Find all functions $f: \mathbb{N}_{0}\rightarrow \mathbb{N}_{0}$ such that for all $n\in \mathbb{N}_{0}$: \[f(f(n))+f(n)=2n+6.\]
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Tags: function, Functional Equations, pen
Peter
27.08.2007 18:12
lyxuansang wrote: The general problem of K9 is Find all fucntions $ f:\mathbb{N}_{0}\rightarrow\mathbb{N}_{0}$ such that for all $ n\in\mathbb{N}_{0}$ : $ f(f(n))+f(n)=2n+3k$ and it on here http://www.mathlinks.ro/Forum/viewtopic.php?t=161084
WalkerTesla
10.05.2016 17:31
Let $f(0) = a$. We have that $0\le a\le 6$ by the definition of the function. By checking each value of $a$, we find that each value of $a$ leads to a contradiction except $a=2$. Therefore $f(0) = 2$. We will now prove that for $x=2n$, $f(x) = x+2$ for all nonnegative integers $n$. The claim is true for $n=0$. Assume the claim is true for $n=y$. This means that $f(f(2y))+f(2y) = 4y+6$. The LHS expands to $f(2y+2) + 2y+2 = 4y+6$, or $f(2y+2) = 2y+4$. Therefore, if the claim is true for $n=y$, it is also true for $n=y+1$, and by induction we have shown the claim is true for all $n$. Now let $f(1) = b$. We have that $0\le b\le 8$ by the definition of the function. By checking each value of $a$ and using the fact we just proved, we can show that $a=3$. Finally, we use a very similar inductive argument to show that $f(2n+1) = 2n+3$ for all nonnegative integers $n$. Therefore we have that $f(n) = n+2$ as desired.
achen29
30.05.2018 17:39
This should work: Plug in $n=0$ $$f(f(0))+f(0)=6$$ We have the cases $f(f(0))=0, f(0)=6 .... f(f(0))=6, f(0)=0$ with a total of 7 cases. I will state the contradictions for each case except case 5, which is indeed the valid answer: Case 1: $f(f(0))=0 f(0)=6$ which translates to $f(6)=0$ Plug $n=6$ into the given conditions yields $f(6)=12$ CONTRADICTION Case 2: $f(f(0))=1, f(0)=5$ which translates to $f(5)=1$ Plug $n=5$ gives $f(1)=15$, plug $n=15$ then gives $f(15)=-7$ CONTRADICTION (range is non-negative) Case 3: $f(f(0))=2, f(0)=4$ which translates to $f(4)=2$ Follow the same steps as Case 2, we get contradiction for f(12) being negative. Case 4: $f(f(0))=3, f(0)=3$ which gives $f(3)=3$ Plugging in $n=3$ gives $f(3)=6$ CONTRADICTION as a function has no more than one image per input. Case 5: no contradiction can be obtained Case 6: $f(f(0))=1, f(0)=5$ translates to $f(5)=1$ Plugging in n=1, n=5, and n=3 we get a contradiction for $f(13)$ being negative. Case 7: $f(f(0))=6, f(0)=0$ translates to $f(0)=6$ contradiction (same reason as Case 4) Therefore, the only valid case is case 5, which gives us: $f(0)=2, f(2)=4,$ and $ f(4)=6$ Inducting gives $\boxed{f(n)=n+2}$