When R+ -> R+ instead of R -> R:

f(x)=x+1 I fake proved it by comparing both sides' degrees, and claiming the degree of f is one. (I believe this method works if the question is asking a polynomial)

Claiming polynomial gets 0 tho

We claim the only solution is $f(x) = x + 1$. This works. Now we prove its the only one. Let $P(x,y)$ denote the given assertion and let $c = f(1)$. $P(1,x): f(x + c) = f(x) + 2$. $P(x,1): f(x + f(x)) = cx + 2$. $P(x, c/x + 1): f(x + f(x) + c) = xf(c/x + 1) + 2$, so $f(x + f(x)) = x f(c/x + 1)$. This implies $xf(c/x + 1) = cx + 2$, so $f(c/x + 1) = c + 2/x$. Setting $x\to c/x$ here gives $f(x + 1) = c + 2x/c$ for any positive real $x$. Claim: $c\ge 1$. Proof: Suppose $c<1$. Then $P(1, 1 - c)$ gives that $c = f(1)>2$, absurd. $\square$ Now, we have for any $x>0$, $f(x + c) = c + 2(x + c - 1)/c = f(x) + 2$, so $f(x) = c + 2(x + c - 1)/c - 2 = c + (2x - 2)/c$. Thus, $f(x) = ax + b$ for some reals $a,b$. From here we can check that $f(x) = x + 1\forall x>0$.

The answer is $f(x) \equiv x+1$, which is easily verified to be the only linear solution. We show conversely that $f$ is linear. Let $P(x,y)$ be the assertion. Claim: $f$ is weakly increasing. Proof. Assume for contradiction $a>b$ but $f(a)<f(b)$. Choose $y$ such that $ay+f(a)=by+f(b)$, that is $y=\frac{f(b)-f(a)}{a-b}$. Then $P(a,y)$ and $P(b,y)$ gives $af(y)+2=bf(y)+2$, which is impossible. $\blacksquare$ Claim: [Up to an error of $2$, $f$ is linear] We have \[ \left\lvert f(x) - (K x + C) \right\rvert \le 2 \]where $K \coloneqq \frac{2}{f(1)}$ and $C \coloneqq f(f(1))-2$ are constants. Proof. Note $P(1,y)$ gives $\boxed{f(y+f(1)) = f(y)+2}$. Hence, $f(nf(1)) = 2(n-1)+f(f(1))$ for $n \ge 1$. Combined with weakly increasing, this gives \[ 2\left\lfloor \frac{x}{f(1)} \right\rfloor + C \le f(x) \le 2\left\lceil \frac{x}{f(1)} \right\rceil + C \]which implies the result. In fact, the $O(1)$ term is bounded by $2$. $\blacksquare$ Rewrite the previous claim to the simpler $f(x) = Kx+O(1)$. Then for any $x$ and $y$, the above claim gives \[ K\left( xy + Kx + O(1) \right) + O(1) = x f(y) + 2 \]which means that \[ x \cdot \left( Ky + K^2 - f(y) \right) = O(1). \]If we fix $y$ and consider large $x$, we see this can only happen if $Ky+K^2-f(y)=0$, i.e.\ $f$ is linear.

The answer is $f(x)=x+1$ only which clearly works. Let $P(x,y)$ denote the assertion. $P(1,y)$ gives $f(y+f(1))=f(y)+2$ for all $y>0$. This implies that $f(x)>\tfrac{2}{f(1)}x-10^{100}$ for all $x>0$. We now prove that $f$ is bounded on any interval $(0,M]$. Indeed suppose that it was unbounded and take $k \in (0,M]$ such that $f(k)$ is really really big. Then $P(k,y)$ for arbitrary $y$ yields $$kf(y)+2=f(ky+f(k))\geq \frac{2}{f(1)}(ky+f(k))-10^{100}.$$But because $k$ itself is bounded but $f(k)$ is really really big, we obtain a contradiction. Thus we have, in particular, some constant $C$ with $f(x)<C$ for all $x \in (0,f(1)]$, which then implies $f(x)<\tfrac{2}{f(1)}x+C$ for all $x>0$. If we let $\tfrac{2}{f(1)}=a$ for convenience, there exists some bounded function $r : \mathbb{R}^+ \to \mathbb{R}$ such that $f(x)=ax+r(x)$. Then substituting this back in gives $$(a^2-r(y))x=2-ar(x)-r(xy+ax+r(x)).$$If we make $x$ massive (fixing $y$), we obtain a contradiction unless $a^2=r(y)$, since the LHS is unbounded in some direction. This is for $y$ arbitrary, so we in fact have $f(x)=ax+a^2$. On the other hand, $a^2=r(y)$ makes the LHS zero, so the RHS must be zero as well, in particular $$2-a^3-a^2=0 \implies a=1,$$since clearly $a>0$, which yields the desired solution. $\blacksquare$

