TITLE: The greatest positive integer $$, for which $49^+1$ is a factor of the sum $49^{125}+49^{124}+⋯+49^{2}+49 +1$, is $63$.
QUESTION [7 upvotes]: The greatest positive integer $$, for which $49^ +1$ is a factor of the sum $49^{125}+49^{124}+⋯+49^{2}+49 +1$, is $63$.
I can prove that $49^{63} +1$ is a factor of $49^{125}+49^{124}+⋯+49^{2}+49 +1$. But I can not prove that $63$ is the greatest.
Can anyone please help me to prove that ?
Edit  ::  How I proved  $49^{63} +1$ is a factor of $49^{125}+49^{124}+⋯+49^{2}+49 +1$.
$49^{125}+49^{124}+⋯+49^{2}+49 +1  = \frac{49^{126} - 1}{48} = \frac{(49^{63} - 1)(49^{63} + 1)}{48} = (49^{63} + 1) \frac{(49 - 1) (49^{62} + 49^{61} + ..... + 1 )}{48} = (49^{63} + 1) (49^{62} + 49^{61} + ..... + 1) $

REPLY [2 votes]: $\!\begin{eqnarray} 
&{\bf Key\ Idea}\ \  &{\rm Put}\ \ a = 49^{\large K},\ b = 49^{\large N}\ \text{below, so } K\ge 63\Rightarrow\, N\ge 63\\[.6em]
&{\bf Lemma}\quad\  & a\!+\!1\,\mid\, ab-1\ \Rightarrow\  b\ge a\ \ \text{for positive integers }\, a,b\\[.6em]
&{\bf Proof}\qquad  &a\!+\!1\mid \underbrace{(a\!+\!1)b-(b\!+\!1)}_{\textstyle ab - 1}\iff  a\!+\!1\mid b\!+\!1 \end{eqnarray}$