Functional equation

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Amin12

16 posts

Amin12

#1 h Aug 13, 2016, 3:39 PM • 7 Y

Y by K.N, yayitsme, A-Thought-Of-God, megarnie, tiendung2006, Aryan-23, Adventure10

Find all function $f:\mathbb{N}\rightarrow\mathbb{N}$ such that for all $a,b\in\mathbb{N}$ ,
$(f(a)+b) f(a+f(b))=(a+f(b))^2$

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K.N

532 posts

K.N

#3 h Aug 13, 2016, 4:21 PM • 3 Y

Y by skrublord420, Adventure10, Mango247

Amin12 wrote:

Find all function $f:\mathbb{N}\rightarrow\mathbb{N}$ such that for all $a,b\in\mathbb{N}$ ,
$(f(a)+b) f(a+f(b))=(a+f(b))^2$

Let $f(1)=c$ then prove that $c+2|(2c+1)^2$ so $c+2|9$ so $c=1$ or $7$ and then prove $c=1$ and then by easy induction we get $f(x)=x$

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pco

23511 posts

pco

#4 h Aug 13, 2016, 4:33 PM • 9 Y

Y by alibez, MMike, Dilshodbek, vsathiam, Pluto1708, karitoshi, Kanep, Adventure10, Mango247

Amin12 wrote:

Find all function $f:\mathbb{N}\rightarrow\mathbb{N}$ such that for all $a,b\in\mathbb{N}$ ,
$(f(a)+b) f(a+f(b))=(a+f(b))^2$

I suppose that, as usual on this forum, $\mathbb N$ is the set of positive integers ( $0\notin\mathbb N$ )
Let $P(x,y)$ be the assertion $(f(x)+y)f(x+f(y))=(x+f(y))^2$
Let $a=f(1)$ and $b=f(2)$

$P(f(x),1)$ $\implies$ $(f(f(x))+1)=\frac{(f(x)+a)^2}{f(f(x)+a)}$
$P(a,x)$ $\implies$ $(f(a)+x)=\frac{(a+f(x))^2}{f(a+f(x))}$
And so $f(f(x))=x+f(a)-1$

And since $P(x,x)$ $\implies$ $f(x+f(x))=f(x)+x$ , we get $f(a)-1=0$ and $f(f(x))=x$ $\forall x$

Then $P(x,f(y))$ becomes new assertion $Q(x,y)$ : $f(x)+f(y)=\frac{(x+y)^2}{f(x+y)}$

Comparing then $Q(x,2)$ with $Q(x+1,1)$ , we get $f(x+1)=f(x)+b-a$ and so $f(x)=(b-a)(x-1)+a$

Plugging this back in original equation, we get $a=1$ and $b=2$ and so
$\boxed{f(x)=x\text{ }\forall x\in\mathbb N}$

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yojan_sushi

330 posts

yojan_sushi

#5 h Aug 14, 2016, 3:24 AM • 4 Y

Y by vsathiam, J-Enterprise7-math, Adventure10, Mango247

Solution

This post has been edited 2 times. Last edited by yojan_sushi, Aug 14, 2016, 3:26 AM

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Dadgarnia

164 posts

Dadgarnia

#6 h Aug 16, 2016, 6:39 PM • 9 Y

Y by lebathanh, doxuanlong15052000, A_Math_Lover, J-Enterprise7-math, naw.ngs, rashah76, tiendung2006, Adventure10, NuMBeRaToRiC

Amin12 wrote:

Find all function $f:\mathbb{N}\rightarrow\mathbb{N}$ such that for all $a,b\in\mathbb{N}$ ,
$(f(a)+b) f(a+f(b))=(a+f(b))^2$