Cursed (bad) sol that I haven't seen anyone else have (when I say "integer" I mean "positive integer"): $f(x)=x+1$ works. Let $k=f(1)$. $P(1,y)$ gives $f(y+k) = f(y)+2$ (very important, call this Eq. 1), notably $f(1+k)=2+k$. Also, in particular, if $n$ is an integer, $f(y+nk) = f(y)+2n$. (call this "Repeated Eq. 1") $P(1+k,y)$ gives $f(y+ky+k+2) = (1+k)f(y)+2$. With $P(1,y+ky+2)$, we get $f(y+ky+2)=(1+k)f(y)$. By Repeated Eq. 1, if $y$ is an integer, $f(y+2)=(1+k)f(y)-2y$. Similarly with $P(1+2k,y)$ and $P(y+2ky+4)$, $f(y+4) = (1+2k)f(y)-4y$. So, we can get $f(5) = f(1+2+2) = k^3+2k^2-k-8 = f(1+4) = 2k^2 + k - 4 \implies k^3-2k-4 = 0 \implies k=f(1)=2$. Now Eq. 1 becomes $f(y+2)=f(y)+2$; again, we can use this for all $y$. Let $g(x)=f(x)-1$. With $P(n,x)$ and odd $n$, $g(nx)=ng(x)$. Also, if $x>2m$ for some integer $m$, by repeated eq 1 and the codomain being positive, $f(x)>2m$ and $g(x)>2m-1$. So, suppose that for some $x$, $g(x) < x$. Let $\varepsilon = x-g(x) > 0$. Then, take sufficiently large odd $c$ so that $c\varepsilon > 3$, meaning $g(cx) = cg(x) = cx - c\varepsilon < cx-3$. But there exists some even integer $b$ in $[cx-2, cx)$ so this is bad: $g(cx) > b-1 \ge cx-3.$ So for all $x$, $g(x) \ge x$. Suppose for some $x$, $g(x)>x$. Then $P(x,1)$ gives $f(x+f(x)) = 2x + 2 \iff g(x+1+g(x)) = 2x+1$. But $g(x+1+g(x)) \ge x+1+g(x) > 2x+1$, contradiction, so $g(x)=x$ for all $x$ and we are done.

Redacted since wrong

I claim that the only solution is $\boxed{f(x) \equiv x + 1}$. It is easy to check that this works. Let $P(x, y)$ denote the given equation. Let $f(1) = u$. $P(1, y)$ gives \[f(y + u) = f(y) + 2\]By induction, we can show \[f(y + ku) = f(y) + 2k\]for positive integers $k$. $P(1 + \ell u, y)$ gives \begin{align*} f((1 + \ell u)y + f(1 + \ell u)) &= (1 + \ell u)f(y) + 2\\ f(y + \ell u + u + 2\ell) &= f(y) + \ell u f(y) + 2\\ f(y + 2\ell) + 2\ell y + 2 &= f(y) + \ell u f(y) + 2\\ f(y + 2\ell) &= f(y) + \ell u f(y) - 2\ell y \end{align*}Let $Q(y, \ell)$ denote this proposition. $Q(1, 1)$ gives \[f(3) = u + u^2 - 2\]$Q(1, 2)$ gives \[f(5) = u + 2u^2 - 4\]$Q(3, 1)$ gives \[f(5) = f(3) + uf(3) - 6 = u^3 + 2u^2 - u - 8\]Hence \[u + 2u^2 - 4 = u^3 + 2u^2 - u - 8 \implies (u-2)(u^2 + 2u + 2) = 0 \implies u = 2\]Hence $f(1) = 2$ and $f(y + 2\ell) = (1 + 2\ell) f(y) - 2\ell y$. Now, $Q(x, 1)$ gives \[f(x + 2) = 3f(x) - 2x\]$Q(x, 2)$ gives \[f(x + 4) = 5f(x) - 4x\]$Q(x + 2, 1)$ gives \[f(x + 4) = 3f(x + 2) - 2x - 4 = 9f(x) - 8x - 4\]Hence \[5f(x) - 4x = 9f(x) - 8x - 4 \implies f(x) = x + 1\]and we are done.

