-1}+2^{l-1},2^{l-1}-2^{m-l-1})$.

If $(x,y)\neq 1$ or $x=y+1$, we're left with having to know how to solve diophantine equations of the form $(x+1)^n-x^n=p^m$.

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  • $\begingroup$ Did you solve $x^n-y^n=p^m$ when $x-y\neq 1$? $\endgroup$ – Elaqqad Apr 15 '15 at 14:17
  • $\begingroup$ @Elaqqad edited. $\endgroup$ – user89167 Apr 15 '15 at 14:30
  • $\begingroup$ The easier cases you mention are treated here: Solve $x^p+y^p=p^z$, $p$ prime and $x,y>0$ with some answers that easily adapt to solve $x^p+y^p=q^z$, $p,q$ prime and $x,y>0$. See also the related Solve $n^s-(n-1)^t=1$ for $n,s,t>0$ (but so far not much insight is given there). $\endgroup$ – rabota Apr 15 '15 at 15:25
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    $\begingroup$ For what it's worth, $n$ is prime (by Zsigmondy) and hence $p\equiv1\pmod n$ (by considering $\DeclareMathOperator\ord{ord}\ord_p(1+x^{-1})$). Though I think this will be difficult as setting $x=1$ means you're asking to find all Mersenne prime powers. My guess is that there are infinitely many solutions for every $x>0$, but that a proof of this is out of reach. $\endgroup$ – rabota Apr 15 '15 at 15:32
  • $\begingroup$ Lemma: If a-b is odd and both a & b are positive, then for every n greater than 1, then $(a^n-b^n)/(a-b)$ have prime factor which does not divedes $(a-b)$. (It is the special case of a more general lemma) Proof by contradiction, assume that every prime factor of $(a^n-b^n)/(a-b)$ divides $(a-b)$, then by the "lifting the exponent lemma" we can conclude that $(a^n-b^n)=n(a-b)$, but notice that $(a^n-b^n)/(a-b)$ has n positive summands, therefor it is greater than n, contradiction. In the case of this problem $*****\textbf{n must be prime}*****$, otherwise it has a contradiction with the lemma. $\endgroup$ – Davood Jul 21 '16 at 15:40

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