Problem: Find all functions $f:\mathbb{N} \rightarrow \mathbb{N}$ that satisfy $2f(m+n) = f(m)f(n)+1$ for all $m,n \in \mathbb{N}$
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Welcome to stackexchange. To get help here rather than downvotes and votes to close, please [edit] the question to show us the question and your start on trying to answer it. Make the title informative. Don't ;ink to an image. Use mathjax: https://math.meta.stackexchange.com/questions/5020/mathjaxbasictutorialandquickreference – Ethan Bolker Aug 06 '19 at 15:51

Usually the tips people give for these problems is to try inputting $1$ or inputting powers of $2$ or inputting primes and see a pattern. – IntegrateThis Aug 06 '19 at 15:52

Also: try $0$ as an input (if $0 \in \Bbb N$ in your particular world...often this depends on whether you're a mathematician or a computer scientist). – John Hughes Aug 06 '19 at 15:56

Thank you for the kindness of you all!!! – Edward Sakamoto Aug 06 '19 at 16:06
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First of all we notice that for $m=n=0$ we get
$2f(0)=f(0)^2+1\Leftrightarrow (f(0)1)^2=0\Leftrightarrow f(0)=1$
For $n=0$ and arbitrary $m$ we then get:
$2f(m)=f(m)+1\Leftrightarrow f(m)=1$ for every $m\in\mathbb{N}$.
So $f:\mathbb{N}\to\mathbb{N}$, $f(n)=1$ is the only function satisfying this condition.
Cornman
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It's very nice for me to think and understand. I appreciate you;) – Edward Sakamoto Aug 06 '19 at 16:07

2The OP has accepted this correct answer. That said, I'm disappointed that you provided it before he had a chance to respond to comments asking that we see his work on the problem. He will learn less by reading your answer than he would have struggling first. – Ethan Bolker Aug 06 '19 at 16:08

@EthanBolker Even you are correct on that, I am sure that when I have not given an answer, then someone else would have done it anyways. Also seeing the vote close and the amount of downvotes on this question, this would have been closed, or ignored anyways. This is a questionanswer side. And how this works here is not that you can go into a huge discussion with the OP. As I said, normally I like these discussion, but here is simply not the right place for it and you barley see them here. I do not think that this is my 'fault', so you do not have to be disappointed by my 'behaviour'. – Cornman Aug 06 '19 at 16:13

@Cornman No one is at fault and you have done nothing wrong. It is indeed true that if you had not answered someone else might have. That's behavior I would like to discourage. But it's OK to have different opinions about how this site works best. Do keep helping people. – Ethan Bolker Aug 06 '19 at 16:22