The only solution is $f(x) = x+1$. Claim: $f$ is strictly increasing. Proof: Suppose $a<b$ but $f(b) < f(a)$. Then $k=\frac{f(a)-f(b)}{b-a}$ is positive and satisfies $ak+f(a) = bk + f(b)$, so \[ af(k) + 2 = f(ak + f(a)) = f(bk + f(b)) = bf(k) + 2 \]resulting in a contradiction. If $f(b) = f(a)$, then $af(b) + 2 = f(ab + f(a)) = f(ba + f(b)) = bf(a) + 2$, which is again a contradiction. $\blacksquare$ Claim: $f(1) = 2$. Proof: For brevity, let $c=f(1)$. Taking $x=1$ and $y=1$ separately yields $f(x+f(x)) = xc + 2$ and $f(y+c) = f(y) + 2$. Labelling these identites as $(1)$ and $(2)$ respectively, we have for any $z\in \mathbb{R}^+$ that \[ (z+c)c + 2 \stackrel{(1)}{=} f( z + c + f(z+c) ) \stackrel{(2)}{=} f(z + f(z) + 2) + 2\]so $f(z + f(z) + 2) = zc + c^2$. Now if $c > 2$, then \[ zc + 4 \stackrel{(1)}{=} f(z+f(z)) + 2 \stackrel{(2)}{=} f(z + f(z) + c) > f(z+f(z) + 2) = zc + c^2 \]implies that $c^2 < 4$ ,which is a contradiction. If $c<2$, then we analogously obtain that $c^2 > 4$, which is another contradiction. So $c=2$. $\blacksquare$ Now that $f(1) = 2$, we take $y=z+c/x$ in the original equation to obtain \[ x f(z+c/x) + 2 = f(xz + f(x) + c) \stackrel{(2)}{=} f(xz + f(z)) + 2 = x f(z) + 4 \]so that $xf(z+2/x) = xf(z) + 2$. Taking $x=2n$, we then have that $f(z+1/n) = f(z) + 1/n$ for any positive integer $n$ and $z\in \mathbb{R}^+$, and so by induction $f(x) = x+1$ for all $x \in \mathbb{Q}^+$. The fact that $f$ is strictly increasing then implies that $f(x) = x+1$ for all positive reals.

Here's a solution using injectivity. Claim: \(f\) is injective. Proof. If \(f(a)=f(b)\), then \[af(b)+2=f(ab+f(a))=bf(a)+2\]implies \(a=b\). \(\blacksquare\) Claim: For all \(x\) and \(y\), we have \(f(f(x))-f(f(y))=xf(y)-yf(x)\). Proof. By \(P(f(x),y)\) and \(P(f(y),x)\), we have \[f(yf(x)+f(f(x)))=f(x)f(y)+2=f(xf(y)+f(f(y))),\]implying \(yf(x)+f(f(x))=xf(y)+f(f(y))\) by injectivity. \(\blacksquare\) Claim: \(f\) is linear. Proof. For any \(x\), \(y\), \(z\), note that summing \begin{align*} f(f(x))-f(f(y))&=xf(y)-yf(x),\\ f(f(y))-f(f(z))&=yf(z)-zf(y),\\ \text{and}\quad f(f(z))-f(f(x))&=zf(x)-xf(z) \end{align*}gives \[xf(y)+yf(z)+zf(x)=yf(x)+zf(y)+xf(z).\]The above equation uniquely determines \(f(z)\) in terms of \(x\), \(y\), \(z\), \(f(x)\), \(f(y)\), and it is clear \(f(x)=ax+b\), \(f(y)=ay+b\), \(f(z)=az+b\) satisfies the equation, so \((x,f(x))\), \((y,f(y))\), \((z,f(z))\) are collinear. \(\blacksquare\) It is easy to check the only linear solution is \(f(x)\equiv x+1\).

Let $P(x,y)$ denote the assertion. Suppose $f(x_1)=f(x_2).$ Then $P(x_1,x_2)$ and $P(x_2,x_1)$ give $$f(x_1x_2+f(x_1))=x_1f(x_2)+2,$$$$f(x_2x_1+f(x_2))=x_2f(x_1)+2,$$implying that $x_1f(x_2)=x_2f(x_1).$ Hence $x_1=x_2,$ so $f$ is injective. Now consider $a,b,x\in\mathbb{R}^+$ with $x$ very large. $P(a,x)$ and $P(b,\tfrac{ax+f(a)-f(b)}{b})$ yield $$f(ax+f(a))=af(x)+2,$$$$f(b(\tfrac{ax+f(a)-f(b)}{b})+f(b))=bf(\tfrac{ax+f(a)-f(b)}{b})+2.$$The left sides are equal, so the right sides must be as well. This means $$f(x)=\tfrac{b}{a}f(\tfrac{a}{b}x+\tfrac{f(a)-f(b)}{b}).$$Let $k$ be an arbitrary positive real number. By repeating these steps with $a$ replaced by $ka$ and $b$ replaced by $kb,$ we get $$f(x)=\tfrac{b}{a}f(\tfrac{a}{b}x+\tfrac{f(ka)-f(kb)}{kb}).$$Then by injectivity, $$\frac{f(a)-f(b)}{b}=\frac{f(ka)-f(kb)}{kb}\implies f(ka)-k\cdot f(a)=f(kb)-k\cdot f(b).$$Since this is true for all $a,b,$ there exists a function $g:\mathbb{R}^+\to\mathbb{R}$ such that for all $x,$ $$f(kx)-k\cdot f(x)=g(k).$$Next, write \begin{align*} f(k_1k_2) &= k_1f(k_2)+g(k_1)\\ &= k_1(k_2f(1)+g(k_2))+g(k_1)\\ &= k_1k_2f(1)+k_1g(k_2)+g(k_1), \end{align*}and similarly $$f(k_2k_1)=k_2k_1f(1)+k_2g(k_1)+g(k_2).$$Thus, $$k_1g(k_2)+g(k_1)=k_2g(k_1)+g(k_2)\implies \frac{g(k_1)}{k_1-1}=\frac{g(k_2)}{k_2-1}.$$This means that for some constant $c,$ $g(k)=c(k-1)$ for all $k.$ To finish, note that for all $x,$ $$f(x)=xf(1)+g(x)=xf(1)+c(x-1).$$In particular, $f(x)$ is linear, so it is easy to check that the only solution is $\boxed{f\equiv x+1}.$

