Solved with CyclicISLscelesTrapezoid The only solutions are $f(x)=x+1$ and constant functions, which clearly work. From now on assume $f$ is not of this form. Let $P(x,y)$ denote the assertion in the problem statement. If $f$ is periodic with period $d$, $P(x,y+d) - P(x,y)\implies f$ is constant $P(x,0)$ and $P(x,f(0)) - P(x,1)\implies f(x+f(0)-1)=f(x)\implies f(0)=1$ $P(1,y)\implies f(f(y)) = f(y+1)$ $P(x,y)-P(y,x)\implies f(xf(y)) - f(yf(x)) = f(x) - f(y)$ $P(f(x),y) - P(f(y),x)\implies f(x+f(y)) = f(y+f(x))$, call this $Q(x,y)$ $Q(f(x),y) - Q(y,f(x))\implies f(f(x+1)+y) = f(f(y+1)+x)\implies f(f(x+1)+y) = f(f(x)+y+1)\implies f(x+1)=f(x)+1$ Note that $f(n)=n+1$ for all $n\in\mathbb{Z}$ $P(-x,-1)\implies f(x) + f(-1-x) = 1$ $Q(f(x), -1-x)\implies f(f(x)-x)=2$ Now suppose $f(x)\neq x+1$ for some $x$. Then $c=f(x)-x-2$ is not $-1$ and $f(c)=0$. $P(x-c,c)\implies f(x)=1-f(cx-c^2)\implies f(x)=f(-1+c^2-cx)$ Now $f(c^2-cx) = f(x+1) = f(-1+c^2-cx-c) \implies c\in\{0,-1\}\implies c=-1$, a contradiction and we’re done.

We claim the solutions are $\boxed{f(x)=C}$ and $\boxed{f(x)=x+1}$, which both work. Now, assume $f$ is nonconstant. If $f(y)=y$, then $y=f(x+y)$, so $f$ is constant, contradiction. Therefore, $f(y)\neq y$, so $f(f(y))\neq f(y)$. Substituting $x=1$ into the original equation gives $f(f(y))=f(y+1)$, so $f(y+1)\neq f(y)$. Therefore, if $f(a)=f(b)$, then $a\neq b+1$. Substituting $y=0$ gives $f(xf(0))=f(x)$, so $xf(0)\neq x+1$ implies $f(0)=1$. We have \begin{align*} f(f(x)f(y))+f(y)&=f(f(x)+y)+f(f(x)y)\\ &=f(yf(f(x)))+f(f(x))\\ &=f(x+1+y)+f((x+1)y).\end{align*} Swapping $x$ and $y$ gives $f(xy+y)-f(y)=f(xy+x)-f(x)$. Substituting $y=-1$ and $y=1$ gives $$f(-x-1)+f(x)=f(-1)+1$$and $$f(x+1)+f(x)=f(2x)+f(1).$$Therefore, $y=-1$ in the original equation gives $$f(xf(-1))=f(x-1)+f(-x)-f(-1)=1$$so $f(-1)=0$, and $y=1$ in the original equation gives $$f(xf(1))=f(x+1)+f(x)-f(1)=f(2x)$$so $xf(1)\neq2x+1$ implies $f(1)=2$. Therefore, $f(-2)+f(1)=1$ implies $f(-2)=-1$. Now, $$f(f(x)f(y))=f(f(x)+y)+f(f(x)y)-f(y)=f(f(x)+y)-f(x)-f(y)+f(x+y)+f(xy),$$so swapping $x$ and $y$ gives $$f(f(x)+y)=f(f(y)+x).$$Therefore, $y=-f(x)$ gives $f(f(-f(x))+x)=1$. Substituting $x=-1$ in the original equation gives $f(-f(y))=f(y-1)+f(-y)-f(y)=1-f(y)$, so $$f(1+x-f(x))=1.$$Now, $x=-2$ implies $f(y-1)=f(f(y)-2)$ and if $y$ is replaced with $1+x-f(x)$, then $$f(x-f(x))=0.$$Let $a=x-f(x)$. Substituting $y=a$ and $y=-1$ into the original equation gives $$f(x+a)+f(xa)=1$$and $$f(x-1)+f(-x)=1.$$This means $f(x)=1-f(-x-1)$, so $f(x+1)=f(-ax-1)$. However, $x+1\neq -ax$ implies $a=-1$, so $x-f(x)=-1$, which implies $f(x)=x+1$.

