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jlacosta   0
May 1, 2025
May is an exciting month! National MATHCOUNTS is the second week of May in Washington D.C. and our Founder, Richard Rusczyk will be presenting a seminar, Preparing Strong Math Students for College and Careers, on May 11th.

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0 replies
jlacosta
May 1, 2025
0 replies
Incircle in an isoscoles triangle
Sadigly   3
N a minute ago by Diamond-jumper76
Source: own
Let $ABC$ be an isosceles triangle with $AB=AC$, and let $I$ be its incenter. Incircle touches sides $BC,CA,AB$ at $D,E,F$, respectively. Foot of altitudes from $E,F$ to $BC$ are $X,Y$ , respectively. Rays $XI,YI$ intersect $(ABC)$ at $P,Q$, respectively. Prove that $(PQD)$ touches incircle at $D$.
3 replies
Sadigly
Friday at 9:21 PM
Diamond-jumper76
a minute ago
Canadian MO 2021 P4
MortemEtInteritum   24
N a minute ago by Thapakazi
A function $f$ from the positive integers to the positive integers is called Canadian if it satisfies $$\gcd\left(f(f(x)), f(x+y)\right)=\gcd(x, y)$$for all pairs of positive integers $x$ and $y$.

Find all positive integers $m$ such that $f(m)=m$ for all Canadian functions $f$.
24 replies
MortemEtInteritum
Mar 12, 2021
Thapakazi
a minute ago
Probably a good lemma
Zavyk09   3
N 8 minutes ago by MathLuis
Source: found when solving exercises
Let $ABC$ be a triangle with circumcircle $\omega$. Arbitrary points $E, F$ on $AC, AB$ respectively. Circumcircle $\Omega$ of triangle $AEF$ intersects $\omega$ at $P \ne A$. $BE$ intersects $CF$ at $I$. $PI$ cuts $\Omega$ and $\omega$ at $K, L$ respectively. Construct parallelogram $KFRE$. Prove that $A, R, P$ are collinear.
3 replies
Zavyk09
Today at 12:50 PM
MathLuis
8 minutes ago
Bushy and Jumpy and the unhappy walnut reordering
popcorn1   53
N 10 minutes ago by lksb
Source: IMO 2021 P5
Two squirrels, Bushy and Jumpy, have collected 2021 walnuts for the winter. Jumpy numbers the walnuts from 1 through 2021, and digs 2021 little holes in a circular pattern in the ground around their favourite tree. The next morning Jumpy notices that Bushy had placed one walnut into each hole, but had paid no attention to the numbering. Unhappy, Jumpy decides to reorder the walnuts by performing a sequence of 2021 moves. In the $k$-th move, Jumpy swaps the positions of the two walnuts adjacent to walnut $k$.