Let $P(x,y)$ denote the assertion of the functional equation. First, we have that $P(1,y)\implies f(y+f(1))=f(y)+2$. Then, \[P(x+f(1),y)\implies f(xy+f(1)y+f(x)+2)=xf(y)+f(1)f(y)+2,\]\[P(x,y+f(1))\implies f(xy+xf(1)+f(x))=xf(y)+2x+2.\]Hence, whenever $xf(1)=f(1)y+2$, we have that $f(1)f(y)=2x$. In particular, $f(y)=\frac2{f(1)}\left(y+\frac2{f(1)}\right)$ for all $y\in\mathbb R^+$, so $f$ is linear. Setting $f(x)=ax+b$ gives $a(xy+ax+b)+b=x(ay+b)+2$ or $a^2x+ab+b=bx+2$, so $a^2=b$ and $ab+b=2$. The only solution of this that yields a function $f:\mathbb R^+\rightarrow\mathbb R^+$ is $(a,b)=(1,1)$, so $f(x)=x+1$.

Submission for overcomplicated solutions (aparently....) I claim that $f(x)=x+1$ is the only function that works. Denote $P(x,y)$ the assertion of the F.E. Claim 1: $f$ is injective. Proof: Let $f(a)=f(b)$ then by $P(a,b)-P(b,a)$ u get $a=b$ Claim 2: $f$ is surjective over $\mathbb R_{>2}$ Proof: Trivial becuase of the RHS Claim 3: There exists $c$ such that $f(x)>cx$ Proof: By $P \left(\frac{f(x)-2}{f(y)},y \right)$ and injectivity (where $x$ is such that $f(x)>2$, clearly it exists) $$\frac{yf(x)-2y}{f(y)}+f \left(\frac{f(x)-2}{f(y)} \right)=x \implies xf(y)>yf(x)-2y \implies \; \text{ we need our} \; c=\frac{f(x)-2}{x}$$Claim 4: $f$ is strictly increasing Proof: Suppose we have $f(c)>f(\alpha c)$ for $\alpha>1$ then set $\alpha=\frac{a}{b}$ for $a,b>2$. Now by $P(xf(z),y)$ $$f(xyf(z)+f(xf(z))=xf(y)f(z)+2=f(xzf(y)+f(xf(y)) \implies xyf(z)-xzf(y)=f(xf(y))-f(xf(z))$$Now $P(f(x),y)$ and then using symetry gives $$yf(x)-xf(y)=f(f(y))-f(f(x)) \implies f(xf(y))-f(xf(z))=xyf(z)-xzf(y)=xf(f(y))-xf(f(z)) \implies f(xu)-f(xv)=xf(u)-xf(z) \; \forall x \in \mathbb R^+ \; , \; u,v>2$$Call this equation $Q(x,u,v)$ then by $Q \left(\frac{x}{c},c, \alpha c \right)$ implies that $f(x)>f(\alpha x)$ for all positive reals $x$ so now make a chain $f(x)>f(\alpha x)>f(\alpha^2 x)>...>f(\alpha^n x)>c \alpha^n x$ and now set $n \to \infty$ to get contradiction . Claim 5: $f(x)=mx+n$ for $m,n \in \mathbb R^+$ Proof: Now by Lebesgue Differentiable Theorem over $(2, \infty)$ there exists $x_0>2$ so that $f$ is diferentiable at $x_0$ so fix $x$ then by $Q(x,u,x_0)$ and $u \to x_0$ $$\lim_{x u \to x x_0} \frac{f(xu)-f(xx_0)}{xu-xx_0}=f'(x_0) \implies f'(xx_0) \; \text{exists and it is equal to} \; f'(x_0)$$But $x$ was arbitrary this whole time so $f'(xx_0)=f'(x_0) \; \forall x \in \mathbb R^+$ meaning that $f'$ is constant so $f(x)=mx+n$ as desired. Finishing: Its easy to verify that only $m=n=1$ works so $f(x)=x+1$ for all positive reals $x$, yaaay.

