Answer: $f(x)=1$ for all $x\in\mathbb{N}$ and $f(x)=2x-1$ for all $x\in\mathbb{N}$..

Solution: For $a=b=1$ we get $f(1)=1$, so $f(a)+1|2a$ for all $a$. $(*)$ For $a=1$ and $b=2$ we get $f(2)+1|4$ so $f(2)=1$ or $f(2)=3$. Case 1: If $f(2)=1$ we get $f(a)+1|2a+2$ for all $a$ $(**)$. Thus with $(*)$ we get $f(a)+1|2$ for all $a$ and so $f(a)=1$ for all $a$. Case 2: If $f(2)=3$. From $(*)$ we get $f(a)\leq 2a-1$ for all $a$. Let $c$ be minimal number for $f(c) \ne 2c-1$. Thus $c\geq 3$ and $f(c)<2c-1$. Let $a=c-1$ and $b=c$. Then $2c-3 + f(c)|4c-4$. Since we always have $$2c-3+f(c)| 2c-3+f(c)$$we get (substract right sides) $$2c-3+f(c) |\underbrace{2c-1-f(c)}_{> 0}\Longrightarrow f(c)\leq 1\Longrightarrow f(c)=1$$ Now let $a=c$ and $b<c$, we get $1+2b-1|2(c+b -1)\Longrightarrow b|c-1$ for all $b<c$. This is clearly contradiction if $c>3$. We have to think what is happening if $c=3$. In this case we get, for $b=c=3$: $f(a)+1|2a+4$. This with $(**)$ give us $f(a)+1|2$ for all $a$. Contradiction. So in this case $f(a)=2a-1$ for all $a$.

Let $P(m,n)$ the assertion of $f(m)+f(n)|2(m+n-1)$. From $P(1,1)$ we get $2f(1)|2$ $\Longrightarrow$ $f(1)=1$ and $P(2,2)$ we get $2f(2)|6$ $\Longrightarrow$ $f(2)=1\vee 3$. Also that $P(n,n)$ we get $f(n)|2n-1$ $\Longrightarrow$ $f(n)\leq 2n-1$ for all $n\in \mathbb{N}$ If $f(1)=1\wedge f(2)=1:$ From $P(m,2)$ we get $f(m)+1|2m+2...(\star)$, on the other hand from $P(m,1)$ we get $f(m)+1|2m...(\star\star)$ $\Longrightarrow$ by $(\star)$ and $(\star\star)$ we get $f(m)=1$ for all $m\in \mathbb{N}$ If $f(1)=1$ and $f(2)=3:$ Let $p$ be a prime from $P(p,1)$ we get $f(p)+1|2p$ and from $P(p,p)$ we get $f(p)|2p-1$ combining both results we get $f(p)=1\vee 2p-1...(\blacksquare)$. From $P(4,2)$ and $P(4,4)$ we get $f(4)+3|10$ and $f(4)|7$ $\Longrightarrow$ $f(4)=7$. Let $p\equiv 3\pmod 8$ be a prime $\Longrightarrow$ by $P(p,4)$ we get $f(p)+7|2(p+3)$ if $f(p)=1$ $\Longrightarrow$ $8|2(p+3)$ which is a contradiction $\Longrightarrow$ from $(\blacksquare)$ we get $f(p)=2p-1$ for all $p$ prime of the form $p\equiv 3\pmod 8$. Suppose that exist $r,s\in \mathbb{N}$ such that $r\neq s$ and $ f(r)=f(s)=t$ $\Longrightarrow$ from $P(p,r)$ and $P(p,s)$ we get $f(p)+t|2(p+r-1)$ and $f(p)+t|2(p+s-1)$ combining both results we get $f(p)+t|2(r-s)$ then we choose $p$ prime of the form $p\equiv 3\pmod 8$ such that $2p-1>2|r-s|$ which brings us to contradiction hence $f$ is inyective. Now suppose by induction: $f(n)=2n-1$ for all $n\leq k$ $\Longrightarrow$ since $f$ is inyetive and $f(k+1)\leq 2k+1$ we get $f(k+1)=2k\vee 2k+1$. If $f(k+1)=2k$ $\Longrightarrow$ by $P(k+1,k+1)$ we get $2k|2k+1$ which is a contradiction hence $f(k+1)=2k+1$ hence the induction is complete $\Longrightarrow$ $f(n)=2n-1$ for all $n\in \mathbb{N}$ Hence all funtions $f$ are $f:1,2n-1$

