Re: Simple question.

On 26 Jan, 00:20, Kenshin <rurouni_sohj...@xxxxxxxxxxx> wrote:
Let f : |N -> |N satisfy f ( f (m) + f (n) ) = m + n for all m, n in |
N. What is f (2008) ?

I've conjectured f(n) = n, but I can't prove the uniqueness.

I've proved f is bijective.

How can I go further?

Thank you so much for your considering.

Nice question. First suppose that f(0) = a. Then f(f(0) + f(0)) =
f(2a) = 0. From this we have f(f(2a) + f(0)) = f(a) = 2a. Therefore
f(f(a) + f(2a)) = f(2a) = 3a = 0 from before, so a = 0. An immediate
corollary of this is that, for any n, f(f(n)) = f(f(n) + f(0)) = n.

Now suppose that f(1) = b. Let's try to prove by induction that f(n) =
nb. So suppose that this is the case for some n, and consider f(n +
1). We know that n = f(f(n)) = f(nb) and 1 = f(f(1)) = f(b), so f(n +
1) = f(f(nb) + f(b)) = (n + 1)b as required.

Therefore f(n) = nb and f(f(n)) = nb^2 = n, so b = 1.
.

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