Answer $f(x)=x+1$, clearly works. Fact 0: $f$ is injective since if $f(a)=f(b)$ then $af(b)+2=bf(a) +2\implies a=b$. Fact 1: Plugging $x=1$ into the given says $f(y+f(1))=f(y)+2$. Fact 2: Fact 1 says $f(x) \ge 2\lfloor \tfrac{x}{f(1)}\rfloor$. Claim 1: $f(x) \ge \frac{2}{f(1)}x+\frac{4}{f(1)^2}$. Proof: By injectivity, whenever $af(b)=cf(d)$ we must have $ab+f(a) = cd+f(c)$. Plug in $a=1,b=224+Nf(1),c=f(224+Nf(1))/f(x),d=x$; by $(\spadesuit)$ we have $c=(f(224)+2N)/f(x)$, so we get $$224+Nf(1)+f(1)=\frac{f(224)+2N}{f(x)}x+f\left(\frac{f(224)+2N}{f(x)}\right)$$Use Fact 2 to bound $f\left(\frac{f(224)+2N}{f(x)}\right) \ge 2\left\lfloor \frac{f(224)+2N}{f(x)f(1)} \right\rfloor$. Then divide both sides of the above by $N$ and take the limit as $N \to \infty$, $$f(1) \ge 2\frac{x}{f(x)} + \frac{4}{f(x)f(1)} \implies f(x) \ge\frac{2}{f(1)}x+\frac{4}{f(1)^2}.$$ Claim 2: $f(x) \le \left(\frac{f(1)^2}{2}-f(1)\right)x+f(1)-\frac{2}{f(1)}$. Proof: Use Claim 1 on the original statement to get $$\frac{2}{f(1)}\left(xy+f(x)\right)+\frac{4}{f(1)^2} \le xf(y)+2$$$$\implies f(x) \le x\left(\frac{f(1)}{2}f(y)-y\right)+f(1)-\frac{2}{f(1)}$$for any $x,y$. Plug in $y=1$. By Claim 2 we have that $f(x)$ is bounded on the range $x \in (0,f(1)]$ below some constant $C$ (in particular $\frac{f(1)^3}{2}-f(1)^2+f(1)-\frac{2}{f(1)}$ or something), so by Fact 1 we have $f(x) \le 2\lfloor \tfrac{x}{f(1)}\rfloor+C$. In addition recall Fact 2 $f(x) \ge 2\lfloor \tfrac{x}{f(1)}\rfloor$, so $\lim_{x \to \infty} \tfrac{f(x)}{x}= \tfrac{2}{f(1)}$. Now recall from the proof of Claim 1 we had the equation $$224+Nf(1)+f(1)=\frac{f(224)+2N}{f(x)}x+f\left(\frac{f(224)+2N}{f(x)}\right).$$Dividing by $N$ and taking the limit as $N \to \infty$ now gives an equality, $$f(1)=2\frac{x}{f(x)}+\frac{4}{f(x)f(1)}.$$So $f(x)$ is linear, and it's clear that this implies $f(x)=x+1$.

Why are (most of) the solutions above so complicated. Let $f(1) = c$. Take $x = 1$ to get $f(y+c) = f(y) + 2$. Replace $y$ with $y+\frac{c}{x}$ to get $f(c+xy+f(x)) = xf \left(y+ \frac{c}{x} \right) + 2 \implies xf(y) + 2 = f(xy+f(x)) = xf \left(y+\frac{c}{x} \right)$. Replace $x$ with $\frac{c}{x}$ to get $\frac{cf(y)}{x} + 2 = \frac{cf(y+x)}{x}$ or $f(x+y) - f(y) = \frac{2x}{c}$, so $f$ is linear. Checking gives $f(x) = ax + a^2$ with $a + a^3 = 2$, so $a = 1$ giving $f(x) = x+1$ is the only solution, and it works. $\blacksquare$

how many points for getting the solution set and proving injectivity?

Note that putting $x=1$ implies $f(x+f(1))-f(x)=2$ for all $x>0$. Inductively, $f(x+nf(1))-f(x)=2n$ for all positive integers $n$. Moreover, taking $y \rightarrow y+f(1)$ gives us that $f(xy+f(x)+xf(1))=xf(y)+2x+2,$ and so $f(xy+f(x)+xf(1))=f(xy+f(x))+2x,$ i.e. $f(y+xf(1))-f(y)=2x$ for all $x,y$ such that $y>f(x)$. We claim the above relation actually holds for all $x,y$. Indeed, for any $x,y$ we may take a large positive integer $n$ such that $nf(1)>f(x),$ and so $y+nf(1)>nf(1)>f(x),$ implying that $f(y+nf(1)+xf(1))-f(y+nf(1))=2x,$ which combined with $f(x+nf(1))=f(x)+2n$ is equivalent to $f(y+xf(1))-f(y)=2x,$ as desired. Thus, $f(y+xf(1))=f(y)+2x$ for all $x,y$, and so $f(xf(1)+1)=f(1)+2x,$ which in turn gives that there exist constants $A,B$ such that $f(x)=Ax+B$ for all $x>1$. It is obvious that $A \geq 0$. Now, take $x,y$ such that $x>1, y>1$ and $xy+Ax+B>1$. Then, the original relation easily implies that $A=B=1$. Thus, $f(x)=x+1$ for all $x>1$. Now, to finish, take $x=2$ in the original equation: $2f(y)+2=f(2y+3)=2y+4,$ and so $f(y)=y+1$, hence $f(x)=x+1$ for all $x>0$, which evidently works.

