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Do you need help with this or is it more of a puzzle?
(I managed to prove that f(3)=3)

More of a puzzle. I know how to get the answer. BTW, how did you prove f(3)=3?

2=f(2)

2 and 2 are not coprime, so that doesn't work.

That's certainly not the solution I have.

I will return to answer questions after dinner.

1 is coprime to every integer, so \(f(n)=n\) by definition.

I think that only proves f(n)=f(n)*f(1)=f(n)

Right, what Joe said.

Anyway, I'm stuck, I'll read KG's solution later.

"and realized you could just force the definition via the identity." What do you mean by that?

f(2*1) = f(2)f(1)
f(1) = 1
f(3*2) = f(3) 2
f(3) = f(6)/2
f(5) = f(10)/2
f(n) = f(2n)/2 for odd n

@experimentX That is correct, but it still doesn't solve the original problem.

I have posted a hint for those who had previously participated.

2=f(2)

2 is not coprime to 2. So \(f(2)f(2)\) does not necessarily equal \(f(4)\)

f(2) = f(1+1) = f(1) + f(1) ???

But this doesn't help either :/

I'll post a second, more helpful hint in half an hour or so if you think you need it.

From the above inequalities, \[5+ k\le f(5) \le 5+2k\]

Have posted 2nd hint. I would also suggest letting \(f(3)=k\). It makes the manipulations easier

If you could convince me that the k's are in fact the same, you would indeed have a proof.

Can I type QED now? :D

I think I see the actual problem now. You're trying to compare an upper bound for 4, and a lower bound for 5. Your claim is that those bounds can't overlap, but I don't see anything that shows those bounds can't overlap. It is true of course, that the best bounds don't overlap, but these bounds are not as small as possible.
If \(f(4)=4+2k\), then of course \(4+2k

I must sleep now. I will return to see any corrections you have made.

Ohh hmm :( I will try again.

How about this?

Shall I type QED now? :D

Oh I knew something was wrong. hmm I will try again. (Y) :D

I can't take \(f(18)=f(3)f(6)\). Can I? I think \(f(2)\cdot f(9)\) is right. Hmm \(2f(9)\).

That's the right track. Now what's the upper bound of\(f(9)\) in terms of \(f(3)\)?

Hint: \[f(9)\leq f(10)-1\]

4k-3 or 4a -3?

Right. So this means that \(f(18) \leq \text{____}\)

8k - 6 \(\ge\) f(18)
How do I solve it for f(17)?

I can only get f(15) using f(3) and f(5)

And from f(15) you can get f(18)

\[2k + k^2 \le f(15) \le 2k^2 -k\]

Just use \[2k + k^2 \le f(15)\]Now this means that \[\text{______}\le f(18)\]

\((k+1)^2 + 2 \le f(18) \le 8k+6\)

\[(k+1)^2 + 2 \le 8k+6\]Now just solve for k

Ohh I can't concentrate at all :(
I see now, \[3\le k \le 3 \implies k=3 \implies f(3) =3\]

and why wasn't f(1) enough for us to use induction? why did we need to solve it for f(3).

ohh, thanks. i couldn't have done this without your help.

You did very well. =D

I appreciate the amount of time you spent on this. Thanks for actually doing this.