hint:f(0)=1,f(f(x))=f(x)+1,f(xf(y))+f(y)=f(x)f(y)+1.so for rational numbers f(x)=x+1 and since f is Continuous f(x)=x+1.

$ f(0) + f(f(0)) = f(0)^2 + 2 $ (1) $ f(0) + f(f(1)) = f(0)f(1) + 2 $ (2) $ 2f(f(0)) = f(0)f(1) + 2 $ (3) $ 2f(f(1)) = f(1)^2 + 2 $ (4) (1)+(4)-(2)-(3)= $ f(f(1))-f(f(0))= (f(0)-f(1))^2 $ (2)-(1)= $ f(f(1))-f(f(0)) = f(0)(f(1)-f(0)) $ so $ f(0)(f(1)-f(0))=(f(1)-f(0))^2$ so we have should $ f(0)=f(1)$ or $ f(1)=2f(0) $ i ) $ f(0)=f(1) $ so $ f(f(0))=f(f(1)) $ and $ f(f(1))=f(f(0))=f(1)=f(0) = x $ with (1) we know $ 2x = x^2 + 2 $ and this means that $ (x-1)^2 + 1 = 0 $ but we konw there isn't real number x that satisfies the equation. so we should have $ 2f(0)=f(1)$ $ f(f(1))-f(f(0)) = f(0)^2 = f(f(1)) - f(0)^2 - 1 $ so $ f(f(1)) = 2f(0)^2 + 1 $ with (2) : $ f(0) + 2f(0)^2 + 1 = 2f(0)^2 + 2 $ so $ f(0)=1$ and $ f(1)=2 $ x=1 so : $ f(f(y)) = f(y)+1 $ and for all natural number $x$ we have : $ f(x) = x+1$ suppose that $x$ in rational number so we can show it as $ x= \frac{p}{q} $ we puy y = q-1 so $ f(p) + f(q) = f(x)q + 2 $ so $ p+1+q+1= f(x)q + 2$ so $p+q=qf(x)$ and we have : $ f(\frac{p}{q})=\frac{p}{q}+1$ so we have for all rational number x : $ f(x) = x+1 $ Because $f$ is Continuous so we have for all real number $x$ : $f(x) = x+1 $

Let $P(x,y)$ the assertion that $f(xf(y))+f(f(y)) = f(x)f(y)+2$, Let $f(0)=c$. Then from $P(0,y)$ we have $f(f(y))=cf(y)+2-c$ It is easy to show that the values take up to arbitrarily large, hence from continuity, function is surjective from $c$ to infinity. If $c=0$, $f(x)=2$ for all $x$ which fails. Hence $c>0$. Now we check $P(x,y)$ where we fix $x$ that is large and take any $y$ even larger. As $c>0$, by expanding both sides with $f(f(y))=cf(y)+2-c$, we have $c(x(cy+2-c))+2-c+c(cy+2-c)+2-c=(cx+2-c)(cy+2-c)\Leftrightarrow c^2y-c^2=2cy-c^2y-4c+c^2+2$ and hence $c=1$. Now for any fixed $x$, we take large enough $y$, and get $xy+x+1+y+2=f(x)(y+1)+2$ and hence $f(x)=x+1$ for all $x$.