The only solutions are $\boxed{f\equiv c}$ for some real constant $c$ and $\boxed{f(x) = x + 1}$. These work. Now we prove they are the only solutions. Let $P(x,y)$ denote the given assertion. Clearly all constant functions work, so we can assume $f$ is not constant. Claim: If $f(a\cdot x) = f(x)$ for all $x\in \mathbb R$, then $a = 1$. Proof: Suppose there was some $a\ne 1$ with $f(a\cdot x) = f(x) \forall x\in \mathbb R$. Now $P(x, ay) - P(x,y)$ gives that $f(x + y) = f(x + ay)$, so $f(x) = f( x + y(a-1))$ for each $x,y$. Since $a\ne 1$, $y(a-1)$ can take any real, so $f$ is periodic with every period, so it is constant, absurd. $\square$ $P(x,0): f(xf(0)) = f(x)$, so $f(0) = 1$. $P(1,y): f(f(y)) = f(y + 1)$. Using this, $P(x, y + 1) - P(x,f(y)): f(x + y + 1) - f(x + f(y)) + f(x(y+1)) - f(xf(y)) = 0$. Now by swapping variables we also have $f(x + y + 1) - f(y + f(x)) + f(y(x + 1)) - f(y f(x)) = 0$. $P(x+1, y) - P(f(x), y): f((x+1) f(y)) - f(f(x) f(y)) = f(x + y + 1) - f(y + f(x)) + f(y(x + 1)) - f(yf(x)) = 0$, so $f((x+1) f(y)) = f(f(x)f(y))$. However, this also implies that $f((y+1) f(x)) = f(f(y) f(x))$, so $f((x+1) f(y)) = f( (y + 1) f(x))$. Plugging $x = -1$ here gives $f(0) = f((y+1) f(-1))$ for each real $y$. If $f(-1)\ne 0$, then $(y+1) f(-1)$ can take any real number, so $f$ is constant, absurd. Therefore $f(-1) = 0$. $P(x,-1): f(x-1) + f(-x) = 1$. $P(-1,x): f(-f(x)) + f(x) = f(x-1) + f(-x) = 1$, so $f(-f(x)) = 1 - f(x)$. Since $f(x -1) + f(-x) = 1$, $f(x) + f(-(x+1)) = 1$, so $f(-f(x)) = f(-(x+1))$. Now plugging $y = -2$ into $f((x+1) f(y)) = f((y+1) f(x))$, we see that $f(-f(x)) = f((x+1) f(-2))$, so $f(-(x+1)) = f( f(-2) \cdot (x+1))$. Therefore, $f(x) = f( -f(-2) \cdot x)$ for all $x\in \mathbb R$, so by our earlier claim, $-f(-2) = 1$, so $f(-2) = -1$. $P(-1,-1)$ also gives now that $f(1) = 2$. $P(x,-2): f(-x) - 1 = f(x - 2) + f(-2x)$. Now, $f(-x) = 1 - f(x - 1)$, so $-f(x-1) = f(x-2) + f(-2x)$, which implies $f(x - 1) + f(x - 2) + f(-2x) = 0$. Since $f(-2x) = 1 - f(2x - 1)$ (from $P(2x, -1)$), we have $f(x - 1) + f(x - 2) = f(2x - 1) - 1$ for each $x$. Setting $x \to x + 2$ here gives $f(x) + f(x + 1) = f(2x + 3) - 1$. $P(x,1): f(2x) + 2 = f(x) + f(x + 1)$. So $f(2x) + 2 = f(2x + 3) - 1\implies f(2x) + 3 = f(2x + 3)$. Since $2x$ can take any real value, we have $f(x) + 3 = f(x + 3)$ for all reals $x$. Claim: $f$ is injective at $0$. Proof: Suppose $f(c) = 0$ for some $c\ne -1$. $P(x, c): f(x + c) + f(xc) = 1$. $P(x+3, c): f(x + c) + 3 + f(xc + 3c) = 1$. Subtracting the two equations gives that $f(xc + 3c) = f(xc) - 3$. Since $xc$ is surjective over reals, $f(x + 3c) = f(x) - 3 = f(x - 3)$, $f(x) = f(x + 3c + 3)$, which means $f$ is periodic with period equal to some $d\ne 0$. $P(k, y + d) - P(k,y): 0 = f(ky + kd) - f(ky)$, so $f(x) = f(x + kd)$ for each $x \ne 0$. But then $kd$ can take any real value not equal to $0$, so $f(x) = f(y)$ for any real $y$, so $f$ is constant, absurd. $\square$ Now using $f((x+1) f(y)) = f((y+1) f(x))$, if we plug in $y = \frac{-1}{f(x)} - 1$, we see $(y+1) f(x) = -1$, so $f((y+1) f(x)) = 0$. However, since $f$ is injective at $0$, this means that $(x+1) f(y) = -1$, so $f(y) = -\frac{1}{x+1}$. We already know that $f(-f(y)) = -f(y) + 1$ from earlier, so $f \left( \frac{1}{x+1} \right) = \frac{1}{x+1} + 1$. Since $\frac{1}{x+1}$ can take any real except zero, we have $f(x) = x + 1 \forall x\ne 0$, but it also holds for $x = 0$, therefore we are done.

The answer is constant solutions and $f(x)=x+1$, which all work. Assume $f$ is nonconstant going forward. Let $P(x,y)$ be the assertion. $P(x,0)$ means that $f(xf(0))=f(x)$ and $P(1,y)$ means that $f(f(y))=f(y+1)$. Denote this latter relation by $\ast$. $P(x,y)$ and $P(y,x)$ mean that $f(xf(y))+f(y)=f(yf(x))+f(x)$. $P(f(x),y)$ means that $f(f(x)f(y))+f(y)=f(f(x)+y)+f(f(x)y)$. $P(f(y),x)$ means that $f(f(x)f(y))+f(x)=f(f(y)+x)+f(f(y)x)$, so $f(f(x)+y)-f(f(y)+x)=f(y)-f(f(x)y)-f(x)+f(f(y)x)=0$. Thus $f(f(x)+y)=f(f(y)+x)$. Let $n$ be a nonnegative integer. $f(f(n)+y)=f(f(y)+n)$. We use $\ast$ to expand out the RHS. $f(f(y)+n)=f(f(f(y)+n-1))=f(f(f(f(y)+n-2)))=\dots = \underbrace{f(f(\dots f}_{n+2} (y)\dots))$. We then use $\ast$ to un-expand the RHS. $\underbrace{f(f(\dots f}_{n+2} (y)\dots))=\underbrace{f(f(\dots f(}_{n+1} y+1)\dots))=\dots =f(y+n+1)$. Thus $f(y+n+1)=f(y+f(n))$. We now prove that $f$ is non-periodic. Suppose $f$ has a nonzero period $k$. Then $f(0)=f(k)$. Recalling $f(xf(0))=f(x)$, $P(x,k)$ gives that $f(0)=f(kx)$ for all $x$, so $f$ would be constant. Using this fact, we attain that $f(n)=n+1$ for all nonnegative integer $n$. We now use this to solve for $f(-1)$. For positive integer $n$, $P(n,f(-1))$ gives $f(nf(-1))+f(-1)=n+f(-n)$. For $n=1$, we get $f(f(-1))=1$. For nonnegative integer $k$, we have that $f(k)=\underbrace{f(f(\dots f}_{k+1} (0)\dots ))=\underbrace{f(f(\dots f}_{k+1} (f(-1))\dots ))=f(f(-1)+k)$. Thus