We claim that the answer is $f(x) \equiv 1$ or $f(x) \equiv 2x-1$. First, by $P(1,1)$, we have $f(1)=1$, and by $P(x,1)$ and $P(x,x)$, we have $1+f(x)|2x$ and $f(x)|2x-1$. By easy CRT wrt modulo $f(x)$ and $f(x)+1$, we have $2x-1 \equiv f(x) \pmod{ f(x)^2+f(x)}$. Therefore, we have $f(x)=2x-1$ or $2x-1 \ge f(x)^2+2f(x) \implies f(x) \le \lfloor \sqrt{2x} \rfloor -1$. This gives us that for $2 \le x \le 4$, $f(x)=2x-1$ or $f(x)=1$. Assume $f(2)=1$. Then $P(x,2)$ gives us $f(x)+1|2x+2$, so $1+f(x)|2$, which forces $f(x) \equiv 1$, which works. Assume $f(2)=3$. I claim that $f(3)=5$. If $f(3)=1$ and $f(4)=1$, $P(2,4)$ gives $10 \equiv 0 \pmod{4}$, contradiction. If $f(3)=1$ and $f(4)=7$, $P(3,4)$ gives $12 \equiv 0 \pmod{8}$, contradiction. Therefore, we have $f(3)=5$ as well. Now we go by strong induction on $n$ to prove that $f(n)=2n-1$ for all $n$. Assume that for all $1 \le i \le n-1$ that $f(i)=2i-1$. We prove $f(n)=2n-1$. $P(n,i)$ gives $2(n-1+i) \equiv 0 \pmod{f(n)+f(i)}$, so $2n-2+2i = f(n)+f(i)$ or $f(n)+f(i) \le n-1+i$. Since $i$ can vary from $1$ to $n-1$ and $f(i)=2i-1$, we either have $f(n)=1$ or $f(n)=2n-1$. Assume that $f(n)=1$. Then $P(n,i)$ gives $2(n-1+i) \equiv 0 \pmod{2i}$ for $1 \le i \le n-1$. This forces $i|n-1$ for all $1 \le i \le n-1$, but $(n-2,n-1)=1$ and $n-2 > 1$, so this is a contradiction. Therefore, $f(n)=2n-1$ for all $n$. This works. We are done.

Firstly, $a=b$ implies that $f(a) \mid 2a-1$ for all $a \in \mathbb{N}$. Let $p>2$ be a prime number ($2$ is the worst prime after all ) and see that $f\left(\frac{p+1}{2}\right) \mid p$. Consider the set \begin{align*} S=\left\{p | f\left(\frac{p+1}{2}\right)>1\right\} \end{align*}We have two possibilities. Case 1. if $S$ is finite. In this case, there exists a constant $M>0$ such that for all $p >M$, $f\left(\frac{p+1}{2}\right)=1$. Fix $a$ and choose $2b=p+1$ for a sufficiently large prime $p$. We have $1+f(a) \mid 2(a+b-1)$. In particular, for all $p,q>M$, $1+f(a) \mid p-q$. However, if $f(a)>1$, then this fails to hold by the Dirichlet's theorem. We get that $f(n)=1$ for all natural numbers $n$. This also satisfies our conditions. Case 2. $S$ is infinite. In this case, for any $M>0$, there exists a prime $p>M$ such that $p \in S$. Fix $a$ and we see that for $2b=p+1$, $f(b)=p$ and $p+f(a) \mid p+(2a-1)$, therefore, $p+f(a) \mid (2a-1)-f(a)$ which fails to hold for a sufficiently large prime $p$ unless $f(a)=2a-1$. We get $f(n)=2n-1$ for all $n$, and this clearly works.

