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$7^{2n} -48n - 1$ 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

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  • $\begingroup$ It's 7^2n as in power of 2n no idea how to do that $\endgroup$ – Joseph Dec 10 '13 at 0:37
  • $\begingroup$ Do you mean $7^{2n} - 4^{8n} - 1$? $\endgroup$ – aschepler Dec 10 '13 at 0:38
  • $\begingroup$ @aschepleryes how to do that? just the 7^2n and next is 48n $\endgroup$ – Joseph Dec 10 '13 at 0:38
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$$7^{2n} -48n - 1|2304$$ $$7^{2n+2} -48(n+1) - 1=49\cdot7^{2n}-48n-49$$ $$=49(7^{2n} -48n - 1)+49\cdot48n-48n$$ $$=49(7^{2n} -48n - 1)+48\cdot48n$$ $$=49(7^{2n} -48n - 1)+2304n$$

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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 $49(7^{2k}-48k-1)+48^2k$.

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