by $P(k,0)$ and $P(k,f(-1))$, $f(kf(-1))=1$. $P(n,f(-1))$ for any positive integer $n$ now gives $f(-1)+1=n+f(-n)$, so $f(-n)=f(-1)-n+1$. $P(-2,1)$ gives $f(-4)+2=f(-1)+f(-2)$, so $f(-1)-1=f(-1)+f(-1)-1$, so $f(-1)=0$. We now show $f$ is injective at $0$. Suppose there is a $c$ such that $f(c)=f(-1)=0$. $P(x,-1)$ and $P(x,c)$ give $f(x-1)+f(-x)=f(0)$ and $f(x+c)+f(xc)=f(0)$. Call these assertions $Q(x)$ and $R(x)$ respectively. $R(-c)$ gives $f(-c^2)=0$. Thus $P(x,-c^2)$ gives $f(x-c^2)+f(-xc^2)=f(0)$. $R(-xc)$ gives $f(-cx+c)+f(-xc^2)=f(0)$, so $f(x-c^2)=f(-cx+c)$. $R(1-x)$ gives $f(1-x+c)+f(-cx+c)=f(0)$, and $Q(x-c-1)$ gives $f(x-c-2)+f(1-x+c)=f(0)$. Thus $f(x-c^2)=f(x-c-2)$ for all $x$. If $c^2\neq c+2$ then $f$ is periodic, which is impossible. Thus $c^2=c+2$, so $c=-1,2$. $f(2)\neq 0$, so $c=-1$ as desired. We now finish the problem. Recall $f(f(x)+y)=f(f(y)+x)$ for all $x,y$. Then letting $y=-1-f(x)$, $f(f(-f(x)-1)+x)=0$. Thus $f(-f(x)-1)=-x-1$. Then if $f(a)=f(b)$, this relation proves that $a=b$, so $f$ is injective. By $\ast$, $f(y)=y+1$ for all $y$, as desired.

thanks to dotted and circleinvert for helping me The answers are $f$ constant and $f(x)=x+1$, both of which clearly work. Assume that $f$ is nonconstant, we'll prove that $f(x)=x+1$. Let $P(x,y)$ be the assertion. We need the following results: Property 1: $f(f(x))=f(x+1)$. This is easy from $P(1,y)$. Property 2: $f(y)\neq y$. Otherwise, taking $P(x,y)$ for that $y$ yields $f(y)=f(x+y)$ for all $x$, a contradiction. In particular $f(x)\neq f(x+1)$. Property 3: $f(x+f(y))=f(y+f(x))$, which follows from $P(f(x),y)$ and $P(y,x)$. Property 4: $f(xy+x)+f(y)=f(xy+y)+f(x)$, which follows from $P(x,f(y))$. Claim: $f(0)=1$. Proof: From $P(x,0)$ we get $f(xf(0))=f(x)$, hence if $f(0)\neq 1$ we can choose $x$ where $xf(0)=x+1$, a contradiction. Claim: $f(-1)=0$. Proof: From $y\to -1$ in Property 4, realize that $1+f(-1)=f(-x-1)+f(x)$. Call this Property 5. From $P(-x,-1)$ we then obtain $f(-xf(-1))=1$, thus $f(-1)=0$. Claim: $f$ is injective at $0$. Proof: Otherwise we can take $a\neq -1$ where $f(a)=0$. Then $P(x,a)$ gives $f(xa)+f(x+a)=1$, hence \[f(x+a)=1-f(xa)=f(-xa-1).\]If $a\neq -1$ then we can choose $x$ where $x+a+1=-xa-1$, a contradiction. Claim: $f(x)=x+1$. Proof: From $y\to -f(x)-1$ in Property 3, we get $f(x+f(-f(x)-1))=0$ hence $f(-f(x)-1)=-x-1$. In particular, we have $f(f(x))=x+2$ from Property 5, and yet Property 1 gives $f(x+1)=x+2$, done.

