Amin12 wrote: Find all function $f:\mathbb{N}\rightarrow\mathbb{N}$ such that for all $a,b\in\mathbb{N}$ , $(f(a)+b) f(a+f(b))=(a+f(b))^2$ Let $f(1)=c$ then prove that $c+2|(2c+1)^2$ so $c+2|9$ so $c=1$ or $7$ and then prove $c=1$ and then by easy induction we get $f(x)=x$

Amin12 wrote: Find all function $f:\mathbb{N}\rightarrow\mathbb{N}$ such that for all $a,b\in\mathbb{N}$ , $(f(a)+b) f(a+f(b))=(a+f(b))^2$ I suppose that, as usual on this forum, $\mathbb N$ is the set of positive integers ($0\notin\mathbb N$) Let $P(x,y)$ be the assertion $(f(x)+y)f(x+f(y))=(x+f(y))^2$ Let $a=f(1)$ and $b=f(2)$ $P(f(x),1)$$\implies$ $(f(f(x))+1)=\frac{(f(x)+a)^2}{f(f(x)+a)}$ $P(a,x)$ $\implies$ $(f(a)+x)=\frac{(a+f(x))^2}{f(a+f(x))}$ And so $f(f(x))=x+f(a)-1$ And since $P(x,x)$ $\implies$ $f(x+f(x))=f(x)+x$, we get $f(a)-1=0$ and $f(f(x))=x$ $\forall x$ Then $P(x,f(y))$ becomes new assertion $Q(x,y)$ : $f(x)+f(y)=\frac{(x+y)^2}{f(x+y)}$ Comparing then $Q(x,2)$ with $Q(x+1,1)$, we get $f(x+1)=f(x)+b-a$ and so $f(x)=(b-a)(x-1)+a$ Plugging this back in original equation, we get $a=1$ and $b=2$ and so $\boxed{f(x)=x\text{ }\forall x\in\mathbb N}$

Amin12 wrote: Find all function $f:\mathbb{N}\rightarrow\mathbb{N}$ such that for all $a,b\in\mathbb{N}$ , $(f(a)+b) f(a+f(b))=(a+f(b))^2$ This problem has solved for $f:\Bbb{R}^+\rightarrow \Bbb{R}^+$. Let $P(x,y)$ be the assertion $(f(a)+b)f(a+f(b))=(a+f(b))^2$. We get: $P(a,a)\rightarrow f(a+f(a))=a+f(a)$ So there exist $x_0 \in \Bbb{R}^+$ such that $f(x_0)=x_0$. Now we get: $P(x_0,x_0)\rightarrow f(2x_0)=2x_0$ $P(a+x_0,x_0)\rightarrow (f(a+x_0)+x_0)f(a+2x_0)=(a+2x_0)^2$ $P(a,2x_0)\rightarrow (f(a)+2x_0)f(a+2x_0)=(a+2x_0)^2$ $\Rightarrow f(a+x_0)=f(a)+x_0$ $P(a,x_0)\rightarrow (a+x_0)^2=(f(a)+x_0)f(a+x_0)=(f(a)+x_0)^2 \Rightarrow f(a)=a$ So we are done.

from f(1)=1 we sub a=1 in F.E: (b+1)f(1+f(b))=(f(b)+1)^2. from f(1)=1. sub b=1 ==>f(2)=2 . assume f(t)=t. sub b=t==> f(t+1)=t+1

open : we need what codition to f:R-->R (f(0)=0 )

Quite strange solution: Let $P(x,y)$ be the assertion $(f(a)+b)f(a+f(b))=(a+f(b))^2$ $P(a,a) \rightarrow (f(a)+a)=f(f(a)+a)$ So there exist $n_0 \in \Bbb{R}^+$ such that $f(n_0)=n_0$. $P(n_0,n_0) \rightarrow f(2n_0)=2n_0$ By easy induction we get: $f(kn_0)=kn_0$ for all positive integer $k$ Assume that there exist positive integer $b$ such that $f(b) \ne b$ $P(kn_0,b)$ we get $(kn_0+b)f(kn_0+f(b))=(kn_0+f(b))^2$ Hence: $kn_0+b|(kn_0+f(b))^2$ By a well-known fact, $kn_0+b$ has infinitely many prime factors when $k$ runs, consider one of them is $p$ We have $p|kn_0+b$ and $p|kn_0+f(b)$ so $p|f(b)-b$. But there are infinitely many $p$ and $f(b)-b \ne 0$, contradiction! So, $\boxed{f(x)=x\text{ }\forall x\in\mathbb N}$

