Suppose $a^n + b^n = a+b$ for all $n \in \mathbb{N}$.
Is it true that the only solutions are:
$$a=1, b=0$$ $$a=0, b=1$$ $$a=b=0$$
?
If so, how do you prove it?
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1$\begingroup$ The right-hand side is bounded. $\endgroup$ – saulspatz Mar 20 '18 at 3:58
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$\begingroup$ No, there is also $a=b=1$. $\endgroup$ – Benjamin Dickman Mar 20 '18 at 5:30
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Let $A:=\{(1,0),(0,1),(0,0),(1,1)\}$
It is clear that if $(a,b) \in A$, then $a^n+b^n=a+b$.
If $(a,b)=(1,-1)$ or $=(-1,1)$, then $a^2+b^2=2 \ne 0=a+b$.
Let $a^n+b^n=a+b$ for all $n$. If $|a|>1$ or $|b|>1,$ then the sequence $(a^n+b^n)$ is unbounded, a cotradiction , since this sequence is constant.
Hence, by 1., 2. and 3. , we have only to investigate the case $|a| < 1$ and $|b| < 1$.
From $a^n+b^n=a+b$ for all $n$ we get with $ n \to \infty$ that $ a=-b$.
Hence $0=a+b=a^2+a^2=2a^2$, therefore $a=b=0$.
Conclusion: $a^n+b^n=a+b \iff (a,b) \in A$.