Prove that there exists a value of $k$ such that, on the $k$-th move, Jumpy swaps some walnuts $a$ and $b$ such that $a<k<b$.
53 replies
popcorn1
Jul 20, 2021
lksb
10 minutes ago
Ez comb proposed by ME
IEatProblemsForBreakfast   1
N 4 hours ago by n1g3r14n
A and B play a game on two table:
1.At first one table got $n$ different coloured marbles on it and another one is empty
2.At each move player choose set of marbles that hadn't choose either players before and all chosen marbles from same table, and move all the marbles in that set to another table
3.Player who can not move lose
If A starts and they move alternatily who got the winning strategy?
1 reply
1 viewing
IEatProblemsForBreakfast
Today at 9:02 AM
n1g3r14n
4 hours ago
geometry
luckvoltia.112   0
4 hours ago
ChGiven an acute triangle ABC inscribed in circle $(O)$ The altitudes $BE, CF$ , intersect
each other at $H$. The tangents at $B$ and $C $of $(O)$ intersect at $S$. Let $M $be the midpoint of $BC$. $EM$ intersects $SC$
at $I$, $FM$ intersects $SB$ at $J.$
a) Prove that the points $I, S, M, J$ lie on the same circle.
b) The circle with diameter $AH$ intersects the circle $(O)$ at the second point $T.$ The line $AH$ intersects
$(O)$ at the second point $K$. Prove that $S,K,T$ are collinear.
0 replies
luckvoltia.112
4 hours ago
0 replies
Exponents of integer question
Dheckob   4
N 4 hours ago by LeYohan
Find the smallest positive integer $m$ such that $5m$ is an exact 5th power, $6m$ is an exact 6th power, and $7m$ is an exact 7th power.
4 replies
Dheckob
Apr 12, 2017
LeYohan
4 hours ago
ISI 2025
Zeroin   0
4 hours ago
Let $\mathbb{N}$ denote the set of natural numbers and let $(a_i,b_i),1 \leq i \leq 9$ denote $9$ ordered pairs in $\mathbb{N} \times \mathbb{N}$. Prove that there exist $3$ distinct elements in the set $2^{a_i}3^{b_i}$ for $1 \leq i \leq 9$ whose product is a perfect cube.
0 replies
Zeroin
4 hours ago
0 replies
Inequalities
sqing   3
N 5 hours ago by sqing
Let $ a,b>0   $ . Prove that
$$ \frac{a}{a^2+a +2b+1}+ \frac{b}{b^2+2a +b+1}  \leq  \frac{2}{5} $$$$ \frac{a}{a^2+2a +b+1}+ \frac{b}{b^2+a +2b+1}  \leq  \frac{2}{5} $$
3 replies
sqing
May 13, 2025
sqing
5 hours ago
Pertenacious Polynomial Problem
BadAtCompetitionMath21420   5
N 6 hours ago by BadAtCompetitionMath21420
Let the polynomial $P(x) = x^3-x^2+px-q$ have real roots and real coefficients with $q>0$. What is the maximum value of $p+q$?

This is a problem I made for my math competition, and I wanted to see if someone would double-check my work (No Mike allowed):

solution
Is this solution good?
5 replies
BadAtCompetitionMath21420
Yesterday at 3:13 AM
BadAtCompetitionMath21420
6 hours ago
Max and min of ab+bc+ca-abc
Tiira   5
N 6 hours ago by sqing
a, b and c are three non-negative reel numbers such that a+b+c=1.
What are the extremums of
ab+bc+ca-abc
?
5 replies
Tiira
Jan 29, 2021
sqing
6 hours ago
2017 DMI Individual Round - Downtown Mathematics Invitational
parmenides51   14
N Today at 11:39 AM by SomeonecoolLovesMaths
p1. Compute the smallest positive integer $x$ such that $351x$ is a perfect cube.


p2. A four digit integer is chosen at random. What is the probability all $4$ digits are distinct?


p3. If $$\frac{\sqrt{x + 1}}{\sqrt{x}}+ \frac{\sqrt{x}}{\sqrt{x + 1}} =\frac52.$$Solve for $x$.


p4. In $\vartriangle ABC$, $AB = 13$, $BC = 14$, and $AC = 15$. Let $D$ be the point on $BC$ such that $AD \perp BC$, and let $E$ be the midpoint of $AD$. If $F$ is a point such that $CDEF$ is a rectangle, compute the area of $\vartriangle AEF$.


p5. Square $ABCD$ has a sidelength of $4$. Points $P$, $Q$, $R$, and $S$ are chosen on $AB$, $BC$, $CD$, and $AD$ respectively, such that $AP$, $BQ$, $CR$, and $DS$ are length $1$. Compute the area of quadrilateral $P QRS$.


p6. A sequence $a_n$ satisfies for all integers $n$, $$a_{n+1} = 3a_n - 2a_{n-1}.$$If $a_0 = -30$ and $a_1 = -29$, compute $a_{11}$.


p7. In a class, every child has either red hair, blond hair, or black hair. All but $20$ children have black hair, all but $17$ have red hair, and all but $5$ have blond hair. How many children are there in the class?


p8. An Akash set is a set of integers that does not contain two integers such that one divides the other. Compute the minimum positive integer $n$ such that the set $\{1, 2, 3, ..., 2017\}$ can be partitioned into n Akash subsets.