i guess the creators didnt see this dumb solution Let $P$ denote the assertion. $P(1, y)$ gives $f(y + f(1)) = f(y) + 2$ so $f(y + c) = f(y) + c$ for some constant $c > 0$, for all $y > 0$. Now, note that $f(x(y + f(1)/x) + f(x)) = (xf(y) + 2) + 2 = xf(y + f(1)/x) + 2$ which reduces to\[f(y + f(1)/x) = f(y) + 2/x\]for all $x, y > 0$. Now, let $f(1)/x = z$ be another variable which is ok since $f(1)/x$ is bijective from positive reals to itself. We now have $f(y + z) = y + Cz$ for some constant $C > 0$. Note $f(y + z) = f(z + y) = f(y) + Cz = f(z) + Cy$ hence $f(y) = Cy + f(z) - Cz$, and plugging in $z = 1$ gives $f(y) = Cy + f(1) - Cz$ is linear, hence $f$ linear, and it is easy to get $f = x+1$ afterwards. Edit: I didn't realize this was identical to Lcz's solution in post 17

$x = 1$ gives $f(y + f(1)) = f(y) + 2$ Now, substitute $y = y + \frac{f(1)}{x}$ to get $f(xy + f(x)) + 2 = xf(y + \frac{f(1)}{x})$. Subtract this from original FE to get $\frac{2}{x} = f(y + \frac{f(1)}{x}) - f(y)$ implying linearity from which we find $f(x) = x + 1$. Motivation: Substituting x = 1 is natural since over positives and eliminates the coefficient. Afterwords, getting something of the form $f(y + f(1))$ makes it promising to get a f(1) inside an f. My first thought was $y = y + f(1)$ however this wasn't too useful because the LHS was complicated. Afterwards, trying $y = y + \frac{f(1)}{x}$ insta-solved the problem.

$P(x,1): f(x+f(x))=xf(1)+2; P(1,y): f(y+f(1))=f(y)+2$. Let $y=x+f(x), f(x+f(x)+f(1))=f(x+f(x))+2=xf(1)+4$ $f(x, \frac{x+f(1)}{x}): f(x+f(x)+f(1))=xf(\frac{x+f(1)}{x})+2=xf(1)+4, f(\frac{x+f(1)}{x})=f(1)+\frac{2}{x}$ Substitue $t=\frac{x+f(1)}{x}, \frac{1}{x}=\frac{t-1}{f(1)}$. Thus, we have $f(t)=f(1)+\frac{2t-2}{f(1)}$ which is a linear function. Let $f(x)=ax+b, f(xy+ax+b)=x(ay+b)+2, axy+a^2x+ab+b=axy+bx+2, a^2=b, ab+b=2, a=b=1\implies \boxed{f(x)=x+1}$

It is easy to check that $f(x)=x+1$ works, now we show that this is the only solution. The main tool is when subbing $x=1$, we have $f(y+c)=f(y)+2$ where $c=f(1)$. Call this $(*)$.

$f(1)=2$ finishes the problem.