I have a solution which doesn't involve prime numbers at all. In fact, it also works if we remove, say, the set of all prime numbers.

Did a huge mistake at the beginning. Edit: Okay, just say that I messed up again. Luckily, it was not very hard to fix. Neat problem. Plug in $a=b=1$, that gives $f(1)=1$. Now plug in $b=1$ to get $f(a)+1 \mid 2a$. Plugging in $a=b$ gives \[ f(a) \mid 2a-1 \implies f(a) \mid 2a-(f(a)+1). \]But $f(a)+1 \mid 2a-(f(a)+1)$ as well. Plugging in $a=2$ gives $f(2),f(2)+1 \mid 3-f(2)$. Therefore, $f(a)=1$ or $f(a)=3$. Case 1: Let $f(a)=1$. Then plug in $b=2$. That gives $f(a)+1 \mid 2(a+1)$. But remember $f(a)+1 \mid 2a$. As $\gcd(2(a+1),2a)=2$, we can easily conclude $f(a)=1$, since else $f(a)+1>2$ divides both $2(a+1)$ and $2a$. It's easy to verify that $f(a)=1$ for all $a \in \mathbb{N}$ is a solution. Case 2: Let $f(a)=3$. We'll proceed by strong induction. Base case: We've already shown $f(1)=1$ and $f(2)=3$. Induction Hypothesis: Assume $f(k)=2k-1$ for all $k \leq n$. Induction step: Assume $f(n+1)=1$. Then plug in $b=n+1$. That gives us \[ f(a)+1 \mid 2(a+n). \]Now take $a=k \leq n$. \[ 2k \mid 2(k+n) \implies k \mid n \]If $n \geq 3$, then we're done, we may choose any $k>1$, such that $\gcd(k,n)=1$, e.g. in particular, we may choose the largest prime smaller than $n$ by Bertrand's Postulate. Thus $n<3$. If $n=1$, then $f(2)=1$, contradiction. So $f(3)=1$. Plug in $b=3$. That gives \[ f(a)+1 \mid 2(a+2). \]Remember $f(a)+1 \mid 2a$. With $\gcd(2(a+2),2a) \mid 4$, we can conclude $f(4) =1$ or $f(4) = 3$. But plugging in $a=2$ and $b=4$ gives a contradiction in both cases. Now plug in $a=n+1$ and $b=n$. That givs \[ f(n+1)+2n-1 \mid 4n .\]Thus $f(n+1)=1$ or $f(n+1)=2n+1=2(n+1)-1$. The first one contradicts what we've just done, thus it's the second one. Therefore, we're done with the induction step. To sum up: We have $f(n)=1$ for all $n \in \mathbb{N}$ or $f(n)=2n-1$ for all $n \in \mathbb{N}$. It's easy to see that both are solutions.

Answer: $\boxed{f(x)=1 \; \forall x \in \mathbb{N}}$ and $\boxed{f(x)=2x-1 \; \forall x \in \mathbb{N}}$ Let $P(a,b)$ denote the problem statement. Claim 1: $f(1)=1$

Claim 2: For all primes $p>2$ , $f(p)=1$ or $f(p)=2p-1$.

Claim 3: Either $f(p)=1$ for all odd primes $p$ or $f(p)=1$ for all odd primes $p$.