Let $A(a,b)=(f(a)+b)f(a+f(b))=(a+f(b))^2$ be assertion of given FE. \begin{align*} A(a,b)\cdot (f(a)+b) \Longleftrightarrow (a+f(b))^2\cdot (f(a)+b)& =(f(a)+b)^2f(a+f(b))\\ &\stackrel{A(b,a)}{=} (f(b)+a)\cdot f(a+f(b))\cdot f(a+f(b))\\ &= (f(b)+a)\cdot f(a+f(b))^2\\ &\Longleftrightarrow (a+f(b))\cdot (f(a)+b)=f(a+f(b))^2\ . \ . \ . \ \bigstar\end{align*} Similarly from $A(b,a)\cdot (f(b)+a)$ we get $(a+f(b))\cdot (f(a)+b)=f(b+f(a))^2\ . \ . \ . \ \spadesuit$ \begin{align*} A(a,b)\cdot f(a+f(b))& \Longrightarrow (a+f(b))^2\cdot f(a+f(b))=(f(a)+b)\cdot f(a+f(b))^2\\ & \stackrel{\bigstar}{=}(f(a)+b)^2\cdot (a+f(b))\\ &\Longleftrightarrow (a+f(b))\cdot f(a+f(b))=(f(a)+b)^2\\ &\stackrel{\cdot (f(a)+b)}{\Longleftrightarrow} (f(a)+b)^3=(f(a)+b)\cdot (a+f(b))\cdot f(a+f(b))\\ & \stackrel{\bigstar}{=}f(a+f(b))^3\\ & \Longleftrightarrow f(a)+b=f(a+f(b))\ . \ . \ . \ \clubsuit\end{align*} Similarly we get from $A(b,a)\cdot f(b+f(a))$ and $\spadesuit$: $$f(b)+a=f(b+f(a))\stackrel{\bigstar\text{ and } \spadesuit}{=}f(a+f(b))\stackrel{\clubsuit}{=}f(a)+b:= Q(a,b)$$

) and note that in previous step we've also proved:$$f(b+f(a))\stackrel{\bigstar\text{ and } \spadesuit}{=}f(a+f(b))\ .\ .\ . \ \bigstar \bigstar$$. \begin{align*} Q(x+y,y)& \Longrightarrow f(y)+x+y=f(x+y)+y\\ &\Longleftrightarrow f(x+y)=f(y)+x\\ & \stackrel{\bigstar \bigstar\text{ and }\clubsuit}{=} f(f(y)+x)\\ &\stackrel{\text{f is injective}}{\Longleftrightarrow}x+y=f(y)+x\\ & \Longleftrightarrow \boxed{f(y)=y,\ \forall y\in \mathbb{N}} \end{align*}

Let $P(x,y)$ denote the assertion that $(f(a)+b)f(a+f(b))=(a+f(b))^2$ for all naturals $a$ and $b$ . $P(1,1)$ gives us that $f(1)+1$ is a fixed point . Letting $a$ be any fixed point , observe that $P(a,a)$ gives us that $2a$ is also a fixed point . As $f(1)+1 \geq 2$ , we can take as large fixed points as we want . At this stage , fix a $b$ arbitrarily and choose a fixed point $x \geq |f(b)-b|^2+2017^{2017}$ . Then, $P(x,b)$ gives us that $(b+x)|(x+f(b)^2)$ which is equivalent to $(b+x)|(f(b)-b)^2$ . Now our tact choice of $x$ forces us that $b=f(b)$ . Now , as our choice of $b$ was arbitrary , we are done . EDIT: Happy New Year 's Eve to all fellow aops'ers

