$7^{2n} -48n - 1$ is divisible by 2304 for all $n \in N$
so I did, P(n) : $7^{2n}-48n-1=2304k$ (k meaning there is an integer which will depend on n)
Prove base case $P(1): 7^2 - 48(1)-1 = 0$, proving that $k=0$ in the meaning of 'divides'
$n=k+1$
$7^{2(k+1)}-48(k+1)-1 = 2304l$ (l being for some integer)
I don't know how to proceed any further. could someone show me whats next?
thanks alot
$$7^{2n} -48n - 1|2304$$ $$7^{2n+2} -48(n+1) - 1=49\cdot7^{2n}-48n-49$$ $$=49(7^{2n} -48n - 1)+49\cdot48n-48n$$ $$=49(7^{2n} -48n - 1)+48\cdot48n$$ $$=49(7^{2n} -48n - 1)+2304n$$
We want to show that $7^{2k+2}-48k-48-1$ is divisible by $2304$, which is $48^2$.
Our expression is equal to $$49(7^{2k}-48k-1)+(49)(48k+1)-48k-48-1.$$ Now a little manipulation will do it. We get $49(7^{2k}-48k-1)+48^2k$.