Now, we can divide this problem into two cases: Case 1: $f(p)=1$ $\forall$ odd primes $p$. $$P(1,n): \; 1+f(n)|2n$$$$P(p,n): \; 1+f(n)|2n+2(p-1) \implies 1+f(n)|2(p-1)$$Again, by Dirichlet it is possible to choose a prime p such that $p \not \equiv 1 \pmod{1+f(n)}$ $$\implies 1+f(n)|2 \implies f(n)=1 \; \forall n$$ Case 2:$f(p)=2p-1$ $\forall$ odd primes $p$. $$P(p,n): 2p-1+f(n)|2p-1+2n-1 \implies 2p-1+f(n)|2n-1-f(n)$$Since we can make $2p-1+f(n)$ infinitely large, we deduce $$2n-1-f(n)=0 \implies f(n)=2n-1 \; \forall n$$

$P(1,1): f(1) \mid 1 \implies f(1)=1$. $P(a,1): f(a) \mid 2a-1$. $P(\frac{p+1}2,1): f(\frac{p+1}2)=1$ or $p$ for prime $p \geq 3$. Say, for some $k$, $f(k)=2k-1$. Thus $P(a,k): f(a)+2k-1 \mid 2a-1 +2k-1 \implies f(a) +2k-1 \mid 2a-1-f(a)$. If we can make $k$ arbitrarily large, we can conclude $f(a)=2a-1$. But, say for some $a$, $f(a) \neq 2a-1$. This would imply we can't make $k$ arbitrarily large, hence for all large $p$, $f(\frac{p+1}2)=1$. $P(a,\frac{p+1}2): f(a)+1 \mid 2a-1+p$, for all large primes $p$. But by Dirichlet, we know infinitely many primes occupy each modular class with respect to $x$ where the class is co-prime to $x$. In other words, if we even have a single $p$ satisfying $p \not \equiv 1-2a \pmod{1+f(a)}$, we'll get a contradiction, or we need $\phi (f(a)+1)=1 \implies f \equiv 1$. Thus our 2 solutions are $f \equiv 1$ and $f(a)=2a-1$ for all naturals $a$.

MEMO 2016 I4 wrote: Find all $f : \mathbb{N} \to \mathbb{N} $ such that $f(a) + f(b)$ divides $2(a + b - 1)$ for all $a, b \in \mathbb{N}$. Remark: $\mathbb{N} = \{ 1, 2, 3, \ldots \} $ denotes the set of the positive integers. To start, if $f$ is constant then clearly $f \equiv 1$. From now on, we look only for non-constant solutions. Let $P(a,b)$ be the given assertion. $P(a,a)$ implies $f(a) \mid (2a-1) \,\, (*)$, hence $a=1$ in $(*)$ implies $f(1)=1$. Thus, $P(a,1) \Rightarrow (f(a)+1) \mid 2a \,\, (**)$. We prove now the following key Claim. Claim: For all odd primes $p$, $f(\frac{p+1}{2})=p$. Proof: Suppose there existed a $p$ such that $f(\frac{p+1}{2}) \neq p$. Take $a=\frac{p+1}{2}$ in $(*)$ to obtain $f(\frac{p+1}{2}) \in \{1, p \}$. Hence $f(\frac{p+1}{2})=1$. Thus, $$P(a,\frac{p+1}{2}) \Rightarrow (f(a)+1) \mid 2a+p-1,$$which combined with $(**)$ implies that $(f(a)+1) \mid p-1 \,\, (***)$. Take now $a=p-2$ in both $(**)$ and $(***)$ to conclude that $$(f(p-2)+1) \mid \gcd (2p-4,p-1)=2 \Rightarrow f(p-2)=1$$for all odd primes. Hence, $P(a,p-2)$ implies $$(f(a)+1) \mid 2a+2p-6 \Rightarrow (f(a)+1) \mid 2p-6,$$therefore $$(f(a)+1) \mid \gcd(2p-6,p-1) \mid 4$$hence $f(a) \in \{1,3 \}$ for all $a$. Note that $$f(10) \mid (2 \cdot 10-1)=19 \not\equiv 0 \pmod 3$$hence $f(10)=1$. Thus, $P(a,10) \Rightarrow (f(a)+1) \mid (2a+18) \Rightarrow (f(a)+1) \mid 18$, therefore $f(a)$ can never be equal to $3$. Thus, $f(a)=1$ always, which is a contradiction since $f$ is non-constant $\blacksquare$ To the problem, consider $P(a,\frac{p+1}{2})$ with $p$ an odd prime to obtain $$(f(a)+p) \mid (2a+p-1) \Rightarrow (f(a)+p) \mid (2a-1-f(a))$$hence if $2a-1-f(a)$ is non-zero and we take $p$ to be large enough we obtain a contradiction. Hence $2a-1-f(a)=0$, therefore $f(a)=2a-1$ for all $a$, which satisfies. To conclude, the only solutions are $f \equiv 1$ and $f \equiv 2a-1$.