good problem thanks a lot for solutions

Let $P(x,y)$ be the assertion. $P(a,a) \to f(a+f(a))=a+f(a)---@ $ $P(a,b+f(b)) \to (f(a)+b+f(b))f(a+b+f(b))=(a+b+f(b))^2---(1)$ $P(a+b,b) \to (f(a+b)+b)f(a+b+f(b))=(a+b+f(b))^2---(2)$ By $(1)$ and $(2)$, we have $f(a+b)=f(a)+f(b)$ Then $@$ becomes $f(f(a))=a$ The original equation becomes $(f(a)+b)^2=(a+f(b))^2 \Rightarrow f(a)-a=constant$ But $f(a+b)=f(a)+f(b) \Rightarrow f(a)=a \forall a \in \mathbb{N}$

Let $a,b$ be two fixed points $P(a,b):f(a+b)=(a+b)$ and hence the set of fixed point is closed under addition.$P(f(a),b): (f(f(a)+b)(f(a)+f(b))=(f(a)+f(b))^2$ switching $a\rightarrow b$ in the previous we have that $f(f(a))-a=\Delta$.$P(a,a) \implies$ $f(a)+a$ is fixed point and so $\Delta=0$ $\implies$ f is an involution.This allows a re-write ($b\rightarrow f(b)$) $$f(a+b)=\frac{(a+b)^2}{f(a)+f(b)}$$.Plugging $a=b=1$ gives $f(2)f(1)=2$ and so $f(1)+f(2)=3$.Plugging a=2,b=1 $f(3)(f(2)+f(1))=9$ and hence $f(3)=3$,plugging a=3,b=1 gives $f(4)(f(3)+f(1))=16$.If $f(1)=2$ this would imply $5\mid 16$ a clear contradiction and so $f(1)=1$ and $f(2)=2$.But as linear combination of fixed points is a fixed point and $(2,1)=1$ by Bezout's we get the $f(n)=n$ $\forall n$.

My solution. Let $a=f(1)$, and let $P(a,b)$ denote the FE. Firstly observe f is injective, for if $f(x)=f(y)$, then $P(z,x), P(y,x)$ give $x=y$. $P(z-a,1)$ gives $$f(z-a) +1 = z^2/f(z) \ \ \ *$$. If $z$ is a prime, we can observe that $f(z)$ can only be $1$ or $z$ or $z^2$. For large $z$, the first and the last dont hold true, the first one by injectivity and the second one by $*$. Call such primes as a-primes. By $*$ again, for a-primes $p$, $f(p-a)=p-1$. Let $q,r$ be a-primes, $q>r$. By $P(q-r, r)$, $f(q-r)=q-r$. $P(p-a, q-r)$ for an a-prime $p$ gives that $p+q-r-1|(p-a+q-r)^2$, or what is equivalent, $(a-1)^2 \equiv 0 \pmod { (p+q-r-1)}$. So by making $q$ large enough, $a=1.$ By $*$ and induction, we are done.

put a,a; then f(a+f(a))=a+f(a) (1) put a,b+f(b); then (f(a)+b+f(b))*f(a+b+f(b))=(a+b+f(b))*(a+b+f(b)) (2) put a+b,f(b); then (f(a+b)+f(b))*f(a+b+f(b))=(a+b+f(b))*(a+b+f(b)) (3) (2),(3) then f(a+b)=f(a)+b and then easy

Let $P (a,b) $ denote the assertion in the question and let $f (1)=c $. $P (f (a),1) $ & $P (c,a) $ $\implies f (f (a))+1=a+f (c) $ .....$(1) $ $P (1,1): f (c+1)=c+1$ ..... $(2) $ $(1) $ & $(2) $ $\implies f (c)=1$. $P (1,c): cf (2)=2$. If $c=2$, then by computing $f (6) $, we get a contradiction. If $c=1$, then induction gives $f (n)=n $ $\forall n\in\mathbb {N} $, which is indeed the required function.