This problem has solved for $f:\Bbb{R}^+\rightarrow \Bbb{R}^+$ . Let $P(x,y)$ be the assertion $(f(a)+b)f(a+f(b))=(a+f(b))^2$ . We get:
$P(a,a)\rightarrow f(a+f(a))=a+f(a)$
So there exist $x_0 \in \Bbb{R}^+$ such that $f(x_0)=x_0$ . Now we get:
$P(x_0,x_0)\rightarrow f(2x_0)=2x_0$
$P(a+x_0,x_0)\rightarrow (f(a+x_0)+x_0)f(a+2x_0)=(a+2x_0)^2$
$P(a,2x_0)\rightarrow (f(a)+2x_0)f(a+2x_0)=(a+2x_0)^2$
$\Rightarrow f(a+x_0)=f(a)+x_0$
$P(a,x_0)\rightarrow (a+x_0)^2=(f(a)+x_0)f(a+x_0)=(f(a)+x_0)^2 \Rightarrow f(a)=a$
So we are done.

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lebathanh

464 posts

lebathanh

#7 h Aug 21, 2016, 1:19 PM • 2 Y

Y by Adventure10, Mango247

from f(1)=1 we sub a=1 in F.E: (b+1)f(1+f(b))=(f(b)+1)^2. from f(1)=1. sub b=1 ==>f(2)=2 . assume f(t)=t. sub b=t==> f(t+1)=t+1

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lebathanh

464 posts

lebathanh

#8 h Aug 21, 2016, 1:37 PM • 2 Y

Y by Adventure10, Mango247

open : we need what codition to f:R-->R (f(0)=0 )

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kun1417

11 posts

kun1417

#9 h Sep 22, 2016, 4:38 PM • 2 Y

Y by Adventure10, Mango247

Quite strange solution:
Let $P(x,y)$ be the assertion $(f(a)+b)f(a+f(b))=(a+f(b))^2$
$P(a,a) \rightarrow (f(a)+a)=f(f(a)+a)$
So there exist $n_0 \in \Bbb{R}^+$ such that $f(n_0)=n_0$ .
$P(n_0,n_0) \rightarrow f(2n_0)=2n_0$
By easy induction we get: $f(kn_0)=kn_0$ for all positive integer $k$
Assume that there exist positive integer $b$ such that $f(b) \ne b$
$P(kn_0,b)$ we get $(kn_0+b)f(kn_0+f(b))=(kn_0+f(b))^2$
Hence: $kn_0+b|(kn_0+f(b))^2$
By a well-known fact, $kn_0+b$ has infinitely many prime factors when $k$ runs, consider one of them is $p$
We have $p|kn_0+b$ and $p|kn_0+f(b)$ so $p|f(b)-b$ . But there are infinitely many $p$ and $f(b)-b \ne 0$ , contradiction!
So, $\boxed{f(x)=x\text{ }\forall x\in\mathbb N}$

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mihajlon

374 posts

mihajlon

#12 h Nov 25, 2016, 3:17 PM • 1 Y

Y by Adventure10

Let $A(a,b)=(f(a)+b)f(a+f(b))=(a+f(b))^2$ be assertion of given FE.

$\begin{align*} A(a,b)\cdot (f(a)+b) \Longleftrightarrow (a+f(b))^2\cdot (f(a)+b)& =(f(a)+b)^2f(a+f(b))\\ &\stackrel{A(b,a)}{=} (f(b)+a)\cdot f(a+f(b))\cdot f(a+f(b))\\ &= (f(b)+a)\cdot f(a+f(b))^2\\ &\Longleftrightarrow (a+f(b))\cdot (f(a)+b)=f(a+f(b))^2\ . \ . \ . \ \bigstar\end{align*}$
Similarly from $A(b,a)\cdot (f(b)+a)$ we get $(a+f(b))\cdot (f(a)+b)=f(b+f(a))^2\ . \ . \ . \ \spadesuit$