Let $P(x,y)$ be the assertion of $f(xy+f(x))=xf(y)+2$. Claim 1. $f$ is injective Proof. If $f(a)=f(b)$ for some $a,b \in \mathbb{R^+}$, then from $P(a,b)$ and $P(b,a)$ we have $$P(a,b) \ : \ f(ab+f(a))=af(b)+2$$$$P(b,a) \ : \ f(ba+f(b))=bf(a)+2$$Notice that the $LHS$ of both equations above are equal, it follows that $$af(b)+2=bf(a)+2 \Rightarrow af(b)=bf(a) \Rightarrow a=b$$As desired. Claim 2. $f(1) \neq 1$ Proof. Assume $f(1)=1$, then from $P(1,1),P(1,2),P(1,3),P(1,4)$ successively we get $f(2)=3,f(3)=5,f(4)=7,f(5)=9$ respectively. But from $P(2,1)$ we have $f(5)=4$, a contradiction. As desired. From $P\left(x ,\frac{f(1)}{x} \right)$ and $P(1,f(x))$, we have $$P\left(x ,\frac{f(1)}{x} \right) \ : \ f(f(1)+f(x))=xf\left( \frac{f(1)}{x} \right)+2$$$$P(1,f(x)) : f(f(x)+f(1))=f(f(x))+2$$Since the $LHS$ of both equations above are equal, the $RHS$ must also be equal. That is, $$xf\left( \frac{f(1)}{x} \right)+2=f(f(x))+2 \Rightarrow xf\left( \frac{f(1)}{x} \right)=f(f(x)) \ (*)$$ From $P(f(x),1)$ and $P(f(1),x)$, we have $$P(f(x),1) \ : \ f(f(x)+f(f(x)))=f(x)f(1)+2$$$$P(f(1),x) \ : \ f(xf(1)+f(f(1)))=f(1)f(x)+2$$Notice that the $RHS$ of both equations above are equal, with injectivity it follows that $$f(x)+f(f(x))=xf(1)+f(f(1))$$Substituting $(*)$ to the equation above gives $$f(x)+xf\left( \frac{f(1)}{x} \right)=xf(1)+f(f(1)) \ (1)$$Changing $x = \frac{f(1)}{x}$ to $(1)$ gives $$f\left( \frac{f(1)}{x} \right)+\frac{f(1)}{x}f(x)=\frac{f(1)^2}{x}+f(f(1))$$Multiplying both sides by $x$ in the equation above gives $$xf\left( \frac{f(1)}{x} \right)+f(1)f(x)=f(1)^2+xf(f(1)) \ (2)$$Subtracting $(2)$ with $(1)$ gives $$f(x)(f(1)-1)=x(f(f(1))-f(1))+f(1)^2-f(f(1)) \ (3)$$ Since $f(1)-1 \neq 0$, $(3)$ implies that $f(x)=ax+b \ \forall \ x \in \mathbb{R^+}$ for some constants $a,b \in \mathbb{R}$. Substituting this to $P(x,y)$ and pairing the coefficients gives $a=b=1$, hence $f(x)=x+1 \ \forall \ x \in \mathbb{R^+}$. Which clearly satisfies. Conclusion. The only function that satisfies is $f(x)=x+1 \ \forall \ x \in \mathbb{R^+}$.

JingheZhang wrote: f(x)=x+1 I fake proved it by comparing both sides' degrees, and claiming the degree of f is one. (I believe this method works if the question is asking a polynomial) Bruh this is what I did this year for J5 :sadge:

Solution removed because $f$ cannot be assumed to be differentiable. I agree with previous solutions that $f(x) = x+1$, however.

bCarbon wrote: New solution using derivatives. Since the equality is true for all $x$ and $y$ we can take partial derivatives with respect to either $x$ or $y$. First, differentiating with respect to $y$ gives $$x\ f'(xy + f(x)) = x\ f'(y)$$Dividing through by $x$, $$f'(xy + f(x)) = f'(y)$$In the limit as $x \to 0$ and any $y > 0$ this implies $$f'(\alpha) = f'(y)$$where $\alpha = \lim_{x \to 0^+} f(x)$ so $f'$ is constant and f is linear, i.e. of the form $f(x) = ax + b$. Substituting, we have $$a(xy + ax + b) + b = axy + bx + 2$$$$\iff$$$$a^2x + ab + b = bx + 2$$$$\iff$$$$b = a^2\ and\ (a-1)(a^2 + 2a + 2) = 0$$for which the only solution is $(a, b) = (1, 1)$. Thus $f(x) = x + 1$. wait how do you know it's continuous?

On a second readthrough, I find ABCDE's solution straightforward. My mistake was assuming f is differentiable itself

The answer is $f\equiv x+1.$ We can easily prove that it's the only linear solution. Conversly, we prove that $f$ is linear. claim:$f$ is injective. Proof. Suppose that $f(a)=f(b)=c,$ then $P(a,b)$ and $P(b,a)$ combined together give : \[ac+2= bc+2 \iff a =b.\] claim: $f$ is linear . Proof. $$P(f(x),y)) : f(f(x)y+ff(x)) =f(x)f(y)+2 .$$ Similarily, $f(f(y)x+ff(y))=f(x)f(y)+2.$ Since, $f$ is injective, we have $$f(x)y+f(f(x)) = xf(y)+f(f(y)) \iff f(f(x))- f(f(y)) = xf(y)-yf(x)\\[0.1cm]$$ item $f(f(1))-f(f(x))=f(x)-xf(1)$ $f(f(x)) -f(f(2)) = xf(2)-2f(x)$ Summing these equalities, we get $f(f(1))-f(f(2)) = x(f(2)-f(1)) - f(x).$ Which is equivalent to $f(x) = x(f(2)-f(1)) + [f(f(2))-f(f(1))] = ax +b$

Thanks @below for pointing out the mistake

ihatemath123 wrote: Subtracting $P(x,y)$ from $P(x,y+f(1))$ gives us \[\frac{f(Y + xf(1)) - f(Y)}{xf(1)} = \frac{2}{f(1)},\]where $Y = xy+f(x)$ can be any real greater than $f(x)$. Letting $C$ be the lowerbound of $f$, it follows that $f$ is differentiable (with a value of $\tfrac{2}{f(1)}$) at all inputs strictly greater than $C$. (This is where we use that $f$ is strictly increasing.) Actually wait, here's a fix to the left/right differentiability issues. Let $t > C$. By our equation, we see that for sufficiently small $x$, $f(x) < t$. Therefore $f(mx/n + C + \text{ small}) = \frac{2}{f(1)}(mx/n) + f(C + \text{ small})$ for integers $m, n$. Now use density of $mx/n$ to argue that the lemma is proved.