Answers: $f$ constant and $f(x)=x+1$; it is easy to check that both of these work. Now assume $f$ is nonconstant. Step 1: We can make $xf(y)=xy$ if $f(y)=y$ for some $y$, and then the FE gives $f(y)=f(x+y)$ for any $x$, which gives $f$ is constant, so we assume that $f(y)$ is never equal to $y$. Step 2: Plug in $x=1$ to get $f(f(y))=f(y+1)$; then $f(y)\ne f(y+1)$ by the previous step, and now we can apply this to plugging in $y=0$: the FE becomes $f(xf(0))=f(x)$, and note that if $f(0)$ were not $1$ then we could make $xf(0)$ and $x$ differ by $1$ and get a contradiction, so $f(0)=1$. Step 3: LHS is fixed if you fix $x$ and change $y$ but fix $f(y)$, and RHS is symmetric so you can also change $x$ but fix $f(x)$; hence, both sides are dependent on only $f(x)$ and $f(y)$, and in particular $f(xf(y))$ is dependent only on $f(x)$ and $f(y)$ so $f(a)=f(b)$ implies $f(xa)=f(xb)$ for any $x$ in the range of $f$. We also observe that from $f(x+y)+f(xy)$ being dependent on only $f(x)$ and $f(y)$, and that if $y$ is in the range of $f$ then $f(xy)$ is dependent only on $f(x)$ if we fix $y$, we have $f(x+y)$ is also dependent only on $f(x)$ if we fix $y$. Step 4: $f(f(x))=f(x+1)$ so $f((x+1)f(y))=f(f(x)f(y))$, hence if we replace $x$ with $x+1$ in the FE we get $f(f(x)f(y))+f(y)=f(x+y+1)+f((x+1)y)$. Hence, $f((x+1)y)-f(y)$ is equal to something symmetric in $x$ and $y$. By doing the switching trick we then get $f(x(y+1))-f(y(x+1))=f(x)-f(y)$. Now $y=-1$ implies $f(-(x+1))+f(x)=1+f(-1)$. Now, using this, we prove $f(a)=f(b)$ iff $f(a+1)=f(b+1)$. The forward direction follows from $f(f(x))=x+1$. For the backward direction, observe that $f(a+1)=f(b+1)$ iff $f(-a)=f(-b)$ since $f(-(x+2))+f(x+1)=1+f(-1)$ and similarly $f(-a-1)=f(-b-1)$ iff $f(a)=f(b)$, so we can use the forward direction here to prove the backward direction. Step 5: Observe that $f$ is injective on its range: if $f(f(x))=f(f(y))$ then $f(x+1)=f(y+1)$ so $f(x)=f(y)$. Step 6: We would like to prove that in fact $f(x)=x+1$ holds over its range. To do this, return to the thing we got from symmetry: $f(x(y+1))-f(y(x+1))=f(x)-f(y)$. We could have also done the same symmetry trick on the original equation without prior manipulation, which gives $f(xf(y))-f(yf(x))=f(x)-f(y)$, so $f(x(y+1))-f(y(x+1))=f(xf(y))-f(yf(x))$. Now, if $y$ is in the range, then for any $x$ we have that f(f(x))=f(x+1) and so $f(y(x+1))=f(yf(x))$. Then $f(x(y+1))=f(xf(y))$ for any $x$ so we get the inputs differ by 1 contradiction unless $f(y)=y+1$. Step 7: Now we have $f(x+1)=f(f(x))=f(x)+1$ so the range of $f$ is closed under addition by an integer. In particular, $f$ is bijective on its range. Now we make the substitution $g(x)=f(x)-1$, and we know the range stays the same, and the other conditions translate easily. We know that $g$ is the identity on integers since $g(x+1)=g(x)+1$ and $g(0)=0$. We also have $g(g(x))=g(x)$, and $g(x)+g(-x)=0$ so $g$ is odd. Also we want to prove that $g$ is the identity to finish the problem. Step 8: We can evaluate $g(g(x)-x)$ as we can change the $-x$ term to anything that evaluates to $g(-x)=-g(x)$, namely $-g(x)$, so this experession turns out to be $g(0)=0$. Hence, if there exists $a$ such that $g(a)\ne a$ then there exists $b$ such that $b\ne 0$ but $g(b)=0$ by $b=g(a)-a$. Hence, it suffices to prove that the preimage of 0 contains only 0. Step 9: If $g(x)=0$, the LHS of the FE in $g$ evaluates to $g(y)$, so we get $g(x+y)+g(xy)=g(y)$ for all $y$. Now, observe that $g(-x)=-g(x)=0$ if $g(x)=0$, so we can replace $x$ with $-x$ and $y$ with $-y$ to get $-g(x+y)+g(xy)=-g(y)$, and averaging the two equations we just got we get $g(xy)=0$ for all $y$. Hence, either $g$ and thus $f$ is constant at 0 (already assumed away earlier) or we must have $g(x)=0$ implies $x=0$, as desired.