$\begin{align*} A(a,b)\cdot f(a+f(b))& \Longrightarrow (a+f(b))^2\cdot f(a+f(b))=(f(a)+b)\cdot f(a+f(b))^2\\ & \stackrel{\bigstar}{=}(f(a)+b)^2\cdot (a+f(b))\\ &\Longleftrightarrow (a+f(b))\cdot f(a+f(b))=(f(a)+b)^2\\ &\stackrel{\cdot (f(a)+b)}{\Longleftrightarrow} (f(a)+b)^3=(f(a)+b)\cdot (a+f(b))\cdot f(a+f(b))\\ & \stackrel{\bigstar}{=}f(a+f(b))^3\\ & \Longleftrightarrow f(a)+b=f(a+f(b))\ . \ . \ . \ \clubsuit\end{align*}$
Similarly we get from $A(b,a)\cdot f(b+f(a))$ and $\spadesuit$ : $f(b)+a=f(b+f(a))\stackrel{\bigstar\text{ and } \spadesuit}{=}f(a+f(b))\stackrel{\clubsuit}{=}f(a)+b:= Q(a,b)$
Now obviously $f$ is injective(Why?) and note that in previous step we've also proved: $f(b+f(a))\stackrel{\bigstar\text{ and } \spadesuit}{=}f(a+f(b))\ .\ .\ . \ \bigstar \bigstar$ .

$\begin{align*} Q(x+y,y)& \Longrightarrow f(y)+x+y=f(x+y)+y\\ &\Longleftrightarrow f(x+y)=f(y)+x\\ & \stackrel{\bigstar \bigstar\text{ and }\clubsuit}{=} f(f(y)+x)\\ &\stackrel{\text{f is injective}}{\Longleftrightarrow}x+y=f(y)+x\\ & \Longleftrightarrow \boxed{f(y)=y,\ \forall y\in \mathbb{N}} \end{align*}$

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PRO2000

239 posts

PRO2000

#13 h Dec 31, 2016, 5:03 AM • 2 Y

Y by Ankoganit, Adventure10

Let $P(x,y)$ denote the assertion that $(f(a)+b)f(a+f(b))=(a+f(b))^2$ for all naturals $a$ and $b$ . $P(1,1)$ gives us that $f(1)+1$ is a fixed point . Letting $a$ be any fixed point , observe that $P(a,a)$ gives us that $2a$ is also a fixed point . As $f(1)+1 \geq 2$ , we can take as large fixed points as we want . At this stage , fix a $b$ arbitrarily and choose a fixed point $x \geq |f(b)-b|^2+2017^{2017}$ . Then, $P(x,b)$ gives us that $(b+x)|(x+f(b)^2)$ which is equivalent to $(b+x)|(f(b)-b)^2$ . Now our tact choice of $x$ forces us that $b=f(b)$ . Now , as our choice of $b$ was arbitrary , we are done .

EDIT: Happy New Year 's Eve to all fellow aops'ers

This post has been edited 1 time. Last edited by PRO2000, Dec 31, 2016, 5:05 AM
Reason: Happy New year

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Dilshodbek

115 posts

Dilshodbek

#14 h Jan 13, 2017, 6:19 AM • 2 Y

Y by Adventure10, Mango247

good problem thanks a lot for solutions

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ltf0501

191 posts

ltf0501

#15 h Jan 15, 2017, 3:41 AM • 2 Y

Y by Adventure10, Mango247

Let $P(x,y)$ be the assertion.
$P(a,a) \to f(a+f(a))=a+f(a)---@$
$P(a,b+f(b)) \to (f(a)+b+f(b))f(a+b+f(b))=(a+b+f(b))^2---(1)$
$P(a+b,b) \to (f(a+b)+b)f(a+b+f(b))=(a+b+f(b))^2---(2)$
By $(1)$ and $(2)$ , we have $f(a+b)=f(a)+f(b)$
Then $@$ becomes $f(f(a))=a$
The original equation becomes
$(f(a)+b)^2=(a+f(b))^2 \Rightarrow f(a)-a=constant$
But $f(a+b)=f(a)+f(b) \Rightarrow f(a)=a \forall a \in \mathbb{N}$

