\BUMP \BUMP

parmenides51 wrote: Determine all triples of natural numbers $(a,b, c)$ with $b> 1$ such that $2^c + 2^{2016} = a^b$. Let $a=2^mk$ with $(k,2)=1$. Then $2^c + 2^{2016} = 2^{mb}k^b$. So $mb=min(c, 2016)$. Then use Catalan's theorem.

Lemma: $2^b+1=x^a$ with $x>0, a,b>1$, has only one solution $x=3, a=2, b=3$. Prove: Case 1: $a = 2m$. $2^b=(x^m-1)(x^m+1)$ so $x^m=3$ so $m=1, a=2, x=3, b=3$ Case 2: $a= 2m+1$. $2^b=(x-1)(x^{2m}+x^{2m-1}+\cdots+x+1)$ so $x-1=2^b$ so contradiction because $a>1$ Solution: Case 1: $c> 2016$. $2^{2016}(2^{c-2016}+1)=a^b$. So from lemma $c=2019, a=3*2^{1008}, b=2$ Case 2: $c< 2016$. $2^c(2^{2016-c}+1)=a^b$. So from lemma $c=2013, b=2$ but $2^{2013}$ is not square so no solutions Case 3: $c=2016$. $2017$ is prime so no solutions