Let $P(x,y)$ be the statement that $f(xf(y))+f(f(y)) = f(x)f(y)+2$. We shall prove in a sequence of claims that the only solution is $f(x)=x+1$. Claim 1: $f(0)=1,f(1)=2$. Proof. Clearly no constant function works. Note that $P(0,x)\implies f(f(x))=f(x)f(0)-f(0)+2$. Again, $P(1,x)\implies f(f(x))=\tfrac{f(1)}{2} f(x)+1$. Equating these expressions for $f(f(x))$, we conclude that $f(x)\left(\frac{f(1)}{2}-f(0)\right)=1-f(0)$. Since we are looking for a non constant function, we must have $\frac{f(1)}{2}-f(0)=1-f(0)=0$, so that $f(0)=1,f(1)=2$, as desired. $\square$ Claim 2: $f(n)=n+1\quad\forall n\in\mathbb{N}_0$. Proof. We shall use induction. The base case is already proved, so suppose that the statement is true for some $n$, i.e., $f(n)=n+1$. Then $P(1,n)$ gives $f(f(n))=f(n)+1\implies f(n+1)=n+2$ , proving our claim. $\square$ Claim 3: $f(x)=x+1\forall x\ge 1$ Proof. $P(1,x)$ gives $f(f(x))=f(x)+1$, so $f(x)=x+1$ holds for all $x$ in the range of $f$. We have already proved that all natural numbers are in its range, so invoking continuity, we have by IVT that all numbers in $[1,+\infty)$ are in its range. This establishes our claim. $\square$ Claim 4: The "weird" finishing. Proof. We shall say an interval is "covered" to mean that we have proved $f(x)=x+1$ for all $x$ in that interval. If we take $x$ as a number in the now-covered interval, which is at this moment $[1,+\infty)$, then $P\left(\frac{1}{x+1},x\right)$ gives $$f(1)+x+2=f\left(\frac{1}{x+1}\right)(x+1)+2\implies f\left(\frac{1}{x+1}\right)=\frac{1}{x+1}+1 \quad (1)$$Since $\frac{1}{x+1}$ ranges through $\left(0,\frac12\right]$, we have $f(x)=x+1$ for all $x\in \left(0,\frac 12\right]$. Now we can use these values of $x$ in $ \left[0,\tfrac 12\right]$ in $(1)$ to settle the interval $\left[\tfrac 23,1\right]$. This interval, in turn, "covers" the interval $\left[\tfrac 12,\tfrac 35\right]$. This again covers $\left[\tfrac 58,\tfrac 23\right]$. The numerators and denominators of the bounds of these "covered" intervals are in fact Fibonacci numbers, which is easily proved by induction. The "inner" bounds of these intervals converge; in fact they approach the golden ratio conjugate $\phi=\frac{\sqrt{5}-1}{2}$. So every number, however close to $\phi$, is eventually "covered". By continuity, $\phi$ itself is "covered" as well, so we are done. $\blacksquare$ Note: The proof of Claim 4 is quite non-rigorous. However, it can easily formalized, and we leave the task to the reader .

Let $P(x,y)$ be the FE. $P(1,x)\implies f(f(x))=\dfrac{f(1)}{2}f(x)+1$ $P(f(x),y)\implies f(f(x)f(y))+f(f(y))=f(f(x))f(y)+2$. Combining this with $P(f(y),x)$ yields $f(f(x))-f(f(y))f(x)=f(f(y))-f(f(x))f(y)\implies \dfrac{f(f(x))}{f(x)+1}=\dfrac{f(f(y))}{f(y)+1}=k\implies f(f(x))=k(f(x)+1)$ Combining with the first observation gives $\dfrac{f(1)}{2}f(x)+1=k(f(x)+1)\implies \left(\dfrac{f(1)}{2}-k\right)f(x)=k-1$. If $k\ne \dfrac{f(1)}{2}$, then $f$ is a constant function $f(x)=c$. But $c+c=c^2+2$ yields no positive real solution for $c$ so $k=\dfrac{f(1)}{2}\implies \dfrac{f(1)}{2}-1=0\implies f(1)=2$. $P(0,1)\implies f(0)+f(f(1))=f(0)f(1)+2\implies f(0)=f(f(1))-2=f(1)-1=1$. Then, a simple induction on the first observation $f(f(n))=f(n)+1$ gives $f(n)=n+1\, \forall n\in\mathbb{Z}_{\ge 0}$. $P\left(\dfrac{m}{n+1}, n\right)$ where $m,n\in\mathbb{Z}_{\ge 0}$ yields $f\left(\dfrac{m}{n+1}\right)=\dfrac{m}{n+1}+1$ so $f(q)=q+1\, \forall q\in\mathbb{Q}_{\ge 0}$. But $f$ is continuous so $\boxed{f(x)=x+1}$ for all non-negative real $x$.

