Proving that 5^n-4n-1 is divisible by 16

by **jinakin** » Wed Nov 02, 2011 10:45 pm

For all natural numbers [tex]n[/tex], prove that
[tex]5^n-4n-1[/tex] is divisible by [tex]16[/tex].

by **perfectmath** » Fri Jun 29, 2012 4:01 am

The problem wrong with [tex]n=0[/tex]. I think [tex]n[/tex] must be positive integer number, not natural number.

by **perfectmath** » Fri Jun 29, 2012 4:05 am

Here is my prove:

For [tex]n=1[/tex], the problem is right.
If the problem right with [tex]n=k[/tex], it mean [tex]16 \mid 5^k-4k-1[/tex]. We will prove the problem true with [tex]n=k+1[/tex].
So we have [tex]5^{k+1}-4(k+1)-1=5(5^k-4k-1)+16k[/tex] is divisible by [tex]16[/tex].

We have Q.E.D

by **Himansh** » Thu Feb 26, 2015 3:35 am

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__________________
Johni

by **Guest** » Mon Sep 05, 2016 4:28 pm

Actually, the statement is still true when n=0. All numbers divide zero.

by **Guest** » Mon Sep 05, 2016 6:46 pm

jinakin wrote:For all natural numbers [tex]n[/tex], prove that
[tex]5^n-4n-1[/tex] is divisible by [tex]16[/tex].