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Find all injections $f: \mathbb N \to \mathbb N$ such that:

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

I have an idea to substitute $n=m$, since if $f(n)=f(m)$, then $n=m$ (the rule for injective functions).

If I do that, I get $f(2m)+f(0)=f(f(m)+m)$ and after that I have no idea what to do.

Any help?

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    If you mean $\mathbb N=\{1,2,3,\ldots\}$ then you have a problem taking $n=m$ since $f(0)$ is meaningless ($0\notin\mathbb N$). – 2017-01-03
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    What is the source of this problem? – 2017-01-03
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    at least the identity satisfies the equality – 2017-01-03
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    @MPW By some conventions, $0 \in \mathbb N$. We need to know what the OP means in this context by $\mathbb N$. – 2017-01-03
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    @wythagoras Even with that convention, what happens when $m>n$? – 2017-01-03
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    @wythagoras : Sometimes, but not usually. But that's why I said "if". I'm really asking for a clarification from OP regarding this. – 2017-01-03
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    @Austin Usually, in such problems, there is an additional condition stated, like $m \leq n$ or $m < n$. Otherwise, it should be assumed. (Unfortunately, the OP doesn't say anything about it). – 2017-01-03
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    @wythagoras : I think we are all in agreement – 2017-01-03
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    In this particular case, I believe that N={0,1,2,3...} and there aren't any of the conditions that u mentioned. Actually, there aren't any conditions, it just says to find all injections for this equation. This is a problem that is given to me from my teacher in discrete math in college and I have no idea how to solve it. – 2017-01-03
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    My school uses N={0,1,2,3..}, so I guess there are no problems taking n=m. – 2017-01-03
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    One set of functions is $f(n)=n+a$ for any $a\in\mathbb{N}$ – 2017-01-03

1 Answers 1

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Setting $m=0$ we get:

$f(n)+f(n)=f(n)-f(0)+f(f(0)+n)$, this implies $f(n)=f(f(0)+n)-f(0)$.

This gives us: $f(n+f(0))=f(n)+f(0)$.

Now, recall that we have $f(2m)+f(0)=f(f(m)+m)$, but the left side is equal to $f(2m+f(0))$.

So we have $f(2m+f(0))=f(f(m)+m)$, we conclude $2m+f(0)=f(m)+m$ with injectivity.

and so $f(m)=m+f(0)$.

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    Very nice, +1. Relation $f(n+f(0))=f(n) + f(0)$ looks so much like definition of addition within Peano axioms, so the end result is not surprising at all, $f$ is acting similarly to successor function. – 2017-01-03
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    So what exactly are you proving with this? – 2017-01-04
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    Which are those injections that satisfy this? – 2017-01-04
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    @Ana-Marija Did you understand? I'm in FINKI as well, by the way :D – 2017-01-04
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    @Ana-Marija the only injections that satisfy it are $f(n)=n+a$ where $a$ is some non-negative integer. – 2017-01-04
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    @FateMetric I understand now, yeah :DD – 2017-01-04