fun ct ional equation. Claim: $f$ is aperiodic. Proof. Increasing $y$ by the period $d$ implies that $f(xy) = f(x(y+d))$, giving constancy. $\blacksquare$ By $P(x, 0)$ and $P(1, y)$ it follows that $f(xf(0)) = f(x)$ and $f(f(y)) = f(y+1)$. Claim: $f(0) = 1$. Proof. By $P(x, f(0))$ it follows that \[ f(xf(1))+f(1)=f(x+f(0))+f(xf(0)) \]which becomes $f(x+1) = f(x + f(0))$. Due to being aperiodic, it follows that $f(0) = 1$. $\blacksquare$ Claim: $f(f(x) + y) = f(x + f(y))$. Proof. By $P(f(x), y)$ we get that \[ f(f(x)f(y)) + f(y) = f(f(x) + y) + f(yf(x)) \]so \[ f(f(x)f(y)) + f(x) + f(y) = f(f(x) + y) + f(x + y) + f(xy) \]As such, $f(f(x) + y) = f(x + f(y))$. $\blacksquare$ It thus follows that $f(f(-1) + y) = f(-1 + f(y)) = f(y)$ for all $y$. As such, $f(-1) = 0$. Claim: If $f(r) = 0$, then $r = -1$. Proof. By $P(x, r)$ it follows that $1 = f(x + r) + f(rx)$. By $P(r, x)$ it follows that $f(rf(x)) + f(x) = 1$. As such, $f(a \cdot x) = f(x)$ for $a = -r$. Then, by $P(x, a)$ it follows that \[ f(xf(1)) + f(a) = f(x + a) + f(x) \]Comparing this to $P(ax, a)$, it follows $f(x + a) = f(ax + a) = f(x + 1)$, which implies $r = -1$. $\blacksquare$ Note that by comparing $P(x, y), P(y, x)$ we get that \[ f(xf(y)) - f(x) = f(yf(x)) - f(y) \]Denote this assertion $Q(x, y)$. Claim: $f(n) = n + 1$ on the integers. Proof. It follows by $Q(x, -1)$ that $1 - f(x) = f(-f(x))$. By $P(x, -1)$, we get that $1 - f(x - 1) = f(-x)$ as well. Combining these two, we get that $f(-f(x)) = f(-(x+1))$. For positive integer $x$, it follows that $f(-f(x) + x) = f(-1) = 0$ so $f(x) = x + 1$. We can then get the result holds for negative integers. $\blacksquare$ Claim: $f(y) = y + 1$ for all $y$. Proof. Note that by $Q(x, y), Q(f(x-1), y)$ it follows that $f(xy) = f(f(x-1)y)$. As such, \[ f(n + ay) = f(n + af(y-1)) \]for integers $n$ follows by comparing $P(x, ay)$ and $P(x, af(y-1))$ (As integers are in the domain). Choose $a$ such that $ay = -1434 - n$. Then it follows that $f(-1) = f(n + 1433 + af(y-1)) = 0$, which implies $y = f(y-1)$. $\blacksquare$

Nice F.E. but i think it is missplaced for TST6, maybe its becuase i dont have test pressure ig (solved in 25 minutes (i was typing sol while solving xD)) Let $P(x,y)$ the assertion of the following F.E., i claim that $f(x)=x+1$ and $f$ constant both are the only solutions, thus now we might assume $f$ non-constant since it clearly works. If $f$ had a period $t$ then by $P(x,y+t)-P(x,t)$ we get that $f$ is constant. $P(x,0)$ gives $f(xf(0))=f(x)$, $P(1,x)$ gives $f(f(x))=f(x+1)$, now $P(x,f(0))-P(x,1)$ gives that $f(x+f(0))=f(x+1)$, which by the claim about $f$ having no period, this