This post has been edited 1 time. Last edited by ltf0501, Jan 15, 2017, 3:41 AM

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nikolapavlovic

1246 posts

nikolapavlovic

#16 h Mar 4, 2017, 10:20 AM • 2 Y

Y by Adventure10, Mango247

Let $a,b$ be two fixed points $P(a,b):f(a+b)=(a+b)$ and hence the set of fixed point is closed under addition. $P(f(a),b): (f(f(a)+b)(f(a)+f(b))=(f(a)+f(b))^2$ switching $a\rightarrow b$ in the previous we have that $f(f(a))-a=\Delta$ . $P(a,a) \implies$ $f(a)+a$ is fixed point and so $\Delta=0$ $\implies$ f is an involution.This allows a re-write ( $b\rightarrow f(b)$ )
$f(a+b)=\frac{(a+b)^2}{f(a)+f(b)}$ .Plugging $a=b=1$ gives $f(2)f(1)=2$ and so $f(1)+f(2)=3$ .Plugging a=2,b=1 $f(3)(f(2)+f(1))=9$ and hence $f(3)=3$ ,plugging a=3,b=1 gives $f(4)(f(3)+f(1))=16$ .If $f(1)=2$ this would imply $5\mid 16$ a clear contradiction and so $f(1)=1$ and $f(2)=2$ .But as linear combination of fixed points is a fixed point and $(2,1)=1$ by Bezout's we get the $f(n)=n$ $\forall n$ .

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WizardMath

2487 posts

WizardMath

#17 h Mar 4, 2017, 1:42 PM • 1 Y

Y by Adventure10

My solution.
Let $a=f(1)$ , and let $P(a,b)$ denote the FE.
Firstly observe f is injective, for if $f(x)=f(y)$ , then $P(z,x), P(y,x)$ give $x=y$ .
$P(z-a,1)$ gives $f(z-a) +1 = z^2/f(z) \ \ \ *$ .
If $z$ is a prime, we can observe that $f(z)$ can only be $1$ or $z$ or $z^2$ . For large $z$ , the first and the last dont hold true, the first one by injectivity and the second one by $*$ . Call such primes as a-primes.
By $*$ again, for a-primes $p$ , $f(p-a)=p-1$ . Let $q,r$ be a-primes, $q>r$ . By $P(q-r, r)$ , $f(q-r)=q-r$ . $P(p-a, q-r)$ for an a-prime $p$ gives that
$p+q-r-1|(p-a+q-r)^2$ , or what is equivalent, $(a-1)^2 \equiv 0 \pmod { (p+q-r-1)}$ . So by making $q$ large enough, $a=1.$ By $*$ and induction, we are done.

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JaN0

10 posts

JaN0

#18 h Jan 17, 2018, 6:02 PM • 1 Y

Y by Adventure10

put a,a; then f(a+f(a))=a+f(a) (1)
put a,b+f(b); then (f(a)+b+f(b))*f(a+b+f(b))=(a+b+f(b))*(a+b+f(b)) (2)
put a+b,f(b); then (f(a+b)+f(b))*f(a+b+f(b))=(a+b+f(b))*(a+b+f(b)) (3)
(2),(3) then f(a+b)=f(a)+b and then easy

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absoluteunit

60 posts

absoluteunit

#46 h Jan 5, 2021, 7:04 AM

Y by

when you solve FEs how do you know you've found all the solutions? In an olympiad is just finding and proving ONE solution enough for full marks?

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A-Thought-Of-God

454 posts

A-Thought-Of-God

#48 h Apr 23, 2021, 6:49 AM • 1 Y

Y by Aimingformygoal

Seems pretty short to me.