$P(0,y): f(f(y))=cf(y)-c+2$ where $c:=f(0)$ $P(1,y):2f(f(y))=f(y)f(1)+2$ $\implies$ $f(y)=\frac{2c-2}{2c-f(1)}$ and so $f$ is either constant which would imply it's equal to the solution $(c-1)^2+1=0$ on the whole domain a clear contradiction.And hence $c=f(0)=1$ and $f(1)=2c=2$.Switching this into the first line $f(f(y))=f(y)+1$(1).Now plug $y:=f(y)$ we get : $$f(f(f(y)))=f(f(y))+1=f(y)+2=f(f(y)+1)$$And by repeatedly plugging $y:=f(y)$ we get thru the induction that $f(f(y)+n)=f(y)+n+1$ $\forall n\in \mathbb{N}$.Now the previous implies $f(n)=n+1$ for all naturals.Let $x:=\frac{p}{q}$ be an arbitrary rational than $P(x,q-1):f(p)+q+1=f(x)q+2$ which implies $f(x)=x+1$ $\forall x\in \mathbb{Q}$.Now let $a_1,a_2,...$ be a sequence of rationals converging to a real $r$.We have: $$\lim_{n\rightarrow \infty} f(a_n)=f(\lim_{n\rightarrow \infty} a_n)=f(r)=r+1$$and hence $f(x)=x+1$ is the only solution.

Edited.............

another way is to find an infinite sequence of $f$ which is boundless . and so the function is surjective since its continuous . and the rest is trivial

$P(1,x)\rightarrow ff(x)=\frac{f(1)}{2}f(x)+1 P(f(x),y)\rightarrow f(f(x)f(y))+\frac{f(1)}{2}f(y)=f(1)f(x)f(y)+f(y)+1....(1).$ switch between$ x,y$ in $(1) \rightarrow (1-\frac{f(1)}{2})f(x)=(1-\frac{f(1)}{2})f(y).$ If $f(1)\neq 2 \rightarrow f(x)=f(y)$ so$ f$ is constant and there is no solution in this case. So $f(1)=2$. Easy induction gives $f(n)=n+1 \forall n\in N.$ $P(\frac{1}{n},n-1)\rightarrow f(\frac{1}{n})=\frac{1}{n}+1 \forall n\in N.$ $\P(\frac{p}{q},q-1) \rightarrow f(\frac{p}{q})=\frac{p}{q}+1$. So $f(x)=x+1$for all rational numbers x and since f is continous so $f(x)=x+1$.

you can fully avoid finding $f$ on rationals. You just need it on the integers. Taking $x=1$ and $x=0$ gives $$2f(f(y))=f(1)f(y)+2\text{ and } f(0)+f(f(y))=f(0)f(y)+2.$$Solving for $f(f(y))$, we see that $$f(1)f(y)+2=2f(0)f(y)+4-2f(0).$$If $f(1)\neq 2f(0)$, then $f$ must be constant. However, this gives a complex solution for $f$, so we must have $f(1)=2f(0)$. Additionally, we have $4-2f(0)=2\implies f(0)=1$. Hence, $f(f(y))=f(y)+1$. But since $f(1)=2f(0)=2$, we may use induction to show that $f(n)=n+1$ for all integers $n$. Now plugging $(x,y)=\left( \frac{t}{n+1},n\right)$ gives $$f(t)+f(f(n))=f\left(\frac{t}{n+1}\right)f(n)+2\implies f(t)-t=(n+1)\left(f\left(\frac{t}{n+1}\right)-\frac{t}{n+1}\right)-n.$$Letting $n\to\infty$, we see that $f(t)-t=(n+1)(f(0)-0)-n=1$, or that $\boxed{f(t)=t+1}$ for all $t\ge 0$.

