To find all natural triples $(a, b, c)$ such that $2^c-1 \mid 2^a+2^b+1$

Question

Find all positive integer triples $(a, b, c)$ such that $2^c-1 \mid 2^a+2^b+1$.

I have no important result on this one!

Answer 1

Appetizers from the comments:

• $c=1$ is left as an exercise
• $c=2$ works when both $a$ and $b$ are even
• $c=3$ works when $a,b$ and $0$ are pairwise non-congruent modulo $3$.

Then the main course:

Claim. There are no solutions with $c>3$.

Proof. We have trivially that if $a\equiv a'\pmod c$, then $$ 2^a\equiv2^{a'}\pmod{2^c-1}. $$ Therefore without loss of generality we can assume that $0\le a,b<c$. But in that case there are no solutions for larger $c$:

• If $a\neq b$ are distinct, then $2^a+2^b+2^0<2^c-1$ (observe that $2^c-1$ is the sum of $c>3$ distinct powers of two).
• If $a=b$, then it won't work. Either the sum $2^a+2^b+1$ is still $<2^c-1$, or $a=b=c-1$ and $2^a+2^b+1=2^c+1$.