a nice functional equation. 

Lol, yeah!

LOL...yeah obviously....big error although the final answer came out correct

That was what I thought as well.

yeah lol

A lucky error indeed!

the existence of such solutions is not trivial at all tho, it requires the use of axiom of choice and knowledge about its various equivalent forms.. how would you approach this?

I was thinking about just saying that if there's a period, then there would be infinite z s.t. f(z)=0, which contradicts our assumption from case 2

Well, you are not able to do this. When changing x your new variable (n := n-x) becomes negative, and we are working in nonnegative numbers.

@@andrewulinov3920 That is why we need to select n to be sufficiently large and then intuitively when we let n approach infinity, we get infinitely large image of f.

Nah we're only talking about numbers bigger than/equal to zero so you can't just take -n. Also you're using injectivity in your argumentation which means that it doesn't make the proof any shorter since showing injectivity was the only thing taking a lot of time

@@janekgroe4304 You're right about the first thing, but I think he wasn't using injectivity. He argued that, since the value of the function is zero, by the assumption that we made in the second case, the only possible way to get zero is for the argument of the function to be zero.

@@andrycraft69 oh yes right I forgot we assumed "injectivity at 0"

Wrong step again leads to correct solution😂

The identity function

That's what...the period was b-a which is 0. This by definition of periodicity, f(x) = f(0+x) which is an identity.

is the same, b-a=0 means no periodicity but also means injectivity is true

f has to be injective and surjective to have an inverse

Yes, because f(x)=f(x*1)=x*f(1), that is, we have a=f(1).

There must be smarter or more elegant ones, but consider f(x+1)=f(x)^2+ 2(f(x)-x)/(x-1) +1. It seems to me (it's hard to pay full attention to tedious calculations though) that the only polynomials satisfying that equation are ax^2+bx for a+b=1. I didn't try to find an argument discarding more general functions. Here's how I generated that equation. I started with f(x+1)=x+1, obviously satisfied by f(x)=x. For the square function, we have f(x+1)=f(x)+2x+1, which is rather similar. To cram both identities in one single expression we write f(x+1)=f(x)+1+g(x) and look for some g that becomes 0 for f(x)=x and 2x for f(x)=2x. The first requirement is easy, we'll just put a factor (f(x)-x) in the definition of g to make sure it is 0. That yields x^2-x when f is the square function, while we want 2x. We just multiply by 2x/(x^2-x), that is, 2/(x-1) and we are done: both x and x^2 are solutions to f(x+1)=f(x)+1+ (f(x)-x)*2/(x-1).

@@pedroteran5885 awesome, thanks!

In the problem statement, the function is defined as having its codomain be the non-negative integers, so f(x) can never be negative.

It could, but the original equation asks us to only look at functions from the non-negative integers to the non-negative integers.

Brill thanks guys

a q p

w

Bruh

You have to prove that the function is injective if you claim f(n) = n

x>=a, x=/=0 x=0 satisfies f(0)=0 but not satisfies f(x) =f(x+(b-a)) because it states that f(x) =/=0 for x=/=0

In other words it means image of f cannot have a repeating value of zero

However, I still think m=/=0 because m=0 has only one solution that is f(0)=0 where x=0. In this way we can prove f(1), where x=1, x>=a, a=0, or a=1. When a=1, m=x-a=1-1=0 which is not accepted. So, a=0, the repeated value of f(1)=f(1+(b-a)) =f(1+b-0)=f(b+1). Last value of the first non-repeated series=f(b+1-1)=f(b)

"from which it was pretty straightforward to prove that f(n) = C n, for some constant C" I don't see how this follows straightforward. Could you please explain?

@@bjornfeuerbacher5514 I was a bit inaccurate in my statement. It's straightforward to prove it, _assuming_ f(n) is a polynomial, which is a bit sloppy. With this assumption, f(n) = a[0] + a[1] n + a[2] n^2 + ... + a[k] n^k. Since f(0) = 0, it must be that a[0] = 0. Expanding the functions on both sides of the equation f(n^2) = n f(n) gives: a[1] n^2 + a[2] n^3 + ... + a[k] n^(k+1) = a[1] n^2 + a[2] n^4 + ... + a[k] n^(2 k) Since these are identically equal, and the expansion on the right hand side has no terms of odd degree, the coefficients of the terms of odd degree on the left hand side must all equal zero, which gives a[2 j] = 0. Equating the remaining terms gives a[2 j-1] = a[j], for all j > 1. When j is even, a[2 j-1] = 0. When it is odd, since j = 2 ((j+1)/2) - 1, j can be replaced by (j+1)/2, giving a[2 j-1] = a[(j+1)/2]. Since (j+1)/2 < j for all j > 1, and there are a finite number of integers between 0 and any finite number j, this descending sequence must eventually terminate. Further, for all odd integers j > 1, the least possible value of (j+1)/2 is 2 (this can be proved by assuming that (j+1)/2

No, at this point polynomials bother me more. Hopefully that changes.

Your step (3) is not well justified. You simply assume that f(.) is a polynomial, rather then prove it.

@@fullfungo he didn't assume that it was a polynomial he assumed it has a series representation which is still not allowed for functions from uncountable sets to uncountable sets but because ℕ∪{0} is countable he can assume there is a series representation...

This solution also kinda works even if the domain was the reals

m can not be negative

@@dicksonchang6647 why not?

@@larspos8264 It is in the problem statement that m and n are non-negative.

@@kimbel2804 ofcouce, I only saw the Z, my bad

The function f is only defined for integers.

f is only defined for Z^nonneg \cup {0} or Z_{\geq 0}, therefore y, z are bound to be either 0 or 1 to have a sum of 1.

The question states that the domain and co domain of f is the set of non negative integers