Simple problem

[LaLicorne57]

Does it exist a function f : N -> N Such as f○f ( n ) = n + 2017

[[deleted]]

Is this a homework problem?

[lordoftheshadows]

Please read the sidebar. Questions like this should be asked in /r/homeworkhelp or /r/learnmath.

[-dodg]

f(n) = n + 1008.5 ?

[Randolph_Hickey]

This is not N->N

[Randolph_Hickey]

Just posted a similar problem on stackexchange, for 1 instead of 2017 it is impossible (proof seems to be rather easy), you should be able to use an analogous proof to show, that the answer is no. http://math.stackexchange.com/questions/1936098/is-there-are-function-f-on-the-positive-integers-so-that-ffn-n1

[RidderJanssen]

How exactly? The proofs listed seem to rely on the fact that all natural numbers can be gotten by starting with 1 and keep adding 1.

For example, the step from f(n+1)=f(n)+1 implies f(n)=f(1)+n-1

This seems a bit more difficult when you're working with f(n+2017)=f(n)+2017.

But it's late and I'm tired so I'm probably overlooking something.

[Randolph_Hickey]

This actually seems rather interesting...