Solution

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hakN

429 posts

hakN

#49 h Apr 23, 2021, 2:34 PM

Y by

Let $f(1)=c$ and let $P(a,b)$ denote the assertion.
Let $f(x)=f(y)$ for some $x,y \in \mathbb{N}$ .
Comparing $P(a,x)$ and $P(a,y)$ gives $x=y$ so $f$ is injective.
$P(1,a)\implies (a+f(1))\cdot f(f(a)+1) = (f(a)+1)^2 \implies a+f(1) \mid (f(a)+1)^2$ . $(1)$
$P(1,1)\implies f(c+1) = c+1$ .
Plugging $a=c+1$ in $(1)$ we get $2c+1\mid (c+2)^2 \implies c\in \{1,4\}$ .
If $c=4$ , then we get $f(5)=5$ .
$P(1,5) \implies f(6)=4$ but this contradicts the injectivity of $f$ .
So $f(1)=1$ .
$P(a,1)\implies f(a+1) = \frac{(a+1)^2}{f(a)+1}$ and by a simple induction, we get $f(n)=n \forall n \in \mathbb{N}$ .

This post has been edited 1 time. Last edited by hakN, Apr 23, 2021, 2:41 PM

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Keith50

464 posts

Keith50

#51 h Jul 6, 2021, 9:17 AM

Y by

Answer: $f(n)=n \ \ \forall n\in \mathbb{N}.$
Proof: It's easy to see that the identity function works. Let $P(a,b)$ denote the given assertion,
Claim 1: There is a fixed point.
Proof: $P(a,a)\implies f(a+f(a))=a+f(a). \quad \square$ Claim 2: $f(f(n))=n \ \ \forall n \in \mathbb{N}.$
Proof: We have $P(f(a),1)\implies (f(f(a))+1)f(f(a)+f(1))=(f(a)+f(1))^2$ and $P(f(1),a)\implies (f(f(1))+a)f(f(a)+f(1))=(f(a)+f(1))^2$ which imply $f(f(a))=a+C$ where $C$ is an integer. But from Claim 1, we know there is a fixed point, so $C=0$ and we have proved the claim. $\square$
Claim 3: $f(1)=1.$
Proof: We have $P(1,f(1)) \implies f(1)f(2)=2.$ If $f(2)=1$ and $f(1)=2,$ then $P(2,2) \implies f(3)=3$ but $P(1,3) \implies f(4)=\frac{16}{5}$ which is impossible. So, $f(2)=2$ and $f(1)=1$ .
Claim 4: $f(n)=n \ \ \forall n \in \mathbb{N}.$
Proof: We have $P(a,1) \implies f(a+1)=\frac{(a+1)^2}{f(a)+1}$ and since $f(1)=1,$ by induction, we have $f(a)=a \ \ \forall a\in \mathbb{N}. \quad \blacksquare$

This post has been edited 1 time. Last edited by Keith50, Jul 6, 2021, 9:17 AM

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rama1728

800 posts

rama1728

#52 h Aug 6, 2021, 6:09 PM

Y by

redacted

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Reason: .

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oVlad

1746 posts

oVlad

#53 h Aug 7, 2021, 12:11 PM

Y by

rama1728 wrote:

Solution

@above, I think your solution is wrong.

Indeed, for any $n\in\mathbb{N}_k$ one can find $x$ and $y$ such that $n=x+f(y).$

However, in order for any pair $x,y\in \mathbb{N}_k$ to satisfy $xf(y)=y^2$ then for any $x,y\in \mathbb{N}_k$ there must exist $a$ and $b$ which simultaneously satisfy $b+f(a)=x$ and $a+f(b)=y.$

This clearly does not always hold. For example, take $x=k+1$ and $y=c.$ Since $b+f(a)=x$ and $\forall n, \ f(n)\geq k$ then we must have $b=1.$ But we also have $a+f(b)=a+f(1)=c.$ Thus, any $c\in\mathbb{N}_k$ must be greater than $f(1)$ which doesn't have to hold.

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Hemlock

75 posts

Hemlock

#56 h Aug 7, 2021, 12:22 PM

Y by

oVlad wrote:

wer wrote:

Nice one!

Nice... what?

I think they mean the problem.