Ans:$f(x)=x+1$ for all $x\in \mathbb{R}^{\geq 0}$. Let $P(x,y)$ be the given assertion, we have \[P(1,x)\implies f(f(x))=\frac{f(1)}{2}f(x)+1 \ \ \ (1).\]Comparing $P(f(x),y)$ and $P(f(y),x)$ yields \[f(f(y))-f(f(x))=f(f(x))f(y)-f(f(y))f(x) \implies \frac{f(f(x))}{1+f(x)}=\frac{f(f(y))}{1+f(y)}\implies f(f(x))=C(1+f(x))\]where $C\in\mathbb{R}^{\geq 0}$. Comparing this with $(1)$, we have $(2C-1)f(x)=2-2C$, if $C\ne \frac{1}{2}$, then we have $f$ is a constant function, $f(x)=k$, $k\in \mathbb{R}^{\geq 0}$ but plugging into the original FE yields $k^2-2k+2=0$ which has no real solution. So, when $C=\frac{1}{2}$, we have $f(1)=2$, also \[P(0,x) \implies f(0)+f(f(x))=f(0)+f(x)+1=f(0)f(x)+2\implies (1-f(0))f(y)=1-f(0)\]since $f$ is not a constant function, $f(0)=1$. Now, by induction on $f(f(x))=f(x)+1$ with $f(0)=1, f(1)=2$, we can prove that $f(x)=x+1$ for non-negative integers. For positive rationals, use \[P(\frac{1}{x+1},x)\implies f(\frac{1}{x+1})=\frac{1}{x+1}+1\]where $x$ is a non-negative integer. Then, let $p, q\in \mathbb{Z}^{+}$ and \[P(\frac{1}{q},p-1)\implies f(\frac{p}{q})=\frac{p}{q}+1.\]Finally, since the function is continuous, we have $f(x)=x+1$, $\forall x\in \mathbb{R}^{\geq 0}$. $\blacksquare$

Let $P(x,y)$ denote the assertion $f(xf(y))+f(f(y))=f(x)f(y)+2$ $P(0,y): f(0)+f(f(y))=f(0)f(y)+2$ Which implies $f(f(y))=f(0)(f(y)-1)+2$ $P(1,y): 2f(f(y))=f(1)f(y)+2$ Equating the two, we get $f(0)(f(y)-1)+2=\frac{f(1)}{2}f(y)$ Which reduces to $f(0)f(y)-f(0)+1=\frac{f(1)}{2}f(y)$ Claim: There must exist at least two unique values of $f$ Proof: FTSOC, suppose otherwise, then Let $f(x)=a$ for all $x$ Then $2a=a^2+2$ $0=(a-1)^2+1 \geq 1$ contradiction This implies that $f(0)=1$, $f(1)=2f(0)=2$ Therefore, $P(1,y)$ reduces to $f(f(y))=f(y)+1$ This implies that for all non negative integers $x$, we get $f(x)=x+1$ Claim: $f(x)=x+1$ for all non negative rational numbers $x$ Proof: Note that our original equation reduces to $f(xf(y))+f(y)=f(x)f(y)+1$ For any arbitrary rational number $x$, choose a value $y$ such that $x(y+1)$ is an integer Therefore, it would reduce to $x(y+1)+(y+1)=f(x)(y+1)+1$ which implies $f(x)=x+1$ Now that we have $f(x)=x+1$ for all rational $x$, and that $f$ is a continuous function, by "sandwiching" $f(x)$ where $x$ is an arbitrary real number with a rational number that is infinitesimally smaller and infinitesimally larger than $x$, we get that $f(x)=x+1$ for all values of $x$ Checking, it works Q.E.D

Can some you explain how : $P\left(\dfrac{m}{n+1}, n\right)$ where $m,n\in\mathbb{Z}_{\ge 0}$ yields $f\left(\dfrac{m}{n+1}\right)=\dfrac{m}{n+1}+1$ so $f(q)=q+1\, \forall q\in\mathbb{Q}_{\ge 0}$.

tony88 wrote: Can some you explain how : $P\left(\dfrac{m}{n+1}, n\right)$ where $m,n\in\mathbb{Z}_{\ge 0}$ yields $f\left(\dfrac{m}{n+1}\right)=\dfrac{m}{n+1}+1$ so $f(q)=q+1\, \forall q\in\mathbb{Q}_{\ge 0}$. What part is confusing? It was already shown that $f(n) = n+1$ for all non-negative integers. Any non-negative rational number can be written as $\frac{m}{n+1}$ where $m,n$ are non-negative integers. The rest explains itself.