Let denote the assertion $a=b \implies f(a+f(a))=a+f(a)$ so there exist $f(x_0)=x_0$ $p(f(a),x_0) \implies f^2(a)+x_0=f^2(x_0)+a \implies f^2(a)=a$ so the function is surjective and injective! $p(a,f(b)) \implies (f(a)+f(b)).f(a+b)=(a+b)^2$ and then putting a=b=1 implies that $f(2).2.f(1)=2$ so $f(1)=1$ or $2$ and not let $f(1)=2, f(2)=1$ and then we can easily see that $f(3)=3$ and then putting $a=3, b=1$ in $(f(a)+f(b)).f(a+b)=(a+b)^2$ is contradiction! so $f(1)=1, f(2)=2$ let $f(t)=t, f(k)=k \implies f(t+k)=t+k$ so because of $f(1)=1, f(2)=2$ we're done with induction!

Nice problem in my opinion. Amin12 wrote: Find all function $f:\mathbb{N}\rightarrow\mathbb{N}$ such that for all $a,b\in\mathbb{N}$ , $(f(a)+b) f(a+f(b))=(a+f(b))^2$ $ $ We claim that $f(n) \equiv n \forall n \in \mathbb{N}$ is the only solution to the givn functional equation. $ $ Begin by noticing that $f(n)=n$ trivially satisfies the equation. Now let $P(a,b)$ denote the proposed assertion. $ $ ($\bigstar$) $P(f(a) , b)$: \begin{align*} f(f(a) + b ) \cdot f(f(a)+f(b)) & = (f(a) + f(b))^2\\ & = (f(f(b))+a) \cdot f(f(a) + f(b))\\ \end{align*}Since $f(f(a)+f(b)) \neq 0 \implies f(f(a)) + b = f(f(b)) + a \implies f(f(n)) = n + k \forall n \in \mathbb{N}$ and some integer $k$. ($\bigstar \bigstar$)$ P(a,a)$: $f(a+f(a)) \cdot (a +f(a)) = ( a + f(a))^2 \iff f(a +f(a)) = a +f(a) \forall a \in \mathbb{N}$. Applying $f$ on both sides yields: $$f(f(a +f(a))) = f(a +f(a)) \iff a + f(a) + k = f(a+f(a)) = a + f(a) \iff k = 0$$$ $ Therefore $f$ is a bijective function. ($\bigstar \bigstar \bigstar$)$P(a, f(b))$: $(f(a)+f(b))\cdot f(a+f(f(b))) = (a + f(f(b))^2 \implies (f(a)+f(b)) \cdot \underbrace{f(a+b) = (a+b)^2}_{f(f(n))=n} \forall a,b \in \ \mathbb{N}$ $ $ Now, consider the last equation and call it $Q(a,b)$. $Q(1,1): f(1) \cdot f(2) = 2$. So we have two cases: (i) $f(1)=1$ and (ii)$f(2)=1$. Suppose $f(1)=2$. Then, $Q(1,3) : f(4) \cdot 4 = 16 \implies f(4) = 4$. But $Q(2,2) : f(4) \cdot 2 = 16 \implies f(4) = 8$. Contradiction. $ $ Hence $f(1) = 1$ and $f(2) = 2$ and by an easy induction we see that $\boxed{f(n) = n \forall n \in \mathbb{N}}$ is indeed a solution.

Amin12 wrote: Find all function $f:\mathbb{N}\rightarrow\mathbb{N}$ such that for all $a,b\in\mathbb{N}$ , $(f(a)+b) f(a+f(b))=(a+f(b))^2$ I think it's a very simple problem. By the way, we will show that the only possible answer is: $\boxed{f(n) = n \forall n \in \mathbb{N}}$ Firstly, let $P(a,b)$ the assertion: $(f(a)+b) f(a+f(b))=(a+f(b))^2$ for each natural $a$ and $b$. By $P(1,1)$ we will conclude that: $f(f(1)+1)=f(1)+1$. Also by $P(1,f(1)+1)$ we will deduce that $(2f(1)+1)(f(1)+2)=((f(1)+2)^{2})$ and since $f(f(1)+1)=f(1)+1$, we will reach to $2f(1)+1=f(1)+2$ so $f(1)=1$ Finally, by an easy induction we will be done by the problem. Provided that: $f(1)=1,...,f(k)=k$ And by $P(k,1)$ kill(!) the problem. $(f(k)+1).f(k+1)=(k+1)^{2}$ As $f(k)=k$, it follows that $f(k+1)=k+1$.