PS. You should use hide for answers. Collected here.
14 replies
parmenides51
Oct 2, 2023
SomeonecoolLovesMaths
Today at 11:39 AM
Range of a function
Pscgylotti   1
N Today at 9:24 AM by Mathzeus1024
Try to get the range of function $f(x)=cosx+\sqrt{cos^{2}x-4\sqrt{2}cosx+4sinx+9}$ :
1 reply
Pscgylotti
Jul 22, 2019
Mathzeus1024
Today at 9:24 AM
Inequalities
sqing   17
N Today at 9:05 AM by sqing
Let $ a,b,c>0 , a+b+c +abc=4$. Prove that
$$ \frac {a}{a^2+2}+\frac {b}{b^2+2}+\frac {c}{c^2+2} \leq 1$$Let $ a,b,c>0 , ab+bc+ca+abc=4$. Prove that
$$ \frac {a}{a^2+2}+\frac {b}{b^2+2}+\frac {c}{c^2+2} \leq 1$$
17 replies
sqing
May 15, 2025
sqing
Today at 9:05 AM
Find all functions
WakeUp   21
N Apr 30, 2025 by CrazyInMath
Source: Baltic Way 2010
Let $\mathbb{R}$ denote the set of real numbers. Find all functions $f:\mathbb{R}\rightarrow\mathbb{R}$ such that
\[f(x^2)+f(xy)=f(x)f(y)+yf(x)+xf(x+y)\]
for all $x,y\in\mathbb{R}$.
21 replies
WakeUp
Nov 19, 2010
CrazyInMath
Apr 30, 2025
Find all functions
G H J
G H BBookmark kLocked kLocked NReply
Source: Baltic Way 2010
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WakeUp
1347 posts
#1 • 4 Y
Y by jhu08, Adventure10, Mango247, and 1 other user
Let $\mathbb{R}$ denote the set of real numbers. Find all functions $f:\mathbb{R}\rightarrow\mathbb{R}$ such that
\[f(x^2)+f(xy)=f(x)f(y)+yf(x)+xf(x+y)\]
for all $x,y\in\mathbb{R}$.
This post has been edited 1 time. Last edited by WakeUp, Nov 19, 2010, 8:18 PM
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pco
23515 posts
#2 • 7 Y
Y by Abdollahpour, jhu08, Adventure10, Mango247, and 3 other users
WakeUp wrote:
Let $R$ denote the set of real numbers. Find all functions $f:\mathbb{R}\rightarrow\mathbb{R}$ such that
\[f(x^2)+f(xy)=f(x)f(y)+yf(x)+xf(x+y)\]
for all $x,y\in\mathbb{R}$.
Let $P(x,y)$ be the assertion $f(x^2)+f(xy)=f(x)f(y)+yf(x)+xf(x+y)$

$P(0,x)$ $\implies$ $f(0)(f(x)+x-2)$

If $f(0)\ne 0$, this implies $f(x)=2-x$ which indeed is a solution.

Let us from know consider that $f(0)=0$

$P(x,0)$ $\implies$ $f(x^2)=xf(x)$
Then : $P(x,y)$ $\implies$ $xf(x)+f(xy)=f(x)f(y)+yf(x)+xf(x+y)$
Same : $P(y,x)$ $\implies$ $yf(y)+f(xy)=f(x)f((y)+xf(y)+yf(x+y)$
Subtracting implies $(x-y)f(x+y)-f(x)-f(y))=0$

and so $f(x+y)=f(x)+f(y)$ $\forall x\ne y$

Plugging this in $xf(x)+f(xy)=f(x)f(y)+yf(x)+xf(x+y)$, we get $f(xy)=f(x)f(y)+yf(x)+xf(y)$ $\forall x\ne y$

$\iff$ $f(xy)+xy=(f(x)+x)(f(y)+y)$

Let then $g(x)=f(x)+x$. We got :
$g(0)=0$
$g(x+y)=g(x)+g(y)$ $\forall x\ne y$
$g(xy)=g(x)g(y)$ $\forall x\ne y$

