Easiest functional equation?

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ZETA_in_olympiad

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ZETA_in_olympiad

#1 h Mar 19, 2022, 3:20 PM

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Here I want the users to post the functional equations that they think are the easiest. Everyone (including the one who posted the problem) are able to post solutions.

This post has been edited 1 time. Last edited by ZETA_in_olympiad, Mar 19, 2022, 3:31 PM
Reason: Making the equation plural

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wamofan

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wamofan

#2 h Mar 19, 2022, 3:21 PM

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$f(x)=1$
(technically it is an FE)

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megarnie

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megarnie

#3 h Mar 19, 2022, 3:25 PM

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Find all functions $f\colon \mathbb{R}\to \mathbb{R}$ such that for all reals $x$ and $y$ , $f(x)=f(y)$

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ZETA_in_olympiad

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ZETA_in_olympiad

#4 h Mar 19, 2022, 3:30 PM

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Find all functions $f: \mathbb{R} \to \mathbb{R}$ such that $f(x+P)=f(x),$ is true.

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brainfertilzer

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brainfertilzer

#5 h Mar 19, 2022, 3:40 PM • 2 Y

Y by wamofan, cj13609517288

Find all functions $f:\{0\}\rightarrow \{ 0\}$ such that
$1 + f(x + f(y + x)) + f(\sqrt{f(f(x))}) = \frac{1}{1 + f(f(f(xy^2\sqrt{f(xy})))}$ for any $x$ and $y$ in the domain of $f$ .

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megarnie

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megarnie

#6 h Mar 19, 2022, 3:43 PM

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haha Let $\mathbb{R}$ be the set of real numbers. Determine all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that, for any real numbers $x$ and $y$ , $f(f(x)f(y)) + f(x+y) = f(xy).$
and Let $\mathbb R$ be the set of real numbers. Determine all functions $f:\mathbb R\to\mathbb R$ that satisfy the equation $f(x+f(x+y))+f(xy)=x+f(x+y)+yf(x)$ for all real numbers $x$ and $y$ .

and Let $R^+$ be the set of positive real numbers. Determine all functions $f:R^+$ $\rightarrow$ $R^+$ such that for all positive real numbers $x$ and $y:$
$f(x+f(xy))+y=f(x)f(y)+1$
and of course Find all functions $f:\mathbb{R} \rightarrow \mathbb{R}$ such that for any $x,y\in \mathbb{R}$ , $f(x+yf(x))=f(xy+1)+f(x-y).$
and also Find all functions $f:\mathbb{R} \to \mathbb{R}$ satisfying the equation $f(x^2+y^2+2f(xy)) = (f(x+y))^2.$ for all $x,y \in \mathbb{R}$ .

This post has been edited 5 times. Last edited by megarnie, Mar 19, 2022, 3:45 PM

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megarnie

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megarnie

#7 h Mar 19, 2022, 3:46 PM

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these ones are really easy like msm level so go ahead and try them \jk

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ZETA_in_olympiad

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ZETA_in_olympiad

#8 h Mar 19, 2022, 3:50 PM

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megarnie wrote:

these ones are really easy like msm level so go ahead and try them

msm level?

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sehgalsh

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sehgalsh

#9 h Mar 19, 2022, 3:53 PM • 1 Y

Y by tennisrules

ZETA_in_olympiad wrote:

megarnie wrote:

these ones are really easy like msm level so go ahead and try them

msm level?

it stands for middle school math.

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ZETA_in_olympiad

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ZETA_in_olympiad

#10 h Mar 19, 2022, 3:54 PM

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Let $f: R \to R$ such that $f(x+y)=f(x)+f(y).$ Prove that there exists $a \in R$ such that $f(x)=ax.$ A no brainer.

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ZETA_in_olympiad

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ZETA_in_olympiad

#11 h Mar 19, 2022, 4:12 PM

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Easiest ISL, just solved rn.

"Find all functions $f$ from the reals to the reals such that

$f\left(f(x)+y\right)=2x+f\left(f(y)-x\right)$
for all real $x,y$ ." [ISL 2002 alg p1]

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CascadeFalls521

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CascadeFalls521

#14 h Mar 21, 2022, 11:02 AM

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$f(x) = lim_{x -> 0} 1/x$ You see, when you take the limit as x approaches 0 in the thing 1/x, it keeps getting bigger, not focusing on any real number, so it is infinity.

So when you plug in any value $a$ with a constant $c$ , $f(a)$ = $f(a+c)$ = $\infty$ as put in other responses, but this is different.

If you were to graph this, there would be nothing to graph! You can't draw a constant of infinity! Therefore, this is the easiest. :weightlift:

It would be different if it was 1/0, because that is undefined, but $lim_{x -> 0} 1/x$ is $\infty$

@above - chill, is MATHCOUNTS for beginners?

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megarnie

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megarnie

#15 h Mar 21, 2022, 11:22 AM

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Let $f: \mathbb{Q} \to \mathbb{Q}$ such that $f(x+y)=f(x)+f(y).$ Prove that there exists $a \in \mathbb{Q}$ such that $f(x)=ax.$ A no brainer.

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rama1728

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rama1728

#16 h Mar 21, 2022, 2:04 PM

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CascadeFalls521 wrote:

$f(x) = lim_{x -> 0} 1/x$ You see, when you take the limit as x approaches 0 in the thing 1/x, it keeps getting bigger, not focusing on any real number, so it is infinity.