let a=b then you can find that f(x)=x

Nice problem $\color{black}\rule{25cm}{1pt}$ The only answer is $f(x)=x$ for all natural numbers $x$. Let $P(a,b)$ be the given assertion. Suppose that we have some $x$ and $y$ such that $f(x)=f(y)$. Then we must have that $P(a,x)=P(a,y)$ Comparing this we have that $f(a)+x=f(a)+y$, which implies that $x=y$, thus we have that $f$ is injective. We choose $a$ and $b$ such that $a+f(b) = p$ is prime. Since we have that $f(a)+b > 1$ this implies two possible values: \[f(p) = p \; \text{or} \; f(p)=1\]Let $S_1$ be the set of primes such that $f(p)=p$ and let $S_2$ be the set of primes such that $f(p)=1$. Since $f$ is injective we have that $\mid S_2 \mid \leq 1$. From $P(a,a)$ we have that $f(a+f(a))=a+f(a)$, so now let's set $a \in S_1$ to get that $f(2a)=2a$. From $P(a,2a)$ we have that $f(3a)=3a$, where $a \in S_1$. By induction we get that $f(na)=na$, where $n$ is an arbitrary natural number and $a \in S_1$. So now let's set $a$ to be the only element of $S_2$ and let $b$ be a composite number divisible by $(a-1)$ then we have that: $$(1+b)f(a+b)=(a+b)^2$$but notice that $(a+b,1+b)=1$, since $a-1 \mid b$. This implies that either all of primes are in $S_1$ or in $S_2$, but since we have that $f$ is injective we have that all the primes are in $S_1$. This implies that for every natural $x > 1$ we have that $f(x)=x$. Let $f(1)=c$. From $P(1,b)$ we have that: $$(b+c)f(1+f(b))=(b+1)^2$$since we have that $f(b)+1 > 1$ we have that $f(f(b)+1)=f(b)+1=b+1$, this implies that: $$b+c=b+1$$thus giving us that $c=1$. This gives us that for all natural numbers $x$ we have that $f(x)=x$.

when you solve FEs how do you know you've found all the solutions? In an olympiad is just finding and proving ONE solution enough for full marks?

Seems pretty short to me.

Let $f(1)=c$ and let $P(a,b)$ denote the assertion. Let $f(x)=f(y)$ for some $x,y \in \mathbb{N}$. Comparing $P(a,x)$ and $P(a,y)$ gives $x=y$ so $f$ is injective. $P(1,a)\implies (a+f(1))\cdot f(f(a)+1) = (f(a)+1)^2 \implies a+f(1) \mid (f(a)+1)^2$. $(1)$ $P(1,1)\implies f(c+1) = c+1$. Plugging $a=c+1$ in $(1)$ we get $2c+1\mid (c+2)^2 \implies c\in \{1,4\}$. If $c=4$, then we get $f(5)=5$. $P(1,5) \implies f(6)=4$ but this contradicts the injectivity of $f$. So $f(1)=1$. $P(a,1)\implies f(a+1) = \frac{(a+1)^2}{f(a)+1}$ and by a simple induction, we get $f(n)=n \forall n \in \mathbb{N}$.

