How can I prove $288\mid 7^{2n+1}-48n-7$? How can I prove $$288\mid 7^{2n+1}-48n-7$$ for all nonnegative integers $n$?
My only thought was to write $$7^{2n+1}-7-48n=7(7^n+1)(7^n-1)-48n.$$ This didn't seem beneficial at all. Please help me understand the problem.
 A: Work with induction on $n$. The base case is easy. Assume that the statement is true for some $n$. We find $7^{2n+3} - 48(n+1) - 7 = 49 \cdot 7^{2n+1} - 48n - 48 - 7 = (7^{2n+1} - 48n - 7) + 48(7^{2n+1} - 1)$. The term $7^{2n+1} - 48n - 7$ is divisible by $288$ per induction hypothesis. As $288 = 48 \cdot 6$ it is left to show that $6 \:|\: 7^{2n+1} - 1$. Can you complete the proof with the hint you were given?
A: Clearly the proposition is true for $n=0$
For integer $n\ge1,$
As $288=48\cdot6$ divides $48^2$ and $7^2=49=1+48,$
we write $\displaystyle7^{2n+1}=7\cdot 49^n=7(1+48)^n$
Now, using Binomial Theorem for positive integer 
$\displaystyle(1+48)^n=1+\binom n148+\binom n248^2+\cdots +\binom n{n-1}48^{n-1}+48^n$$\equiv1+48n\pmod{48^2}$ 
$\displaystyle\implies (1+48)^n\equiv 1+48n\pmod{288}$ as $288=48\cdot6$ divides $48^2$
$\displaystyle\implies 7^{2n+1}=7\cdot49^n\equiv7(1+48n)\pmod{288}$
Now, $\displaystyle 7(1+48n)=7+(1+6)48n\equiv7+48n\pmod{288}$ as $288=48\cdot6$
A: $288=2^53^2$ so it suffices to prove that both $32$ and $9$ divide the expression.  This is normally done via modular arithmetic.
Mod $9$, the expression is $(-2)^{2n+1}+3n+2$, or $-2\cdot 4^n+3n+2$.  You can plug in $n=0,1,\ldots, 8$ to verify that this 0 (mod $9$) each time.
Mod $32$ it is similar.  $7^{2n+1}-7-48n=7(7^{2n}-1)+16n=7(17^n-1)+16n$.  For even $n$, we have $17^n=1$ (mod $32$), and $16n=0$ (mod $32$).  For odd $n$, we have $7\times 16+16=0$ (mod $32$).