From the first, we get $g(-x)=-g(x)$ and so $g(2x+(-x))=g(2x)+g(-x)$ and so $g(2x)=2g(x)$ and so $g(x+y)=g(x)+g(y)$ $\forall x,y$
From the second, we get $g(x(x+1))=g(x)g(x+1)=g(x)^2+g(x)$
But also $g(x(x+1))=g(x^2+x)=g(x^2)+g(x)$
And so $g(x^2)=g(x)^2$ and so $g(xy)=g(x)g(y)$ $\forall x,y$

So :
$g(x+y)=g(x)+g(y)$ $\forall x,y$
$g(xy)=g(x)g(y)$ $\forall x,y$
And so, very classical, $g(x)=x$ and $f(x)=0$

Hence the two solutions :
$f(x)=2-x$
$f(x)=0$
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WakeUp
1347 posts
#3 • 3 Y
Y by jhu08, Adventure10, Mango247
pco wrote:
From the second, we get $g(x(x+1))=g(x)g(x+1)=g(x)^2+g(x)$
But also $g(x(x+1))=g(x^2+x)=g(x^2)+g(x)$
And so $g(x^2)=g(x)^2$ and so $g(xy)=g(x)g(y)$ $\forall x,y$

Hi pco, could you please explain this part of the solution? Note also $f(x)=-x$ is a solution.
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Solving
185 posts
#4 • 3 Y
Y by jhu08, Adventure10, Mango247
SO im right?

x=y
2f(x^2)=f(x)^2+x(f(x)+f(2x))
2f(0)=f(0)^2
f(0)=0
or
f(0)=1/2
let x=x, y=0
f(x^2)+f(0)=f(x)f(0)+xf(x)
f(0)=1/2
f(x^2)+1/2=f(x)(1/2+x)
f(x)=ax+b
ax^2+b+1/2=(ax/2+ax^2+b/2+bx)
b/2=b+1/2
b=-1
a=2
f(x)=2x-1 is the solution
for
f(0)=0
f(x)=x
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pco
23515 posts
#5 • 3 Y
Y by jhu08, Adventure10, Mango247
WakeUp wrote:
pco wrote:
From the second, we get $g(x(x+1))=g(x)g(x+1)=g(x)^2+g(x)$
But also $g(x(x+1))=g(x^2+x)=g(x^2)+g(x)$
And so $g(x^2)=g(x)^2$ and so $g(xy)=g(x)g(y)$ $\forall x,y$

Hi pco, could you please explain this part of the solution?
Yes, :oops:, I wrote too quickly !
First we can see that $g(x)=0$ is a solution (and so $f(x)=-x$ is indeed !
If $g(x)$ is not the all zero function, let then $u$ such that $g(u)\ne 0$. If $u=1$, choose instead $u=-1$.
Then the second equation gives us $g(u)(g(1)-1)=0$ and so $g(1)=1$

Then $g(x(x+1))=g(x)g(x+1)$ (using second equation since $x\ne x+1$) $=g(x)(g(x)+g(1))=g(x)^2+g(x)$
But $g(x(x+1))=g(x^2+x)=g(x^2)+g(x)$
And so, $g(x^2)=g(x)^2$ and so the second equation $g(xy)=g(x)g(y)$ is also true if $x=y$
...

WakeUp wrote:
Note also $f(x)=-x$ is a solution.
Yes, :oops: $g(x)=0$ is also a solution (I forgot it)

And so :
$f(x)=0$
$f(x)=-x$
$f(x)=2-x$
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borntobeweild
331 posts
#6 • 4 Y
Y by Jerry37284, jhu08, Adventure10, Mango247
This is just about as interesting as a FE can get while still dying to the standard strategies of plugging stuff in, taking cases, and testing. Nevertheless, it was a fun problem.