Answer: $f(n)=n \ \ \forall n\in \mathbb{N}.$ Proof: It's easy to see that the identity function works. Let $P(a,b)$ denote the given assertion, Claim 1: There is a fixed point. Proof: \[P(a,a)\implies f(a+f(a))=a+f(a). \quad \square\]Claim 2: $f(f(n))=n \ \ \forall n \in \mathbb{N}.$ Proof: We have \[P(f(a),1)\implies (f(f(a))+1)f(f(a)+f(1))=(f(a)+f(1))^2\]and \[P(f(1),a)\implies (f(f(1))+a)f(f(a)+f(1))=(f(a)+f(1))^2\]which imply $f(f(a))=a+C$ where $C$ is an integer. But from Claim 1, we know there is a fixed point, so $C=0$ and we have proved the claim. $\square$ Claim 3: $f(1)=1.$ Proof: We have \[P(1,f(1)) \implies f(1)f(2)=2.\]If $f(2)=1$ and $f(1)=2,$ then \[P(2,2) \implies f(3)=3\]but \[P(1,3) \implies f(4)=\frac{16}{5}\]which is impossible. So, $f(2)=2$ and $f(1)=1$. Claim 4: $f(n)=n \ \ \forall n \in \mathbb{N}.$ Proof: We have \[P(a,1) \implies f(a+1)=\frac{(a+1)^2}{f(a)+1}\]and since $f(1)=1,$ by induction, we have $f(a)=a \ \ \forall a\in \mathbb{N}. \quad \blacksquare$

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rama1728 wrote:

@above, I think your solution is wrong. Indeed, for any $n\in\mathbb{N}_k$ one can find $x$ and $y$ such that $n=x+f(y).$ However, in order for any pair $x,y\in \mathbb{N}_k$ to satisfy $xf(y)=y^2$ then for any $x,y\in \mathbb{N}_k$ there must exist $a$ and $b$ which simultaneously satisfy $b+f(a)=x$ and $a+f(b)=y.$ This clearly does not always hold. For example, take $x=k+1$ and $y=c.$ Since $b+f(a)=x$ and $\forall n, \ f(n)\geq k$ then we must have $b=1.$ But we also have $a+f(b)=a+f(1)=c.$ Thus, any $c\in\mathbb{N}_k$ must be greater than $f(1)$ which doesn't have to hold.

oVlad wrote: wer wrote: Nice one! Nice... what? I think they mean the problem.

We claim that $f(a)=a$ is the ONLY function that works. Let $P(a,b)$ the assertion of this F.E. Claim 1: $f$ is injective Proof: if there existed $x,y$ such that $f(x)=f(y)$ then by $P(a,x)-P(a,y)$ $$f(a)+x=f(a)+y \implies x=y \implies f \; \text{injective}$$Claim 2:$f(p)=p$ for a prime $p$ big enough. Proof: Set a really big prime $p$ and use $P(p-f(b),b)$ $$(f(p-f(b))+b)f(p)=p^2 \implies f(p) \mid p^2$$Now clearly $f(p) \ne p^2$ so either $f(p)=1$ or $f(p)=p$, but since we have $f$ injective we have that if there existed $p$ big such that $f(p)=1$ then it would be unique, so we ignore thst $p$ and take the rest of big primes $p$ that hold $f(p)=p$. Finishing: Now back to the F.E. by $P(p,a)$ where $p$ is such a big prime $$p+a \mid p^2+2pf(a)+f(a)^2 \implies p+a \mid (f(a)-a)(f(a)+a+2p) \implies p+a \mid (f(a)-a)^2 \implies f(a)=a$$Hence the unique function that works is $\boxed{f(a)=a \; \forall a \in \mathbb N}$ thus we are done

$P(a,a) : f(a+f(a)) = a + f(a)$. $P(a,b+f(b)) : (f(a) + b + f(b))f(a+b+f(b)) = (a + b + f(b))^2$. $P(a+b,b) : (f(a+b) + b)f(a+b+f(b)) = (a + b + f(b)^2$. so we have $(f(a) + b + f(b)) = (f(a+b) + b) \implies f(a+b) = f(a) + f(b)$ so $f$ is a cauchy function which implies $f(a) = ca$. $(ca + b)(a + bc) = (a + bc)^2 \implies a + bc = b + ca \implies a(c-1) = b(c-1) \implies c = 1$ so $f(a) = a$. Answers : $f(a) = a$.