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MathLuis

1525 posts

MathLuis

#57 h Mar 23, 2022, 2:17 AM • 2 Y

Y by mathscrazy, mijail

We claim that $f(a)=a$ is the ONLY function that works.
Let $P(a,b)$ the assertion of this F.E.
Claim 1: $f$ is injective
Proof: if there existed $x,y$ such that $f(x)=f(y)$ then by $P(a,x)-P(a,y)$
$f(a)+x=f(a)+y \implies x=y \implies f \; \text{injective}$ Claim 2: $f(p)=p$ for a prime $p$ big enough.
Proof: Set a really big prime $p$ and use $P(p-f(b),b)$
$(f(p-f(b))+b)f(p)=p^2 \implies f(p) \mid p^2$ Now clearly $f(p) \ne p^2$ so either $f(p)=1$ or $f(p)=p$ , but since we have $f$ injective we have that if there existed $p$ big such that $f(p)=1$ then it would be unique, so we ignore thst $p$ and take the rest of big primes $p$ that hold $f(p)=p$ .
Finishing: Now back to the F.E. by $P(p,a)$ where $p$ is such a big prime
$p+a \mid p^2+2pf(a)+f(a)^2 \implies p+a \mid (f(a)-a)(f(a)+a+2p) \implies p+a \mid (f(a)-a)^2 \implies f(a)=a$ Hence the unique function that works is $\boxed{f(a)=a \; \forall a \in \mathbb N}$ thus we are done

This post has been edited 2 times. Last edited by MathLuis, Mar 23, 2022, 2:19 AM

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Mahdi_Mashayekhi

695 posts

Mahdi_Mashayekhi

#58 h Mar 25, 2022, 6:18 AM

Y by

$P(a,a) : f(a+f(a)) = a + f(a)$ .
$P(a,b+f(b)) : (f(a) + b + f(b))f(a+b+f(b)) = (a + b + f(b))^2$ .
$P(a+b,b) : (f(a+b) + b)f(a+b+f(b)) = (a + b + f(b)^2$ .
so we have $(f(a) + b + f(b)) = (f(a+b) + b) \implies f(a+b) = f(a) + f(b)$ so $f$ is a cauchy function which implies $f(a) = ca$ .
$(ca + b)(a + bc) = (a + bc)^2 \implies a + bc = b + ca \implies a(c-1) = b(c-1) \implies c = 1$ so $f(a) = a$ .
Answers : $f(a) = a$ .

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Mahdi_Mashayekhi

695 posts

Mahdi_Mashayekhi

#59 h Jul 9, 2022, 5:30 PM

Y by

another solution
$P(1,1) : (f(1) + 1)f(1+f(1)) = (1+f(1))^2 \implies f(1+f(1)) = 1+f(1)$
so there exists $k$ such that $f(k) = k$
$P(k,k) : f(2k) = 2k$ and $P(2k,k) : f(3k) = 3k$ and now with induction we have $f(nk) = nk$ .
Claim $: f(1) = 1$ .
Assume $f(1) = c$
$P(f(a),k) , P(f(k),a) : f(f(a)) + k = f(f(k)) + a = k + a \implies f(f(a)) = a$
we have $f(1) = c$ so $f(f(1)) = f(c) = 1$ .
$P(1,c) : 2cf(2) = 4 \implies cf(2) = 2$
if $c = 2$ then $f(f(1)+1) = f(3) = f(1) + 1 = 3$ but $P(1,3) : f(4)$ is not Natural so contradiction so $c = 1$ so $f(1) = 1$ .
we had that if $f(k) = k$ then $f(nk) = nk$ so now that $f(1) = 1$ we have $f(n) = n$ .

