The greatest positive integer $$, for which $49^+1$ is a factor of the sum $49^{125}+49^{124}+⋯+49^{2}+49 +1$, is $63$. The greatest positive integer $$, 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) $
 A: You've already noted that your sum is $\dfrac{49^{126}-1}{49-1}$. Now, choose $0\lt k \lt 63$ and set $x=49^{k+63}+1$; then $$49^{126}=(x-1) \cdot 49^{63-k} \equiv -49^{63-k}\pmod x$$ and therefore $$49^{126}-1\equiv -(49^{63-k}+1)\pmod x$$ But $0<49^{63-k} +1\lt x$, so it's impossible to have $49^{63-k}+1\equiv 0$ and therefore impossible to have $49^{126}-1\equiv 0 \pmod x$. This means that $x$ can't divide $49^{126}-1$ and so can't divide your sum.
A: $\!\begin{eqnarray} 
&{\bf Key\ Idea}\ \  &{\rm Put}\ \ a = 49^{\large K},\ b = 49^{\large N}\ \text{below, so } K\ge 63\Rightarrow\, N\ge 63\\[.6em]
&{\bf Lemma}\quad\  & a\!+\!1\,\mid\, ab-1\ \Rightarrow\  b\ge a\ \ \text{for positive integers }\, a,b\\[.6em]
&{\bf Proof}\qquad  &a\!+\!1\mid \underbrace{(a\!+\!1)b-(b\!+\!1)}_{\textstyle ab - 1}\iff  a\!+\!1\mid b\!+\!1 \end{eqnarray}$
A: $$
\sum_{i=0}^{125} 49^i 
= \frac{49^{126}-1}{49-1}
= \frac{(49^{63}+1)(49^{63}-1)}{48}
$$
Note that $49\equiv1\pmod{48}\\ \Longrightarrow 49^{63}-1\equiv 0 \pmod{48}$
Therefore $\sum_{i=0}^{125} 49^i=(49^{63}+1)(\cdots)$
And there’s no factor in the (...) part which can be multiplied to $49^{63}+1$ to make it a greater $49^k+1$. The next $k$ satisfying $49^{63}+1$ being the factor of $49^k+1$ is $189$, and $$(49^{63}+1)(49^{126}-49^{63}+1)=49^{189}+1$$
Obviously there’s no $49^{126}-49^{63}+1$ in (...), and $49^{189}+1$ is greater than the original sum as well.
A: Hint:
$$
\sum\limits_{k = 0}^n {x^{\,k} }  = {{x^{\,n + 1}  - 1} \over {x - 1}} = \prod\limits_{k = 1}^n {\left( {x - e^{\,ik2\pi /\left( {n + 1} \right)} } \right)} \quad \buildrel {n\,odd} \over
 \longrightarrow \quad  \cdots 
$$
you shall be able and conclude yourself