Clearly $f \equiv 1$ and $f(n) = 2n - 1$ work. We show these are the only ones. Let $P(a,b)$ denote the assertion. $P(1,1) \implies f(1) = 1$. $P(2,1) \implies f(2) \in \{1,3\}$. If $f(2) = 1$, then $P(a,1)$ combined with $P(a,2)$ gives $f(a) + 1 \mid 2 \implies f(a) = 1$ for all $a$, which is a solution. Let $f(2) = 3$. From $P(a,1)$ we have $f(a) \leq 2a - 1$. $P(3,2)$ and $P(3,1)$ gives $f(3) + 1 \mid 6$ and $f(3) + 3 \mid 8$ and these two give $f(3) = 5$. Now we will prove by strong induction that $f(n) = 2n - 1$. Base cases are already done, so assume this is true for all $m \leq n-1$. Now let $m \leq n-1$ be an integer. $P(n,m) \implies f(n) + 2m - 1 \mid 2n + 2m - 2 \implies f(n) + 2m - 1 \mid 2n - f(n) - 1$. If $f(n) = 2n - 1$, we are done. If $f(n) < 2n - 1$, then $2n - f(n) - 1 > 0$ and so $f(n) + 2m - 1 \leq 2n - f(n) - 1 \implies 2f(n) + 2m \leq 2n \implies f(n) + m \leq n$ for all $m \leq n-1$. So plugging in $m = n-1$, we have $f(n) = 1$. But that implies $2m \mid 2n + 2m - 2 \implies m \mid n - 1$ for all $m \leq n-1$. But that is clearly a contradiction. So we are done.

exist $A$ ,any prime $p$,$p >A$ $$f(p)=2p-1$$Answer:$f(x)=2x-1$

\[f(a)+f(b)|2a+2b-2\]All functions which satisfy the divisibility are $f(a)\equiv 1$ for all $a\in \mathbb{Z}^+$ and $f(a)=2a-1$ for all $a\in \mathbb{Z}^+$. $P(a,a)$ gives $f(a)|2a-1$ and putting $a=1$ yields $f(1)=1$. Also putting $a=2$ gives $f(2)=1$ or $f(2)=3$. $\textbf{i)}f(2)=1$ $P(p,1)\implies f(p)+1|2p\implies f(p)=\{1,p-1,2p-1\}$ $P(p,2)\implies f(p)+1|2p+2$ If $f(p)=p-1,$ then $p|2$ which is impossible for $p\geq 3$. If $f(p)=2p-1,$ then $2p|2\iff p|1$ which is impossible. Thus $f(p)=1$ for all primes $p$. $P(a,2)\implies f(a)+1|2a+2$ and $P(a,3)\implies f(a)+1|2a+4$ hence $f(a)+1|2\iff \boxed{f(a)=1, \ \forall a\in \mathbb{Z}^+}$ $\textbf{ii)}f(2)=3$ $P(\frac{p+1}{2},\frac{p+1}{2})\implies f(\frac{p+1}{2})|p\implies f(\frac{p+1}{2})=\{1,p\}$. If there exists some $f(\frac{q+1}{2})=1$ where $q\geq 3$ is a prime, then $P(a,\frac{q+1}{2})\implies f(a)+1|2a+q-1$ Also $P(a,1)\implies f(a)+1|2a$ thus $f(a)+1|q-1$ which yields $f(a)\leq q-2$. Hence $f$ is bounded. Since $f(\frac{r+1}{2})=1,r$ and $f$ is bounded, there exists some $N$ which satisfies that $f(\frac{r+1}{2})=1$ for all $r\geq N$ primes. $P(a,\frac{r+1}{2})\implies f(a)+1|r-1$. Putting $a=2$ and taking $r\equiv 3(mod \ 4)$ prime gives a contradiction. Hence $f(\frac{p+1}{2})=p$ for all $p\geq 3$ primes. $P(a,\frac{p+1}{2})\implies f(a)+p|2a+p-1\implies f(a)+p|2a+p-1-f(a)-p=2a-1-f(a)$ If we take $p$ sufficiently large, then $\boxed{f(a)=2a-1, \ \forall a\in \mathbb{Z}^+}$ as desired.$\blacksquare$

