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Source: OMMC POTM February 2022

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DottedCaculator

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DottedCaculator

#1 h Nov 2, 2023, 11:15 PM • 2 Y

Y by centslordm, ItsBesi

Find all functions $f:\mathbb R \to \mathbb R$ (from the set of real numbers to itself) where $f(x-y)+xf(x-1)+f(y)=x^2$ for all reals $x,y.$

Proposed by cj13609517288

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#2 h Nov 2, 2023, 11:38 PM • 1 Y

Y by centslordm

Plug in $y=1$ to get $(x+1)f(x-1)=x^2-f(1) \implies f(x-1)=\frac{x^2-f(1)}{x+1}$ for $x \neq -1$ . On the other hand if $x=-1$ then the RHS should be zero as well, hence $f(1)=1$ , so $f(x-1)=x-1$ for $x \neq 1$ . Finally plug in $x=0$ to get that $f$ is odd, so $f(x)=x$ for all $x$ : done. $\blacksquare$

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#3 h Nov 5, 2023, 12:36 PM • 1 Y

Y by parmenides51

Let $P(x,y)$ denote the given assertion.

$P(1,1): f(1)=1$ .

$P(x,1): f(x-1)+xf(x-1)+1=x^2\implies (x+1)f(x-1)=x^2-1=(x-1)(x+1)\implies f(x-1)=x-1\forall x\ne -1$ .

$P(-2,-2): -2f(-3)+f(-2)=4\implies 6+f(-2)=4\implies f(-2)=-2$ .

So $\boxed{f(x)=x}$ .

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6051 posts

#4 h Nov 23, 2023, 12:57 AM

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If $x=1,$ and $y=1,$ we have $f(0)+f(0)+f(1)=1,$ hence $2f(0)+f(1)=1,$ and if $x=0,$ and $y=0,$ we have $2f(0)=0,$ hence $f(0)=0,$ so $f(1)=1,$ now if $y=1,$ we have $f(x-1)+xf(x-1)+f(1)=x^2,$ hence $(x+1)f(x-1)=x^2-1,$ so $f(x-1)=\frac{x^2-1}{x+1}=x-1,$ so clearly $f(x)=x,$ is the only solution.

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#5 h Dec 31, 2023, 5:53 AM • 2 Y

Y by ATGY, HACK_IN_MATHS

Bro lmao, the link that was put for this problem in the OTIS unit was an instagram post :rotfl:

:rotfl:

. I changed it to this thread's link now.

The solution is $f(x) \equiv x$ .

Let $P$ denote the assertion.

$P(x,y)\implies f(x-y) + xf(x-1) + f(y) = x^2$ .

$P(0,0) \implies f(0)=0$ .

$P(0,y) \implies f(-y) = -f(y)$ , that is $f$ is odd.

$P(x,0) \implies f(x) + xf(x-1) = x^2 \implies xf(x-1) = x^2 - f(x)$ .

So, $P(x,y) \implies f(x-y) + xf(x-1) + f(y) = x^2 \implies f(x-y) + x^2 - f(x) + f(y) = x^2\implies f(x+y) = f(x) + f(y)$ . Thus we get that $f$ is Cauchy.

$P(1,0) \implies f(1) = 1 \implies f(-1) = -1$ .

So, $P(x,y) = (f(x) - f(y)) + x(f(x) - 1) + f(y) = x^2 \implies xf(x) = x^2 \implies f(x) = x \; \forall x \neq 0$ .

Thus $f(x) = x \; \forall x \in \mathbb{R}$ .

This post has been edited 1 time. Last edited by kamatadu, Dec 31, 2023, 5:58 AM

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#6 h Jan 21, 2024, 12:22 PM

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Solution

Let $P(x,y)$ denote the given assertion.

$P(0,0) \implies 2f(0)=0 \implies f(0)=0$

$P(0,x) \implies f(-x)+f(x)=0 \implies -f(x)=f(-x) \implies f$ -is odd

$P(x,0) \implies f(x)+xf(x-1)=x^2 \implies f(x)=x^2-xf(x-1)$ $...(1)$

$P(1,x) \implies f(1-x)+f(x)=1 \implies f(x)=1-f(1-x)$ Combining with $f$ -is odd we get: $f(x)=1+f(x-1)$ $...(2)$
Now Combining $(1)$ with $(2)$ we get:

$x^2-xf(x-1)=1+f(x-1) \implies x^2-1=xf(x-1)+f(x-1) \implies (x-1)(x+1)=(f(x-1))(x+1) \implies f(x-1)=x-1 \forall x\in \mathbb{R} /{-1}$
$x \rightarrow x+1 \implies f(x+1-1)=x+1-1 \implies f(x)=x$ $\forall x\in \mathbb{R} /{-1}$

$P(1,0) \implies f(1)=1 \implies -f(1)=-1$ Combining with $f$ -is odd we get : $f(-1)=-1$

Since $f(x)=x$ $\forall x\in \mathbb{R} /{-1}$ and $f(-1)=-1$ we get that : $f(x)=x$ $\forall x\in \mathbb{R}$

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AlanLG

#7 h Apr 12, 2024, 6:01 PM

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We claim the solution is $f(x)=x$ , which clearly satisfies. Let $P(x,y)$ be the assertion
$P(x,x)\implies f(0)+xf(x-1)+f(x)=x^2$ By plugging $x=0$ here we get $f(0)=0$ , and plugging $x=1$ gives $f(1)=1$ , now
$P(x,1)\implies f(x-1)+xf(x-1)+1=x^2$ Then $f(x)-f(x-1)=1$ and therefore
$x\cdot \left(f(x)-1\right)+f(x)=x^2$ , and therefore $f(x)=x \forall x\in\mathbb{R}$ (the case where $x=-1$ is trivial as $f(0)-f(-1)=1$ )

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SomeonecoolLovesMaths

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SomeonecoolLovesMaths

#8 h Apr 12, 2024, 6:24 PM

Y by

AlanLG wrote:

We claim the solution is $f(x)=x$ , which clearly satisfies. Let $P(x,y)$ be the assertion
$P(x,x)\implies f(0)+xf(x-1)+f(x)=x^2$ By plugging $x=0$ here we get $f(0)=0$ , and plugging $x=1$ gives $f(1)=1$ , now
$P(x,1)\implies f(x-1)+xf(x-1)+1=x^2$

$(x+1)f(x-1)=(x+1)(x-1)$ For $x \neq -1$ ,
$f(x-1) = x-1 \Longrightarrow \text{Let} a = x-1 \Longrightarrow f(a) = a$ For $x = -1$ ,
It has been shown above.

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#9 h Apr 12, 2024, 8:46 PM

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$f(x)=x$ , $\forall x \in \mathbb R$

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cursed_tangent1434

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cursed_tangent1434

#10 h Apr 12, 2024, 11:36 PM

Y by

The answer is $f(x)=x$ for all $x \in \mathbb{R}$ . It’s easy to see that these functions satisfy the given equation. We now show these are the only solutions.

This is entirely plug and chug. Note that $P(0,0)$ gives us $f(0)=0$ . Further, $f(0,x)$ gives $f(-x)+f(x)=0$ for all $x\in \mathbb{R}$ which implies that $f$ is odd.

Now, $P(1,x)$ gives
$f(1-x)+f(0)+f(x)=1 \implies f(1-x)=1-f(x)$ But, since we established that $f$ is odd, this means $f(x-1)=f(x)-1$ for all $x\in \mathbb{R}$ . Now, $P(x,0)$ gives
$\begin{align*} f(x) +xf(x-1)+f(0) &= x^2\\ f(x)+x(f(x)-1) &= x^2\\ (x+1)f(x) &= x^2+x \end{align*}$ which immediately gives us that $f(x)=x$ for all $x\neq 1\in \mathbb{R}$ . But since $f(-1)=f(0)-1=-1$ , we can conclude that all solutions to the given functional equation are of the claimed forms.

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balllightning37

389 posts

balllightning37

#11 h Aug 30, 2024, 6:36 PM

Y by

I think I have a different approach using Cauchy.

We claim that $f(x)=x$ is the only solution. This trivially works.
First, $P(0,0)$ gives $f(0)=0$ . Then $P(x,0)$ gives $f(x)+xf(x-1)=x^2$ . Putting $x=1$ here gets us $f(1)=1$ . Also, $P(x,x)$ implies that $f(-x)+f(x)=0$ , or $f$ is odd.
Since we also have $f(x-y)+xf(x-1)+f(y)=x^2$ , we can then get that $f(x-y)+f(y)=f(x)$ for all $x$ , $y$ . Substituting $x'=x-y$ , $y'=y$ gives $f(x'+y')=f(x')+f(y')$ .

