I don't understand. So $ c$ and $ n$ are given, and we find $ a$ and $ b$ in terms of $ c$ and $ n$? That's weird.

No we shouldn't find a and b in terms of c and n. (i.e. a=2 b=2 is a solution and we should find the solutions like this) I think we should consider the cases. 1- a=b 2- a>b. I found that in the first case only a=b=2 is a solution and in the second case i got a contradiction but I am not clear.

ridgers wrote: Find all natural numbers a, b such that $ a^{n} + b^{n} = c^{n + 1}$ where c and n are naturals. I dont know how is it possible to find $ a$ and $ b$ if $ c$ and $ n$ are fixed given values. But here is a classical family of solutions : $ (u(u^n+v^n))^n+(v(u^n+v^n))^n=(u^n+v^n)^{n+1}$

pco wrote: ridgers wrote: Find all natural numbers a, b such that $ a^{n} + b^{n} = c^{n + 1}$ where c and n are naturals. I dont know how is it possible to find $ a$ and $ b$ if $ c$ and $ n$ are fixed given values. But here is a classical family of solutions : $ (u(u^n + v^n))^n + (v(u^n + v^n))^n = (u^n + v^n)^{n + 1}$ $ c$ and $ n$ are not fixed value.

If $ c$ and $ n$ are not fixed.The equation $ a^n+b^n=c^{n+1}$ has infinitely many solutions in $ N$.For example, $ a_k=1+k^n,b_k=k(1+k^n),c_k=1+k^n$,where $ k \in N$ satisfies the hypothesis.

SUPERMAN2 wrote: If $ c$ and $ n$ are not fixed.The equation $ a^n + b^n = c^{n + 1}$ has infinitely many solutions in $ N$.For example, $ a_k = 1 + k^n,b_k = k(1 + k^n),c_k = 1 + k^n$,where $ k \in N$ satisfies the hypothesis. Hummm, this is exactly my solution $ (u(u^n+v^n))^n+(v(u^n+v^n))^n=(u^n+v^n)^{n+1}$ where $ u=1$ and $ v=k$

The similar problems of Dorin Andrica and I think they are all very nice: The following equations have infintely many intergral solutions $ \forall n \in N$: i)$ x^n+y^n+z^n+t^n=a^{n-1}$ ii)$ x^n+y^n+z^n+t^n=a^{n+1}$

ridgers wrote: Find all natural numbers a, b such that $ a^{n} + b^{n} = c^{n + 1}$ where c and n are naturals. The exact version is : Find all natural numbers $ a,b$ such that for all $ n\in N$ then there exists $ c\in N$ satisfying $ a^{n} + b^{n} = c^{n + 1}$. And the unique result is $ a=b=2$ and then $ c=2$.

Allnames wrote: Find all natural numbers $ a,b$ such that for all $ n\in N$ then there exists $ c\in N$ satisfying $ a^{n} + b^{n} = c^{n + 1}$. A nice one. Lemma 1. If $ p \in \mathbb{P} \setminus\{2\},p^a ||a-b$ and $ p^b||n$ then $ p^{a+b}||a^n-b^n$. Lemma 2. If $ p=2, p^{a+1}||a^2-b^2, p^b||n$ then $ p^{a+b}||a^n-b^n$. (Both done by induction on b). Let p>2 a prime s.t. $ \upsilon_p(a-(-b))=x \in \mathbb{N}_0$ then for all odd n, by lemma 1: $ n+1 \mid \upsilon_p(c^{n+1})=\upsilon_p(a^n+b^n)=x+\upsilon_p(n)$, contradiction because lhs is greater for enough large n. So a+b must be a power of 2, in particular 2|a-b. Now if $ (a-b)^2>0 \text{ and }\upsilon_2{a^2-b^2}=x+1$ then for all odd n, by lemma 2: $ n+1 \mid \upsilon_2{a^n+b^n}=x+\upsilon_2(n)$, again contradiction. So a=b. So $ 2a^n=c^{n+1}$: if 2<p|a then $ n+1|\upsilon_p(2a^n)=n\upsilon_p(a) \implies n+1 \mid \upsilon_p(a)$ again contradiction. So $ a=2^k$ for suitable choice of $ k \in \mathbb{N}$ and $ 2^{kn+1}=c^{n+1}$, in particular $ n+1|kn+1-k(n-1) \implies n+1 \mid k-1$ for all n, so a=b=c=2.

Allnames wrote: ridgers wrote: Find all natural numbers a, b such that $ a^{n} + b^{n} = c^{n + 1}$ where c and n are naturals. The exact version is : Find all natural numbers $ a,b$ such that for all $ n\in N$ then there exists $ c\in N$ satisfying $ a^{n} + b^{n} = c^{n + 1}$. And the unique result is $ a = b = 2$ and then $ c = 2$. Thank you very much . This is the correct version

Another approach. Lemma1. Denote $ \{c_i\}_{i > 0}$ the sequence of integer such that $ a^i + b^i = c_i^{i + 1}$. Then $ c_{i + 1} \ge c_i$. So it is definitively costant because of is bounded by (a+b). Lemma2. We have $ 2 \equiv a^{p - 1} + b^{p - 1} = c^{p} \equiv c \pmod p$ for all p enough large, so c=2 definitively. Then for all enough large n we have $ (2^{n + 2})^2 = 2^{n + 1} \cdot 2^{n + 3} \implies$ $ (a^{n + 1} + b^{n + 1})^2 = (a^n + b^n)(a^{n + 2} + b^{n + 2}) \implies a = b = c = 2$. Another approach. $ c$ tends to $ \max\{a,b\}$, so for enough large n we'll have $ a^n=(b-1)b^n \implies a=b=c=2$.

bboypa wrote: Another approach. $ c$ tends to $ \max\{a,b\}$, so for enough large n we'll have $ a^n = (b - 1)b^n \implies a = b = c = 2$. It was also my approach. I remember that this nice problem belongs to Lautentiu Panatopol.

Russia Mo: a, b are two given natural numbers such that the sum of their n-th powers for any n is a (n+1)-th power: a^{n}+b^{n}=c^{n+1} Solution: Let p>a+b be a prime. a^{p-1}+b^{p-1}=c^{p} By Fermat's little theorem it follows that p must divide c-2. So c=pk+2. If k>0 then a^{p-1}+b^{p-1}\ge (p+2)^p\ge (a+b+2)^p which is a contradiction. So c=2 and it can be easily shown that a=b=2.

I guess there are infinitely many solutions. Because if n = 1 then search for natural numbers a and b such that a +b equals to a perfect square. There are infinitely many cases that a+b is a perfect square, so find sqrt(a+b) and you will find c.