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ZETA_in_olympiad

2211 posts

ZETA_in_olympiad

#60 h Nov 4, 2022, 6:15 AM • 1 Y

Y by EtHeReAl06

Let $P(x,y)$ denote the assertion $(f(x)+y)f(x+f(y))=(x+f(y))^2$ . We solve this for $f:\mathbb R^+\to \mathbb R^+$ . $P(x,x)$ gives $f(x+f(x))=x+f(x)$ . Comparing $P(x+y,y)$ and $P(x,y+f(y))$ shows $f$ is additive, so $f(x)=kx$ . Checking gives $k=1$ works only.

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VicKmath7

1389 posts

VicKmath7

#61 h Dec 30, 2022, 11:38 PM • 2 Y

Y by teomihai, Tellocan

Firstly, $f$ is injective (otherwise, let $f(x)=f(y)$ and compare $P(a, x), P(a, y)$ ). Now, $P(p-f(x), x)$ implies that $f(p)=p$ for all sufficiently large primes $p$ ( $f(p)=1$ holds for at most one prime due to the injective property) and thus $f(p-f(x))=p-x$ . Take another large prime $q$ and apply this once more: $f(p-q+x)=f(p-f(q-f(x))=p-q+f(x)$ , so $f(x+c)=f(x)+c$ for $p-q=c$ . Now, comparing $P(x+c,y)$ and $P(x, y)$ gives that $(f(x)+y)+f(x+f(y))=2(x+f(y))$ . The last two numbers have product $(x+f(y))^2$ , so $f(x)+y=x+f(y)$ , which implies that $f(x)=x+k$ and substituting back gives that $f(x)=x$ is the only solution.

This post has been edited 4 times. Last edited by VicKmath7, Dec 30, 2022, 11:49 PM

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OZOS

12 posts

OZOS

#63 h Feb 15, 2023, 6:35 PM

Y by

$p(f(a),1),p(f(1),a)\Rightarrow f(f(a))=a+\alpha$ . by comparing $p(a,f(2)),p(a+1,f(1))$ we have $f(a+1)=f(a)+c$ and so f is linear, $f(x)=ax+b(\forall x \in \Bbb{N})$
plugging this in the equation we get $f(x)=x(\forall x)$

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noppi_kun

16 posts

noppi_kun

#64 h Dec 28, 2024, 10:14 AM

Y by

By comparing $P(f(a), b)$ and $P(f(b), a)$ , we deduce that $f(f(a)) + b = a + f(f(b))$ . From this, $f(f(a)) - a$ must take a constant value, which we denote as $c$ .

From $P(a, a)$ , we find $f(a + f(a)) = a + f(a)$ . Therefore, $a + f(a) - c = f(f(a + f(a))) = f(a + f(a)) = a + f(a)$ , which implies $c = 0$ . Thus, $f(f(a)) = a$ , and $f$ is bijective.

Now, consider $P(a, f(b))$ . Notice that $(f(a) + f(b)) f(a + b) = (a + b)^2$ .
From $P(1, f(1))$ , we have $f(1) f(2) = 2$ .
Assume $f(1) = 2$ and $f(2) = 1$ .

Using $P(1, f(2))$ , we find $f(3) = 3$ . Next, from $P(1, f(3))$ , we find $f(4) = \frac{16}{5}$ , which is invalid. Therefore, $f(1) = 1$ .

Assume $f(n - 1) = n - 1$ . From $P(1, f(n - 1))$ , we deduce $n f(n) = n^2$ , which implies $f(n) = n$ .

By induction, we have shown that $f(n) = n$ for all $n$ .

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FredAlexander

22 posts

FredAlexander

#65 h Apr 15, 2025, 1:00 PM • 3 Y

Y by Yunis019, MuradSafarli, Nuran2010

Put in $P(a, a)$ , we get $f(a+f(a))=a+f(a)$ . Then by $P(a+f(a), b)$ it implies $(a+f(a)+b)f(a+f(a)+f(b))=(a+f(a)+f(b))^2$ . This means $a+f(a)+b|(a+f(a)+f(b))^2$ and $a+f(a)+b|(f(b)-b)^2$ take $a$ arbitrarily large we get that $f(b)=b$ , done.

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