The only solutions are $f\equiv 1$ and $f(n) = 2n - 1$. These clearly work. Now we prove they are the only solutions. Let $P(a,b)$ denote the given assertion. $P(a,a): 2f(a) \mid 2 (2a - 1) \implies f(a) \mid 2a - 1$. This implies $f(1) \mid 1 \implies f(1) = 1$. $P(a, 1): f(a) + 1 \mid 2a$. This implies for any prime $p$ that $f(p) + 1 \mid 2p$, so $f(p) + 1 \in \{1, 2, p, 2p\}$. Since $f(p) + 1 > 1$, $f(p)$ must be in $\{1, p-1, 2p - 1\}$. If $f(p) = p - 1$ and $f(p) \ne 1$, then $P(p,p)$ gives that $p - 1 \mid 2p - 1$, so $p - 1 \mid 2(p - 1) - (2p - 1) = 1$, absurd since $p - 1 \ne 1$. Therefore, $f(p) \in \{1, 2p - 1\}$ for all primes $p$. Case 1: There exist infinitely many primes $p$ with $f(p) = 2p-1$. Let $n$ satisfy $f(n) = 2n-1$. $P(a,n): f(a) + 2n-1 \mid 2(a + n - 1)$. But this implies that \[f(a) + 2n - 1 \mid 2(a + n - 1) - (f(a) + 2n - 1) = 2a - 1 - f(a)\]Notice that infinitely many primes $p$ satisfy $f(p) = 2p - 1$, so there exist arbitrarily large prime satisfying this. Choose $p$ to be large enough so that $f(a) + 2p - 1 > |2a - 1 - f(a)|$. We then have that $f(a) + 2p - 1$ must divide $2a - 1 - f(a)$, and since $|f(a) + 2p - 1| = f(a) + 2p - 1 > |2a - 1 - f(a)|$, we must have $2a - 1 - f(a) = 0$, so $f(a) = 2a + 1$ for all positive integers $a$. Case 2: There exist only finitely many primes where $f(p) = 2p - 1$. Then all sufficiently large primes $p$ satisfy $f(p) = 1$. Claim: For all primes $p$, $f(p) = 1$. Proof: Suppose otherwise. Choose $q$ to be a prime where $f(q) = 2q - 1$ and $2q - 1 \ne 1$. Since $2q - 1 \ne 1$, $q > 2$, so choose $p$ to be a prime where $p \equiv 2\pmod q$. $P(p,q): 2q \mid 2(p + q - 1)$, so $q\mid p + q - 1 \implies q\mid 2 + 0 - 1 = 1$, absurd. $\square$ $P(a, 2): f(a) + 1 \mid 2a + 2$. $P(a, 3): f(a) + 1 \mid 2a + 4$. Comparing gives that $f(a) + 1 \mid 2$, so $f(a) = 1$ for all positive integers $a$.