is equivalent to $f(0)=1$. Now applying symetry on the F.E. we get $f(xf(y))-f(yf(x))=f(x)-f(y)$, now $P(f(x),y)$ gives $f(f(x)f(y))-f(f(x)+y)=f(yf(x))-f(y)$ which after applying symetry we get that $f(f(x)+y)=f(f(y)+x)$, we call this $Q(x,y)$, now by symetry on $Q(x,f(y))$ we get: $$f(x+f(y+1))=f(y+f(x+1))=f(y+1+f(x)) \implies f(x+1)=f(x)+1 \implies f(n)=n+1 \; \forall n \in \mathbb Z$$By induction we have $f(x+n)=f(x)+n$ for $n \in \mathbb Z$, now by $P(-x,-1)$ we get $f(-x-1)=1-f(x)$ so $f(x)+f(-x)=2$. And now by $Q(f(x),-x)$ we get $f(f(x)-x+1)=f(2)$ meaning that $f(f(x)-x-2)=0$, now we go with this claim. Claim: If $f(c)=0$ then $c=-1$ must hold. Proof: By $P(x-c,c)$ we get $f(cx-c^2)=1-f(x)$ so $f(x)=f(c^2-cx-1)$ and $f(c^2-cx)=f(x+1)=f(c^2-cx-c-1)$ which means that $f(c^2-c-1-cx)=f(c^2-cx)$ which is equivalent to $f(c^2-c-1+z)=f(c^2+z)$ so $c=-1$ as desired. Now back to the problem we get that $f(x)-x-2=-1$ so $f(x)=x+1$ as desired, thus we are done .

Wrapped FE for TST P6??? Epic. Let $P(x,y)$ denote the given FE. Note that $$\boxed{f(x)=c \ \ \forall x \in \mathbb{R}}$$is a solution for any constant $c$. From now assume $f$ is non-constant. Claim 1: $f(x+f(y))=f(y+f(x))$ for all $x,y \in \mathbb{R}$. Proof: $P(f(x),y)$ followed by $P(y,x)$ gives $$f(f(x)f(y))+f(y)+f(x)=f(f(x)+y)+f(x+y)+f(xy).$$Now switch around $x$ and $y$. $\blacksquare$ Claim 2: $f(x)+f(xy+y)=f(y)+f(xy+x)$ for all $x,y \in \mathbb{R}$. Proof: $P(1,y)$ gives $f(f(y))=f(y+1)$. Now comparing $P(x,f(y))$ and $P(x,y+1)$, we get $$f(x+y+1)+f(xy+x)=f(x+f(y))+f(xf(y))$$$$\implies f(x+y+1)+f(xy+x) = f(x+f(y))+f(x+y)+f(xy)-f(y) \ \ \dots \textbf{(1)}$$Now just switch $x,y$ in $\textbf{(1)}$ and use Claim 1. $\blacksquare$ Call the FE in Claim 2 as $Q(x,y)$. Claim 3: $f(x+y+1)=f(x+f(y))$ for all $x,y \in \mathbb{R}$ with $|y| \geq 2$. Proof: $Q(a,bc+c)$, followed by $Q(b,c)$ gives: $$f(a)+f(abc+ac+bc+c)=f(abc+ac+a)+f(bc+c)$$$$\implies f(a)+f(b)+f(abc+ac+bc+c)=f(abc+ac+a)+f(bc+b)+f(c).$$Now switch $a,b$ in the above and compare to get $$f(abc+ac+a)+f(bc+b)=f(abc+bc+b)+f(ac+a)$$for all reals $a,b,c$. In particular, if $b \neq 0$, taking $c=\frac{1}{b^2}$ we get $$f \left(\frac{a}{b}+\frac{a}{b^2}+a\right)+f \left(b+\frac{1}{b} \right)=f \left(\frac{a}{b}+b+\frac{1}{b} \right)+f \left(\frac{a}{b^2}+a \right).$$Now, given $x,y \in \mathbb{R}$ with $|y| \geq 2$, we can find a $b \neq 0$ such that $y=b+\frac{1}{b}$, and choose $a$ such that $x=\frac{a}{b}$. Then the above gives $$f(xy+x)+f(y)=f(x+y)+f(xy).