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DottedCaculator

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DottedCaculator

#1 h Jan 15, 2024, 5:00 PM • 3 Y

Y by MathLuis, Supercali, Rounak_iitr

Find all functions $f\colon\mathbb R\to\mathbb R$ such that for all real numbers $x$ and $y$ ,
$f(xf(y))+f(y)=f(x+y)+f(xy).$
Milan Haiman

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CT17

#2 h Jan 15, 2024, 5:00 PM • 4 Y

Y by CyclicISLscelesTrapezoid, Mop2018, Rounak_iitr, CooperHqh

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.

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DottedCaculator

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DottedCaculator

#3 h Jan 15, 2024, 5:00 PM • 1 Y

Y by Rounak_iitr

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 $f(x)=x+1$ .

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#4 h Jan 15, 2024, 5:01 PM

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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.

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i3435

#5 h Jan 15, 2024, 5:03 PM

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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.

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#6 h Jan 15, 2024, 6:04 PM • 1 Y

Y by Korean_fish_Kaohsiung

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.

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#7 h Jan 15, 2024, 8:13 PM • 1 Y

Y by asdf334

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.

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#8 h Jan 16, 2024, 7:36 AM • 1 Y

Y by Rounak_iitr

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 $y = f(y-1)$ . $\blacksquare$

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#9 h Jan 16, 2024, 11:15 AM • 1 Y

Y by ATGY

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

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This post has been edited 3 times. Last edited by MathLuis, Jan 16, 2024, 11:18 AM

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#11 h Jan 17, 2024, 8:52 AM • 1 Y

Y by kkloveMinecraft

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
$\boxed{f(x)=x+1 \ \ \forall x \in \mathbb{R}}$ $\blacksquare$

This post has been edited 1 time. Last edited by Supercali, May 5, 2024, 1:16 PM

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#12 h Jan 19, 2024, 10:51 AM • 1 Y

Y by Rounak_iitr

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 $f(x)=x+1$ .

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#15 h Dec 7, 2024, 4:11 AM

Y by

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)

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Korean_fish_Kaohsiung

#16 h Apr 9, 2025, 8:08 AM

Y by

This problem is epic, took me way too long tho

answer: $f(x)=c$ and $f(x)=x+1$ , let $P(x,y)$ be the assertion

step $1$ : $f(0)=1$ or else $f$ is a constant
Proof
step $2$ : plug some things to get some properties
property $1$ : $f(f(x))=f(x+1)$ , this is trivial by $P(1,y)$
property $2$ : if there exists a $t$ such that $f(t)=t$ , then $f(x)$ is a constant, this is trivial by $P(x,t)$
property $3$ : $f(xf(y))+f(y)=f(yf(x))+f(x)$ , this is trivial by swapping $x,y$
property $4$ : $f(x+f(y))+f(xf(y))=f(x+y+1)+f(xy+x)$
Proof
property $5$ : $f(f(x)f(-1))=1$
Proof

step $3$ : $f(-1)=0$ or else $f$ is a constant
Proof
Now by the proof we have two more properties
property $6$ : $f(y)+f(-y-1)=1$

property $7.1$ : $f(f(y)-1)=f(y)$
and property $7.2$ : $f(-f(y))=f(-y-1)$ this is a result of $f(f(y)-1)+f(-f(y))=1$ and $f(-f(y))+f(y)=1$ and property $6$

step $4$ : $f(y-f(y))=0$
Proof

step $5$ : $f$ is injective on $f(x)=0$ or else $f$ is a constant
Proof

Final : by step $4$ and $5$ we get that $y-f(y)=-1$ , so $f(x)=x+1$ and we are finished.

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#17 h May 28, 2025, 7:24 PM • 1 Y

Y by CatalanThinker

Clearly, $f(x)\equiv c$ is a solution. From now on, assume $f$ is nonconstant.

$P(0,x)\implies f(xf(0))=f(x)$
$P(1,y)\implies f(f(y))=f(y+1)$
Let's assume for the sake of contradiction $f(y)=y$ for some $x\in \mathbb{R}$ .
$P(x,y)\implies f(x+y)=f(y)$ , $\Rightarrow \Leftarrow$
So, $f(x)\neq f(f(x))=f(x+1)$ .
However, if $f(0)\neq 1$ , $x=\frac{1}{f(0)-1}$ gives
$f\left(1+\frac{1}{f(0)-1}\right)=f\left(\frac{1}{f(0)-1)}\right), \Rightarrow \Leftarrow$ So, $f(0)=1$ .

$P(x,y)\implies f(xf(y))+f(y)=f(x+y)+f(xy)$
$P(y,x)\implies f(yf(x))+f(x)=f(x+y)+f(xy)$
So,
$f(yf(x))+f(x)=f(xf(y))+f(y), f(yf(x))-f(y)=f(xf(y))-f(x)$
$P(f(x),y)\implies f(f(x)f(y))+f(y)=f(f(x)+y)+f(yf(x))$
Switching $x$ and $y$ gives $f(f(x)f(y))+f(x)=f(f(y)+x)+f(xf(y))$
Subtracting yields $f(f(x)+y)=f(x+f(y))\dots (1)$

$P(x,y+1)\implies f(xf(y+1))+f(y+1)=f(xf(f(y)))+f(f(y))=f(x+y+1)+f(xy+x)$
$P(x,f(y))\implies f(xf(f(y)))+f(f(y))=f(x+f(y))+f(xf(y))$
Subtracting yields $f(x+f(y))+f(xf(y))=f(x+y+1)+f(xy+x)$
Switching $x$ and $y$ gives $f(y+f(x))+f(yf(x))=f(x+y+1)+f(xy+y)$
So, $f(xy+y)+f(x)=f(xy+x)+f(y)\dots (2)$ .

$y=-1\xRightarrow{(2)} f(-x-1)+f(x)=1+f(-1)$
$P(-x,-1)\implies f(-x-1)+f(x)=f(-xf(-1))+f(-1)$
So, $1=f(-xf(-1))$ , $\Rightarrow\Leftarrow$ unless $f(-1)=0$ , since we can put $x=\frac{-1}{f(-1)}$ .

Say $f(a)=0$ , $a\neq -1$ .
Notice that $y=-1\xRightarrow{(2)} 1=f(x-1)+f(-x)\dots (3)$
$x\rightarrow -x\implies 1-f(x)=f(-x-1)$
$y=a\implies f(x+a)+f(ax)=1$
So, $f(ax)=1-f(x+a)=f(-x-a-1)$ .
Putting $x=\frac{-a}{a+1}$ gives a contradiction since $f(x+1)\neq f(x)$
So, $a=-1$ , $\Rightarrow \Leftarrow$ .

Finally, $f(f(-f(x)-1)+x)=0\implies f(-f(x)-1)+x=-1\xRightarrow{(3)} 1-f(x+1)+x=-1$
So, $f\equiv x+1$ .

This post has been edited 1 time. Last edited by Adywastaken, May 29, 2025, 2:18 PM

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