Question

Prove or disprove that \(n^4 + 4^n\) is divisible by 5 when \(n\) is odd,
and \(n > 5\)

Answer 1

@shubhamsrg

Answer 2

Induction?

Answer 3

any method... please note.. i have changed the question to "prove or disprove.. "

Answer 4

clearly, n = 1 is in solution set

Answer 5

Counterexample: \(n=5\)

Answer 6

assume : \(5 | k^4 + 4^k \)

prove : \(5 | (k+1)^4 + 4^{k+1} \)

?

Answer 7

Oh no, so this is not a true statement :/

Answer 8

Nope. The way I thought about it was to consider the digits in the ones place of each odd power of a chosen \(n\), then adding the ones digit of the \(n\)-th power of 4. If they don't add to 0 or 5, the rule doesn't hold.

Answer 9

http://openstudy.com/study#/updates/535aadbae4b05d7c6ff743f1

actually i got this problem when working @Yttrium 's problem^

Answer 10

\(n^4 -1\) is divisible by 5

Answer 11

when n is odd,
4^n leaves a remainder -1

so n^4 + 4^n is always divisible by 5 whenever n is odd

Answer 12

but ur counter example made me cry

Answer 13

For \(n^4-1\), if \(n=5\), don't you have \(5^4-1=624\)?

Answer 14

And sorry to disappoint :P

Answer 15

got the pitfall :)


here is the modified statement :

\(n^4 + 4^n\) is divisible by 5 whenever n > 5 is odd

Answer 16

you cant give a counter example this time @SithsAndGiggles ;)

Answer 17

I think you might have to exclude all \(n\) that are multiples of 5...
\[n=10~~\Rightarrow~~10^4+4^{10}=1,049,576\]

Answer 18

Oh my !! the statement wont look good with \(n \ne 5k\) lol
i give up :( thanks @SithsAndGiggles :)

Answer 19

No problem! If you're trying for a general result, you might want to consider the ones digit of successive powers of \(n\). There seem to be some nice patterns that should help you determine divisibility by 5.

Answer 20

I gave a reply on Yttrium's problem.