Let $a,b\geq 1$ be integers and $k=\frac{a}{b}>1$. Solve $$(n+1)^k=n^k+1,\quad n\in\mathbb{Z}.$$ It is clear that $n=0$ is a solution for such equation. I found that if $a,b$ are odd, then $n=-1$ is also a solution of the equation. Do you have another solutions?
Asked
Active
Viewed 69 times
2
-
Hint: $(n+1)^k-n^k$ strictly increasing – Michael Galuza Jul 26 '15 at 03:14
-
It has only 2 solutions, right? – Worachead Sommanee Jul 26 '15 at 03:51
-
Yes. $(n+1)^k-n^k$ strictly increasing for $n>0$. For $n<0$ it depends on $k$, but if $n+1<0$, function is monotone. So, you should check $n=0$ and $n=-1$; you checked it. – Michael Galuza Jul 26 '15 at 04:07
-
Even if n negative? – Moti Jul 26 '15 at 05:37
-
@Moti if $n$ is nagative, then $n=-1$ is only one solution for the equation (when $a,b$ are odd). Do you have another solution? – Worachead Sommanee Jul 27 '15 at 08:25
-
I need to think if $$n^{(k-1)}+n^{(k-2)}+...+1=0$$ has more than one solution for some k – Moti Jul 28 '15 at 16:38