Reading this won't teach you anything except for what tricks you should have tried
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Onlygodcanjudgeme
18 posts
#7 • 3 Y
Y by jhu08, EHHSW, Adventure10
put $ (x,y) = (0,0) $ then we take that 1) $ f(0) =0 $ 2)$ f(0) =2 $
1) put $ (x,y) = (x,0) $ then we take that $ f(x^2) = x \cdot f(x) $ and this is odd function .
put $ (x,y) = (x,-x) $ then we take that f(-x) = x , f(x) =0
2)put $ (x,y) = (0,x) $ then we take that f(x) = 2-x
so answer is f(x)=0 , f(x) = -x ,f(x) = 2-x
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Jerry37284
46 posts
#8 • 3 Y
Y by jhu08, Adventure10, Mango247
@Onlygodcanjudgeme :I think you got "$\forall x$ : either $f(x)=0$, either $f(x)=-x$" and not "either $f(x)=0$ $\forall x$ , either $f(x)=-x$ $\forall x$ "
This post has been edited 1 time. Last edited by Jerry37284, Dec 10, 2018, 3:34 AM
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Keith50
464 posts
#9 • 3 Y
Y by jhu08, Mango247, Mango247
Let $P(x,y)$ be the given assertion, $P(0,0)\implies 2f(0)=f(0)^2\implies f(0)=0,2$.
If $f(0)=2$, then $P(0,x)\implies 4=2f(x)+2x\implies f(x)=2-x \ \ \forall x\in\mathbb{R}$.
So, if $f(0)=0,$ $P(x,0)\implies f(x^2)=xf(x) \ \ \ (1)$
$P(x,-x)\implies f(x^2)+f(-x^2)=f(x)f(-x)-xf(x) \ \ \ (2)$
Using $P(x,y)$ and $P(-x,-y)$, we can arrive at \[f(x)f(y)+yf(x)+xf(x+y)=f(-x)f(-y)-yf(-x)-xf(-x-y),\]letting $y=0$, we get $f(x)=-f(-x)$ where $x\ne0$, so $f$ is odd.
From $(2)$, using the fact that $f$ is odd, \[f(x)(f(x)+x)=0\implies f(x)=0,-x.\]Now, assume that there exists $a,b\in \mathbb{R}, a,b\ne 0$ such that $f(a)=0$ and $f(b)=-b$, using $(1)$, $f(a^2)=0, f(b^2)=-b^2$,
using $P(a,b)$ we get \[f(ab)=af(a+b).\]If $f(ab)=-ab$, \[-b=f(a+b).\]If $f(a+b)=0$, then $b=0$, a contradiction.
If $f(a+b)=-(a+b)$, then $a=0$, a contradiction. Thus, when $f(ab)=0$, $f(a+b)=0$ as $a=0$ is a contradiction.
Take $P(a,b)$ and $P(b,a)$, subtracting one from another gives \[f(a^2)-f(b^2)=bf(a)+af(a+b)-af(b)-bf(a+b)\]which simplifies to \[b^2=ab\implies b(a-b)=0\]and so $a=b$ since $b\ne 0$ but this means $f(b)=f(a)=0=-b$, a contradiction.
Hence, we have \[f(x)\equiv 0, 2-x, -x\]as solutions and plugging them into the equation, we see that they indeed satisfy. $\blacksquare$
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Kimchiks926
256 posts
#10 • 2 Y
Y by mkomisarova, jhu08
Solved together with @blastoor
Let $P(x,y)$ denote assertion of given functional equation.

Note that $P(0,0)$ gives us $2f(0) = f(0)^2$, which means that $f(0)=0$ or $f(0)=2$. If $f(0) =2$, then $P(0,x)$ gives us:
\begin{align*} 
 f(0) + f(0) = f(0)f(x) + xf(0) \\
4 = 2f(x) + 2x \\
f(x) = 2 -x 
\end{align*}It is easy to check that function $f(x) =2-x$ works.