Now, we can break the given condition down to get $f(x)+xf(x)+xf(-1)=x^2$ or $f(x)+xf(x)=x^2+x$ . This means that $f(x)=x$ for all $x \neq -1$ . As $f(-1)=-1$ , we must have $f(x)=x$ for all $x$ .

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#12 h Aug 30, 2024, 6:47 PM

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Note that $P(-1, 1) \implies f(1) = 1$ .
Now $P(x, 1) \implies f(x) = x \forall x \ne -2$ .
Finally, $P(-1, 0) \implies f(-2) = -2$ .
So $f \equiv id$ , and this works.

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Eka01

#13 h Oct 12, 2024, 5:29 PM

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Put $x=0$ to get that $f$ is odd which implies $f(0)=0$ .
Next put $x=1$ and $y=0$ to get $f(1)=1$ .
Now put $x=1$ to get $f(1-y)+f(y)=1$ which implies $f(y-1)=f(y)-1$ since $f$ is odd.
Now put $y=0$ and use the above result to get $f(x)(x+1)=x^2+x$ which implies $\boxed{f(x)=x \ \forall \ x \in \mathbb{R}}$

This post has been edited 1 time. Last edited by Eka01, Oct 12, 2024, 5:30 PM

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eg4334

#14 h Feb 9, 2025, 10:07 PM

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Let $P(x, y)$ be the assertion. $P(0, y)$ gives $f$ to be odd. $P(0, 0)$ gives $f(0)=0$ . $P(x, 0)$ gives $f(x)=x(x-f(x-1))$ . $P(1, y)$ gives $f(y)+f(1-y)=1$ . Now $f(x)=x(x-f(x-1)) = x(x+f(1-x)) = x(x+1-f(x))$ which gives $\boxed{f(x)=x}$ for all $x \neq -1$ . But $f(-1)=f(0)-1=-1$ so its true for all $x$ .

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#15 h Feb 28, 2025, 4:37 AM

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DottedCaculator wrote:

Find all functions $f:\mathbb R \to \mathbb R$ (from the set of real numbers to itself) where $f(x-y)+xf(x-1)+f(y)=x^2$ for all reals $x,y.$

Proposed by cj13609517288

Let $P(x,y)$ be the assertion $f(x-y)+xf(x-1)+f(y)=x^2$ .
Using either $P(2,1)$ or $P(-1,1)$ or $P(-1,-2)$ gives $f(1)=1$ . Then:
$P(x+1,1)\Rightarrow f(x)=x\forall x\ne-2.$ Choose some $k$ such that $-1-k\ne-2$ and $k\ne-2$ (such as $k=20$ ) then:
$\begin{align*} P(-1,k)&\Rightarrow f(-1-k)-f(-2)+f(k)=1\\ &\Rightarrow -1-k-f(-2)+k=1\\ &\Rightarrow f(-2)=-2, \end{align*}$ so we get $\boxed{f(x)=x}$ for all $x$ which is a solution.

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#16 h Mar 9, 2025, 10:35 PM

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No words needed.

This post has been edited 2 times. Last edited by Marcus_Zhang, Mar 9, 2025, 10:37 PM

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#17 h Mar 12, 2025, 10:50 PM

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Solution

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#18 h Mar 15, 2025, 11:00 PM

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Let $P(x, y)$ denote the assertion. Then $P(x, 1), x \neq -1 \implies f(x-1)=\frac{x^2-f(1)}{x+1}.$ $P(2, 1) \implies f(1) = 1 \implies f(x-1)=x-1 \implies f(x)=x$ for all $x \neq -2.$ Now, observe that $x=0 \implies f(x)=-f(-x)$ so $f(-2)=-2.$

Therefore $f(x)=x$ for all $x.$

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#19 h Apr 25, 2025, 7:55 AM

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I claim the only answer is $f(x) = x.$ Plug in $x = y = 0$ to get that $f(0) = 0.$ Next, plug in $x = 0$ to get that $f(y) + f(-y)= 0,$ and hence $f$ is an odd function. Finally, plug in $y = 0$ to get that $f(x) + xf(x - 1) = x^2.$ Plug in $x = 1 - x$ to get that $f(1 - x) + (1-x)f(-x) = (1 - x)^2.$ Using the fact that it is odd, we have the two equations:
$\begin{align*} f(x) + xf(x - 1) & = x^2 \\ -f(x - 1) + (x - 1)f(x) & = (x - 1)^2 \end{align*}$ By solving for each one, we see that $f(x) = x.$

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