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Find all injective functions $f:N \rightarrow N$ such that

$$f(f(m)+f(n))=f(f(m))+f(n)$$

Where $m,n$ are natural numbers.

FisherDisinformation
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Tommy
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  • What was your attempt at this? Is there a particular aspect you get stuck on? – anak Aug 12 '15 at 13:12
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    Hint: an easier way to say it is that $f(x+y)=f(x)+y$ if $x,y$ are in the range of $f$. – Ian Aug 12 '15 at 13:41

1 Answers1

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An example of an injective function that satisfies this property is $f(x)=x+a$ where $x,a \in \mathbb{N}$ and $a$ constant. If you look at the number line this basically shifts every natural number to the right by $a$. This is obviously injective and a function from $\mathbb{N} \rightarrow \mathbb{N}$. $$f(f(m)+f(n))=f(m+a+n+a)=f(m+2a + n)=m+2a + n + a = f(m+a) + f(n) = f(f(m))+f(n)$$ Another thing to look at for yourself is checking if $f(x)=a \cdot x$ satisfies this equation, because this function is injective on $\mathbb{N}$.

Jan
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