From now we assume that $f(0) =0$. Note that $P(x,0)$ gives us:
$$ f(x^2) = xf(x) \qquad (1) $$In relation $(1)$ replacing $x$ by $-x$ yields:
$$ f(x^2) = xf(x) = -xf(-x) \implies -f(x) = f(-x) $$Also note that $P(x,-x)$ gives us:
\begin{align*}
f(x^2) + f(-x^2) = f(x)f(-x) -xf(x) \\
f(x^2) -f(x^2) = -f(x)^2 - xf(x) \\ 
f(x)^2 = -xf(x) 
\end{align*}We conclude that $f(x) = 0$ or $f(x) =-x$. Now we are left to escape pointwise trap. Assume that there exist nonzero real numbers $a, b$ such that $f(a) =-a$ and $f(b) = 0 $. Note that $f(b^2) = bf(b) = 0$ and that $P(b,a-b)$ gives us:
\begin{align*} 
f(b^2) +f(b(a-b))= f(b)f(a-b) + (a-b)f(b) + bf(a) \\ 
f(b(a-b)) = -ab 
\end{align*}If $f(b(a-b)) = b^2 -ab$, then $b^2 =0 $, which is contradiction since $b \ne 0$. On another hand id $f(b(a-b)=0=ab$, then one of the numbers $a,b$ is zero, which is again contradiction.

We conclude that $f(x)=0$, $f(x) = 2 -x$, $f(x) =-x$ are only solutions.
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508669
1040 posts
#11 • 1 Y
Y by jhu08
WakeUp wrote:
Let $\mathbb{R}$ denote the set of real numbers. Find all functions $f:\mathbb{R}\rightarrow\mathbb{R}$ such that
\[f(x^2)+f(xy)=f(x)f(y)+yf(x)+xf(x+y)\]for all $x,y\in\mathbb{R}$.

Let $P(x, y)$ be the assertion. We claim that all solutions to the given functional equation are of form $\boxed{f(x) = 2-x}$ for reals $x$, $\boxed{f(x) = -x}$ for all reals $x$ and $\boxed{f(x) = 0}$ for all reals $x$. It is not hard to see they work. Now we show that these are the only such functions.

We see that $P(x, 0) \implies f(0)(f(x) + x - 2) = 0$, so if $f(0) \neq 0$, then $\boxed{f(x) = 2-x}$ which is indeed a solution.

Otherwise, let $f(0) = 0 \dots (1)$. Then by $P(x, 0)$, we yield that $f(x^2) = xf(x) = -xf(-x)$ (by replacing $x$ by $-x$) and so $f$ is odd function. Now $P(x, y) - P(y, x)$ along with $f(x^2) = xf(x)$ gives that $(x-y)(f(x+y)-f(x)-f(y)) = 0$ and so if $x \neq y$, definitely $f(x+y) = f(x) + f(y)$ and so $f$ is additive. Now, we re-arrange few terms in $P(x, y)$.

$P(x, y) \implies f(x^2)+f(xy)=f(x)f(y)+yf(x)+xf(x+y) \implies xf(x) + yf(x) = f(x)f(y) + yf(x) + xf(x) + xf(y) \implies 0 = f(x)f(y) + xf(y) = f(y)(f(x) + x)$ which means that $f(x) = -x$ or $f(x) = 0$ for all reals $x$.

Let us say that $A = \{ x \lvert f(x) = -x, x \neq 0 \}$ and $A = \{ x \lvert f(x) = 0, x \neq 0 \}$. Let $a \in A, b \in B$. We see that $f(ab) = bf(a) + af(b)$, and so here in this case, $f(ab) = b \times -a = a \times 0 = 0$, so either of $a$ or $b$ is $0$, a contradiction to definition of elements belonging to sets $A$ and $B$. Hence, $\lvert A \rvert = 0$ or $\lvert B \rvert = 0$. We see that $f(0) = 0 = -0$. Therefore, we see that all solutions to the given functional equation are of form $\boxed{f(x) = 2-x}$ for reals $x$, $\boxed{f(x) = -x}$ for all positive reals $x$ and $\boxed{f(x) = 0}$ for all reals $x$
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jasperE3
11352 posts
#13 • 1 Y
Y by jhu08
Hint