So when you plug in any value $a$ with a constant $c$ , $f(a)$ = $f(a+c)$ = $\infty$ as put in other responses, but this is different.

If you were to graph this, there would be nothing to graph! You can't draw a constant of infinity! Therefore, this is the easiest. :weightlift:

It would be different if it was 1/0, because that is undefined, but $lim_{x -> 0} 1/x$ is $\infty$

@above - chill, is MATHCOUNTS for beginners?

I am actually doubting this Zeta person by a lot.

He doesn't know what is cauchy fe, yet he posted a solution:

And the thing here is, he doesnt know that assuming a function differentiable is wrong, but knows how to proceed after that, which makes it look very sketchy.

And now when you check his status, he says he is in Dhaka college, but his goal is for IMO gold which doubts me even more.

I am not trying to tease him, but I just dont want to see another thinker123.

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IAmTheHazard

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IAmTheHazard

#17 h Mar 21, 2022, 2:20 PM

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Find all functions $f : \mathbb{Z} \to\mathbb{ Z}$ such that
$n^2+4f(n)=f(f(n))^2$ for all $n\in \mathbb{Z}$ .

Very easy and elegant FE!

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rama1728

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rama1728

#18 h Mar 21, 2022, 2:25 PM

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IAmTheHazard wrote:

Find all functions $f : \mathbb{Z} \to\mathbb{ Z}$ such that
$n^2+4f(n)=f(f(n))^2$ for all $n\in \mathbb{Z}$ .

Very easy and elegant FE!

Is this ISL 2014 A6?

This post has been edited 1 time. Last edited by rama1728, Mar 21, 2022, 2:26 PM
Reason: .

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megarnie

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megarnie

#19 h Mar 21, 2022, 2:26 PM

Y by

rama1728 wrote:

IAmTheHazard wrote:

Find all functions $f : \mathbb{Z} \to\mathbb{ Z}$ such that
$n^2+4f(n)=f(f(n))^2$ for all $n\in \mathbb{Z}$ .

Very easy and elegant FE!

Is this ISL 2014 A7?

no it is A6

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DottedCaculator

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DottedCaculator

#20 h Mar 21, 2022, 2:38 PM

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If $f(x)=1$ find $f(665)$

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asdf334

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asdf334

#21 h Mar 21, 2022, 2:45 PM

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DottedCaculator wrote:

If $f(x)=1$ find $f(665)$

what is x?

This post has been edited 2 times. Last edited by asdf334, Mar 21, 2022, 2:45 PM

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megarnie

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megarnie

#22 h Mar 21, 2022, 2:52 PM

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Let $f:\mathbb R\to\mathbb R$ satisfy the equation $f(x+f(x+y))+f(xy)=x+f(x+y)+yf(x)$ for all real numbers $x$ and $y$ .

Now let $g(x)$ be the product of possible values of $f(x)$ .

Find $-g(1+3\sqrt{74})$ .

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rama1728

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rama1728

#23 h Mar 21, 2022, 4:04 PM

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megarnie wrote:

Let $f:\mathbb R\to\mathbb R$ satisfy the equation $f(x+f(x+y))+f(xy)=x+f(x+y)+yf(x)$ for all real numbers $x$ and $y$ .

Now let $g(x)$ be the product of possible values of $f(x)$ .

Find $-g(1+3\sqrt{74})$ .

Why are you considering IMO 2015 P5 as an MSM fe?

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Gumball

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Gumball

#24 h Mar 21, 2022, 6:10 PM

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asdf334 wrote:

DottedCaculator wrote:

If $f(x)=1$ find $f(665)$

what is x?

I think x is f^-1(x)

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megarnie

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megarnie

#25 h Mar 21, 2022, 6:11 PM

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rama1728 wrote:

megarnie wrote:

Let $f:\mathbb R\to\mathbb R$ satisfy the equation $f(x+f(x+y))+f(xy)=x+f(x+y)+yf(x)$ for all real numbers $x$ and $y$ .

Now let $g(x)$ be the product of possible values of $f(x)$ .

Find $-g(1+3\sqrt{74})$ .

Why are you considering IMO 2015 P5 as an MSM fe?

it's not but most problems I post on MSM that have an answer usually have an answer of Click to reveal hidden text

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ZETA_in_olympiad

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ZETA_in_olympiad

#26 h Jun 23, 2022, 12:19 PM

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IAmTheHazard wrote:

Find all functions $f : \mathbb{Z} \to\mathbb{ Z}$ such that
$n^2+4f(n)=f(f(n))^2$ for all $n\in \mathbb{Z}$ .

Very easy and elegant FE!

Solve it then. One of the hardest I have encountered yet.

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jasperE3

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jasperE3

#27 h Apr 3, 2025, 3:39 AM

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ZETA_in_olympiad wrote:

Find all functions $f: \mathbb{R} \to \mathbb{R}$ such that $f(x+P)=f(x),$ is true.

If $P=0$ then $f$ can be any function $\mathbb R\to\mathbb R$ , else:
Let $g:[0,1)\to\mathbb R$ be any function, then $f(x)=g\left(\left\{\frac xP\right\}\right)$ .

ZETA_in_olympiad wrote:

Let $f: R \to R$ such that $f(x+y)=f(x)+f(y).$ Prove that there exists $a \in R$ such that $f(x)=ax.$ A no brainer.

$\mathbb R$

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