$$Putting this in $\textbf{(1)}$, we get $f(x+y+1)=f(x+f(y))$, as required. $\blacksquare$ Now suppose $f(t) \neq t+1$ for some $t \geq 2$. Let $k=t+1-f(t) \neq 0$; then Claim 3 gives $f(x+k)=f(x)$ for all real $x$. Therefore, comparing $P(x,0)$ and $P(x,k)$, we get $f(xk)=f(0)$ for all real $x$. Since $k \neq 0$, this implies that $f$ is constant, contradiction! Therefore $f(t)=t+1$ for all $t \geq 2$. Finally, for arbitrary real $y$, take $x$ large enough so that $y+f(x)=x+y+1 \geq 2$ and $x+f(y) \geq 2$. Then Claim 1 implies $x+y+2=x+f(y)+1$, which in turn gives the solution $$\boxed{f(x)=x+1 \ \ \forall x \in \mathbb{R}}$$$\blacksquare$

Felt a bit... standard? Can someone check this? Let $P(x,y)$ denote the given equation. Note that all constant functions work. Now, assume that $f$ is nonconstant. Step 1: $f(0)=1$. Proof: $P(x,0)$ implies $f(xf(0))=f(x)$, and $P(x,yf(0))$ implies $f(x+y)=f(x+yf(0))$, that is $f(x)=f(x+y(f(0)-1))$. If $f(0) \neq 1$, this implies that $f$ is constant, a contradiction. Step 2: $f(f(x)+y)=f(f(y)+x)$ for all $x,y \in \mathbb{R}$. Proof: Compare $P(f(x),y), P(f(y),x)$ and use $P(x,y)$ and $P(y,x)$. Step 3: $f(f(x+1)-f(x)-1+y)=f(y)$ for all $x,y \in \mathbb{R}$. Proof: Note that $P(1,x)$ gives $f(f(x))=f(x+1)$, and so $f(f(x+1)+y)=f(f(f(x))+y)=f(f(y)+f(x))$, and similarly $f(f(y+1)+x)=f(f(x)+f(y))$, hence $f(f(x+1)+y)=f(f(y+1)+x)=f(f(x)+y+1)$, to which we may put $y \rightarrow y-f(x)-1$ and conclude. Now, if $f$ is periodic with period $T \neq 0$, we may use $P(x,y+T)$ to obtain that $f(xy)=f(xy+xT)$ for all $x,y$, hence $f$ is constant, a contradiction. Therefore $f$ is not periodic, hence by Step 3 we obtain $f(x+1)=f(x)+1$ for all $x,y \in \mathbb{R}$. Step 4: $f(x)+f(y)=f(x+y)+1$ for all $x,y \in \mathbb{R}$. Proof: Note that $f(-1)=f(1)-1=0$, and so $P(x,-1)$ implies $f(x-1)+f(-x)=1$, which implies that $f(-x)=2-f(x)$. Now, we may use $P(-x,y)$ and $P(x,y)$ to finally obtain $f(x+y)+f(y-x)=2f(y)$. Put $y=x$ here to obtain $f(2x)+1=2f(x)$, and so $f(x+y)+f(y-x)=f(2y)+1,$ which rewrites to what we wanted to prove. Step 5: $f(x)=f(x \cdot \dfrac{f(y)}{y+1})$ for all $x \in \mathbb{R}$ and $y \in \mathbb{R} \ \{-1 \}$. Proof: Use Step 4 and $P(x,y)$ to obtain $f(x(y+1))=f(xf(y))$. Now, if for some $y \neq -1$ we had $\dfrac{f(y)}{y+1} \neq 1,$ then $f(x)=f(xA)$ for some $A \neq -1$. Use $P(x,yA)$ to obtain $f(x+y)=f(x+yA)$, which in turn implies that $f(x)=f(x+y(A-1))$, and so $f$ is constant, a contradiction. Therefore, $f(y)=y+1$ for all $y \neq -1$, and we may now conclude that $f(x)=x+1$ for all $x$, which works. To sum up, all solutions are $f(x) = c$ and $f(x)=x+1$.

After getting $f(f(x)+y)=f(f(y)+x)$ the rest is nearly same with this problem (I might not think of the idea if haven't seen that problem)