Matematika wrote: Determine all triplets off positive integers $ (a,b,c)$ for which $ \mid2^a - b^c\mid = 1$ If $ c=1$, we have the trivial solutions $ (a,b,c)=(a,2^a\pm 1,1)$ Consider now $ c>1$ 1) $ b^c=2^a+1$ $ b$ is odd $ >1$ (else $ 2^a=0$) and so $ b=2^ku+1$ with $ u$ odd and $ k>0$ $ (2^ku+1)^c=2^a+1$ $ \implies$ $ 2^kcu=2^a\pmod{2^{k+1}}$ and, since $ k<a$ (since $ c>1$) $ c=2d$ even. The equation becomes $ (b^d-1)(b^d+1)=2^a$ and so $ b^d-1=2$ and so $ b=3$ and $ d=1$ and the unique solution $ (a,b,c)=(3,3,2)$ 2) $ b^c=2^a-1$ If $ b=1$, we have the solution $ (a,b,c)=(1,1,c)$ Else, $ a>1$ and $ b$ is odd $ >1$ and so $ b=2^ku+1$ with $ u$ odd and $ k>0$ $ (2^ku+1)^c=2^a-1$ $ \implies$ $ 2^a=2\pmod{2^{k}}$ and so (since $ u>0$ and $ c>1$ imply $ k<a$) $ k=1$ $ (2u+1)^c=2^a-1$ $ \implies$ $ 2cu=2\pmod 3$ and so $ c=2d+1$ odd. The equation becomes $ b^{2d+1}+1=2^a$ and so $ b+1=2^u$ $ (2^u-1)^{2d+1}=2^a-1$ $ \implies$ $ (2d+1)2^u=2^a\pmod{2^{u+1}}$ and so $ a=u$ and $ c=1$, impossible (since we considered $ c>1$) 3) Synthesis of solutions : $ (a,b,c)=(a,2^a\pm 1,1)$ $ (a,b,c)=(3,3,2)$ $ (a,b,c)=(1,1,c)$

Hi pco, part (1) of your solution is correct. But unfortunately, part (2) is wrong. 2) $ b^c=2^a-1$ If $ b=1$, we have the solution $ (a,b,c)=(1,1,c)$ SO FAR EVERYTHING IS FINE. Else, $ a>1$ and $ b$ is odd >1 and "so $ b=2^ku+1$ with $ u$ odd and $ k>0$" The claim within quotes is wrong. How does one know that $ b$ when divided by $ 2^k$ leaves a remainder of 1. I am including herewith my solution to the second part. We have $ a \geq 2$. Hence, $ 2^a - 1 \equiv 3 (\;\text{mod} \;4)$. If $ c$ is even, then $ b^c \equiv 1 (\;\text{mod} \;4)$. Therefore, $ c$ is odd and we have $ 2^a = b^c +1 \Rightarrow 2^a = (b+1)(b^{c-1} -b^{c-2} +...+1) \Rightarrow (b+1) = 2^{l}$ where $ l \leq a$. $ b = 2^l -1 \Rightarrow b^c = (2^l - 1)^c = 2^{lc} - c 2^{l(c-1)} +...+c 2^{l} - 1 = 2^a -1$ $ \Rightarrow 2^{lc} - c 2^{l(c-1)} +...+c 2^{l} = 2^a$ $ \Rightarrow 2^{l(c-1)} - c 2^{l(c-2)} +...-\frac{c(c-1)}{2}2^{l} + c = 2^{a-l}$ If $ l < a$, then the LHS is an odd integer and the RHS is an even integer. Hence, $ l=a$. This implies $ b = 2^{a} - 1 \Rightarrow c = 1$. Hence, the solutions for $ a \geq 2$ are $ (a,b,c) = (a,2^{a}-1,1)$.

nicetry007 wrote: unfortunately, part (2) is wrong. 2) $ b^c = 2^a - 1$ If $ b = 1$, we have the solution $ (a,b,c) = (1,1,c)$ SO FAR EVERYTHING IS FINE. Else, $ a > 1$ and $ b$ is odd >1 and "so $ b = 2^ku + 1$ with $ u$ odd and $ k > 0$" The claim within quotes is wrong. How does one know that $ b$ when divided by $ 2^k$ leaves a remainder of 1. If $ b$ is odd$ >1$, $ b=v+1$ with $ v$ even$ >0$. Let then $ k$ be the greatest power of $ 2$ which divides $ v$. We can then write $ v=2^ku$ with $ k>0$ and $ u$ odd. So $ b=2^ku+1$ for some integer $ k>0$ and some odd positive integer $ u$. And, fortunately, this part of my demo, at least, is right.

Hi Pco, I'm sorry about that. But I still cannot understand how you were able to conclude "$ (2u+1)^c=2^a-1 \implies 2cu=2\pmod 3$ and so $ c=2d+1 odd$". Could you please explain it.

nicetry007 wrote: Hi Pco, I'm sorry about that. But I still cannot understand how you were able to conclude "$ (2u + 1)^c = 2^a - 1 \implies 2cu = 2\pmod 3$ and so $ c = 2d + 1 odd$". Could you please explain it. No prob. Dont hesitate to ask. I'm always glad when I see that someone reads my demos. Here, there is a typo error : it's mod 4, not mod 3 : $ (2u + 1)^c = 2^a - 1$ $ \implies$ $ 2cu = 2\pmod 4$ : $ (2u+1)^c=2cu+1\pmod 4$ since all other terms of $ (2u+1)^c$ are $ =0\pmod 4$ and $ 2^a=0\pmod 4$ since $ a>1$ so $ 2cu+1=-1\pmod 4$ and $ 2cu=-2=2\pmod 4$ and so $ c = 2d + 1$ odd (else $ 2cu=0\pmod 4$)

The solutions are $(a,2^a\pm1,1),(1,1,c),(3,3,2)$. If $c=1$, then $|2^a-b|=1$, so $b=2^a\pm 1$, this yields the solutions $(a,2^a\pm1,1)$. Assume $c>1$. If $a=1$, then $b=1$. This gives the solutions $(1,1,c)$. Finally, if $a>1$, then we have two consecutive perfect powers. Mihaelescu's theorem then asserts they must be $8$ and $9$, so $a=3,b=3,c=2$. $\blacksquare$ To avoid appealing to Mihaelescu, we can do the following: a) If $b^c-1=2^a$, we have $(b-1)(b^{c-1}+\ldots+b+1)=2^a$. Evidently, $b$ is odd. Also, $b-1$ and $b^{c-1}+\ldots+b+1$ are powers of $2$. Finally, $\gcd(b-1,b^{c-1}+\ldots+b+1)=\gcd(b-1,c)$. Since $b-1,b^{c-1}+\ldots+b+1>1$, it follows $c$ is even. Hence, $(b^{c/2}-1)(b^{c/2}+1)=2^a$, so $b^{c/2}=3$, giving $b=3,c=2$, and $a=3$. b) If $b^c+1=2^a$, we know by Zsigmondy that $b^c+1$ has a primitive prime factor or $b=2,c=3,2^a=9$. However, since the latter case is impossible and $b^c+1$ is a power of $2$, this must mean $b+1$ is odd. However, $b$ is clearly odd, so this case yields no solutions.

Take $a,c>1$ and use catalans conjecture