A different finish after proving $yf(x)+f^2(x)=xf(y)+f^2(y) \ \ (\star)$, without the need to plug in $f(x)=ax+b$ and solve for $a,b$ is as follows: From the original equation plugging in $1$'s we see there exists a $t$ such that $f(t)=t+1$. Then plugging $y:=f(y)$ into $(\star)$ gives \[ f(y)f(x) = xf^2(y)+f^3(y) - f^2(x) = yf^2(x)+f^3(x) - f^2(y), \]and hence \[ (x+1)f^2(y) + f^3(y) = (y+1)f^2(x) + f^3(x). \]Plugging $(x,y):=(f(x),f(y))$ into $(\star)$ gives \[ f(x)f^2(y)+f^3(y) = f(y)f^2(x)+f^3(x). \]Subtracting the last two equations gives \[ (f(x)-x-1)f^2(y) = (f(y)-y-1)f^2(x). \]Then $y:=t$ in the above gives $f(x)\equiv x+1$ which works.

\[f(xy+f(x))=xf(y)+2\]Answer is $f(x)=x+1$ which works. Claim: $f$ is injective. Proof: Suppose that $f(a)=f(b)$ for a pair $a\neq b$. \[P(a,b),P(b,a): \ \ af(b)+2=f(ab+f(a))=f(ab+f(b))=bf(a)+2\implies af(b)=bf(a)\]However this contradicts with $a\neq b$.$\square$ Plug $\frac{x}{f(y)},y$ to get \[f(\frac{xy}{f(y)}+f(\frac{x}{f(y)}))=x+2=f(\frac{xz}{f(z)}+f(\frac{x}{f(z)}))\]By injectivity, we conclude that \[Q(x,y,z): \ \ \frac{xy}{f(y)}+f(\frac{x}{f(y)})=\frac{xz}{f(z)}+f(\frac{x}{f(z)})\]Taking $Q(xf(1),y,1)$ implies \[\frac{xyf(1)}{f(y)}+f(\frac{xf(1)}{f(y)})=x+f(x)\]Set $f(1)=c$. Choose $x=f(y)$ to verify $yc+f(c)=f(y)+f(f(y))\iff f(f(y))=yc-f(y)+f(c)$. \[Q(f(x)f(y),y,x): \ \ yf(x)+f(f(x))=xf(y)+f(f(y))\]\[yf(x)+xc-f(x)+f(c)=xf(y)+yc-f(y)+f(c)\]Thus, $f(x)(y-1)-f(y)(x-1)=cy-cx$. Subsituting $y=2$ gives $f(x)=x(f(2)-c)-(f(2)-2c)$ hence $f$ is linear. Let $f(x)=ax+b$. \[axy+a^2x+ab+b=axy+xb+2\iff (a^2-b)x=2-ab-b\]So $a^2=b$ and $(a+1)a^2=ab+b=2$ subsequently, $(a-1)(a^2+2a+2)=a^3+a^2-2=0$ which yields $a=b=1$ so $f(x)=x+1$ as desired.$\blacksquare$

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cde kills? Let $P(x,y)$ be the assertion. Note that $P(1,y)$ yields $f(y+c)=f(y)+d$ for constants $c,d$. Now, comparing $P(x,y+c)$ with $P(x,y)$, we have \[f(xy+cx+f(x))=xf(y)+dx+2=f(xy+f(x))+dx\]Now, say we want $z=xy+f(x)$, then rearranging, $y=\frac{z-f(x)}{x}$. Thus, for all $z>f(x)$, we have $f(z+cx)=f(z)+dx$. But from $f(y+c)=f(y)+d$ for all reals, say we want $z'=z-nc$ for $z' \leq f(x)$, where $n\in \mathbb{Z}$ and $z$ is some valid real (sufficiently large). Then, \[f(z'+cx)=f(z-nc+cx)=f(z+cx)-nd=f(z)+dx-nd=f(z-nc)+dx=f(z')+dx\]Hence we actually have $f(z+cx)=f(z)+dx$ for all $x,z$ real, without any bounds. Note $f(1+cx)=f(1)+dx$, take $x\to \frac{x-1}{c}$. This clearly implies $f$ is linear, from which we get $f(x)=x+1$. did i fakesolve oops edit: i didnt fakesolve