another solution $P(1,1) : (f(1) + 1)f(1+f(1)) = (1+f(1))^2 \implies f(1+f(1)) = 1+f(1)$ so there exists $k$ such that $f(k) = k$ $P(k,k) : f(2k) = 2k$ and $P(2k,k) : f(3k) = 3k$ and now with induction we have $f(nk) = nk$. Claim $: f(1) = 1$. Assume $f(1) = c$ $P(f(a),k) , P(f(k),a) : f(f(a)) + k = f(f(k)) + a = k + a \implies f(f(a)) = a$ we have $f(1) = c$ so $f(f(1)) = f(c) = 1$. $P(1,c) : 2cf(2) = 4 \implies cf(2) = 2$ if $c = 2$ then $f(f(1)+1) = f(3) = f(1) + 1 = 3$ but $P(1,3) : f(4)$ is not Natural so contradiction so $c = 1$ so $f(1) = 1$. we had that if $f(k) = k$ then $f(nk) = nk$ so now that $f(1) = 1$ we have $f(n) = n$.

Let $P(x,y)$ denote the assertion $(f(x)+y)f(x+f(y))=(x+f(y))^2$. We solve this for $f:\mathbb R^+\to \mathbb R^+$. $P(x,x)$ gives $f(x+f(x))=x+f(x)$. Comparing $P(x+y,y)$ and $P(x,y+f(y))$ shows $f$ is additive, so $f(x)=kx$. Checking gives $k=1$ works only.

Firstly, $f$ is injective (otherwise, let $f(x)=f(y)$ and compare $P(a, x), P(a, y)$). Now, $P(p-f(x), x)$ implies that $f(p)=p$ for all sufficiently large primes $p$ ($f(p)=1$ holds for at most one prime due to the injective property) and thus $f(p-f(x))=p-x$. Take another large prime $q$ and apply this once more: $f(p-q+x)=f(p-f(q-f(x))=p-q+f(x)$, so $f(x+c)=f(x)+c$ for $p-q=c$. Now, comparing $P(x+c,y)$ and $P(x, y)$ gives that $(f(x)+y)+f(x+f(y))=2(x+f(y))$. The last two numbers have product $(x+f(y))^2$, so $f(x)+y=x+f(y)$, which implies that $f(x)=x+k$ and substituting back gives that $f(x)=x$ is the only solution.

$p(f(a),1),p(f(1),a)\Rightarrow f(f(a))=a+\alpha$. by comparing $p(a,f(2)),p(a+1,f(1))$ we have $f(a+1)=f(a)+c$ and so f is linear, $f(x)=ax+b(\forall x \in \Bbb{N})$ plugging this in the equation we get $f(x)=x(\forall x)$

By comparing \( P(f(a), b) \) and \( P(f(b), a) \), we deduce that \( f(f(a)) + b = a + f(f(b)) \). From this, \( f(f(a)) - a \) must take a constant value, which we denote as \( c \). From \( P(a, a) \), we find \( f(a + f(a)) = a + f(a) \). Therefore, \( a + f(a) - c = f(f(a + f(a))) = f(a + f(a)) = a + f(a) \), which implies \( c = 0 \). Thus, \( f(f(a)) = a \), and \( f \) is bijective. Now, consider \( P(a, f(b)) \). Notice that \( (f(a) + f(b)) f(a + b) = (a + b)^2 \). From \( P(1, f(1)) \), we have \( f(1) f(2) = 2 \). Assume \( f(1) = 2 \) and \( f(2) = 1 \). Using \( P(1, f(2)) \), we find \( f(3) = 3 \). Next, from \( P(1, f(3)) \), we find \( f(4) = \frac{16}{5} \), which is invalid. Therefore, \( f(1) = 1 \). Assume \( f(n - 1) = n - 1 \). From \( P(1, f(n - 1)) \), we deduce \( n f(n) = n^2 \), which implies \( f(n) = n \). By induction, we have shown that \( f(n) = n \) for all \( n \).