Seems like I overcomplicated \[f(xf(y))+f(f(y)) = f(x)f(y)+2\]Note that $c^2-2c+2=(c+1)^2+1\neq 0$ hence $f$ is nonconstant. Claim: $f(0)=1$. Proof: First, if $f(0)=0$, then $P(0,y)$ gives $f(f(y))=2$ hence $f(xf(y))=f(x)f(y)$. But plugging $1,y$ implies $2=f(f(y))=f(1)f(y)$ but $f$ is nonconstant so this is impossible. Let $f(0)=c\neq 0$. $P(0,y)$ yields $c+f(f(y))=cf(y)+2$ $\iff f(f(x))=cf(x)-c+2$. Now plug $1,y$ to get $2cf(y)-2c+4=2f(f(y))=f(1)f(y)+2$ or $(2c-f(1))f(y)=2c-2$ Since $f$ is nonconstant, we can see that $f(1)=2c$. Plugging $P(x,0)$ implies \[f(xc)+f(c)=f(x)c+2\iff f(cx)=cf(x)+2-f(c)\]After choosing $x=1$, we conclude that $2f(c)=2c^2+2$ which is equavilent to $f(c)=c^2+1$. By using this, \[f(cx)=cf(x)+1-c^2\]Compare $P(x,y)$ with $P(cx,y)$ to get \[cf(xf(y))+1-c^2+f(f(y))=f(cxf(y))+f(f(y))=f(cx)f(y)+2=cf(x)f(y)+f(y)-c^2f(y)+2\]By taking $y=1$ and using $f(1)=2c$, \[f(2c)-c^2+1+c^2f(2x)+c-c^3=cf(2cx)+1-c^2+f(2c)=2c^2f(x)+2c-2c^3+2\]Thus, $f(2c)+c^2f(2x)=2c^2f(x)+c-c^3+1+c^2$ or $f(2x)-2f(x)=-\frac{f(2c)}{c^2}+\frac{1}{c}-c+\frac{1}{c^2}+1$ If $x=0$, then $\frac{1}{c}+\frac{1}{c^2}+1=\frac{f(2c)}{c^2}$. Hence $f(2x)=2f(x)-c$. Take $x=c$ to see $f(2c)=2f(c)-c=2c^2-c+2$. \[2c^2-c+2=f(2c)=c^2+c+1\implies c^2-2c+1=0\iff c=1\]Which proves $f(0)=1$.$\square$ $P(0,x)$ yields $f(f(x))=f(x)+1$. Also note that if $f(a)=0$, then $P(x,a)$ would give a contradiction. Now we get that $2f(x)=f(2x)+1$. After plugging $P(\frac{f(x)}{f(y)},y)$, by symmetry \[f(\frac{f(y)}{f(x)})f(x)=f(f(x))+f(f(y))=f(\frac{f(x)}{f(y)})f(y)\]Choose $y=0$ which implies $f(\frac{1}{f(x)})f(x)=f(f(x))=f(x)+1$ so $f(\frac{1}{f(x)})=1+\frac{1}{f(x)}$. Plug $P(\frac{1}{f(x)},y)$ to see that \[f(\frac{f(y)}{f(x)})+f(y)=\frac{f(y)}{f(x)}+f(y)+1\iff f(\frac{f(y)}{f(x)})=\frac{f(y)}{f(x)}+1\]Since $2f(x)=f(2x)+1$ and $f(1)=2$, we can observe that $f(2^k)=2^k+1$ by induction with base case $k=0$. So $f$ is surjective over $[1,\infty)$. This implies $f(y)/f(x)$ takes all positive real values hence $f(x)=x+1$ for all positive reals as desired.$\blacksquare$