Let $P(x,y)$ be the assertion $f(x^2)+f(xy)=f(x)f(y)+yf(x)+xf(x+y)$
$P(0,0)\Rightarrow 2f(0)=f(0)^2$
If $f(0)=2$ then:
$P(0,x)\Rightarrow\boxed{f(x)=2-x}$, which works.
Now assume $f(0)=0$.
$P(x,0)\Rightarrow f(x^2)=xf(x)$
$P(1,x)\Rightarrow f(x+1)=f(x)+f(1)-f(x)f(1)-xf(1)$
We use this recurrence to find $f(1)$.
$P(1,1)\Rightarrow f(2)=f(1)-f(1)^2$
$P(1,2)\Rightarrow f(3)=f(2)-f(1)-f(2)f(1)=f(1)^3-2f(1)^2$
$P(1,3)\Rightarrow f(4)=f(3)-2f(1)-f(3)f(1)=-f(1)^4+3f(1)^3-2f(1)^2-2f(1)$
But $f(4)=2f(2)=2f(1)-2f(1)^2$, so we find that $2f(1)-2f(1)^2=-f(1)^4+3f(1)^3-2f(1)^2-2f(1)$. Solving, we have $f(1)\in\{-1,0,2\}$.

$\textbf{Case 1: }f(1)=0$
$P(1,x)\Rightarrow f(x)=f(x+1)$
$P\left(x,\frac yx+1\right)-P\left(x,\frac yx\right)\Rightarrow f(x+y)=f(x)+f(y)$ if $x\ne0$, but since it holds for $x=0$, $f$ is additive.
By USAMO 2002/4, since $f(x^2)=xf(x)$ and $f$ is additive, we must have $f(x)=xf(1)$, hence $\boxed{f(x)=0}$ which works.

$\textbf{Case 2: }f(1)=-1$
$P(1,x)\Rightarrow f(x+1)=2f(x)+x-1$
$P(x,1)\Rightarrow f(x^2)+f(x)=xf(x+1)\Rightarrow xf(x)+f(x)=2xf(x)+x^2-x\Rightarrow\boxed{f(x)=-x}$ since $f(1)=-1$, which works.

$\textbf{Case 3: }f(1)=2$
$P(1,x)\Rightarrow f(x+1)=-f(x)-2x+2$
$P(x,1)\Rightarrow(x-1)f(x)=-x^2+x\Rightarrow f(x)=\begin{cases}-x&\text{if }x\ne1\\2&\text{if }x=1\end{cases}$ which doesn't work.
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RopuToran
609 posts
#14 • 1 Y
Y by jhu08
Here is my way to solve the first case where $f(0)=0$ :D
$P(x,0)$ gives us $$f(x^2) = xf(x),  \forall x \quad (1)$$$P(y,x)$ gives us $$f(y^2) + f(xy)= f(x)f(y)+xf(y)+yf(x+y),  \forall x,y \quad (2)$$By pluging (1) into (2) then subtracting (2) and the original FE, we got $$(x-y) (f(x)+f(y)) = (x-y) f(x+y), \forall x,y$$, which implies $$ f(x) + f(y) = f(x+y), \forall x \neq y \quad (3)$$From $(1)$, we also have $f(x^2)= -x f(x)$ which leads to $$f(x)= f(-x), \forall x \quad (4)$$By $P(x,-x)$ and using $(4)$, we got $$f(x)^2 = -xf(x), \forall x (5)$$Using $(1)$ and $(3)$, from the origina FE, we have $$f(xy) = f(x)f(y)+xf(y) + yf(x), \forall x \neq y$$, which equivalent to $$ f(x)(f(y)+y) = f(xy) - xf(y), \forall x \neq y \quad (6)$$Case 1: There is a number $k \neq 0$ such that $f(k) = -k$. In $(6)$, let $y=k$, we have $f(kx)=-kx, \forall x \neq k$. Thus $f(x) = -x, \forall x \neq k^2$. In the other hand, $f(k^2) = kf(k) = -k^2$. So, $f(x)=-x, \forall x$.
Case 2: There is no number $k$ other than $0$ such that $f(k) = -k$, which means $f(x) \neq -x, \forall x \neq 0$. With $(5)$, we implies $f(x) = 0, \forall x \neq 0$. In the otherhand, $f(0)=0$, thus $f(x) = 0$.

P/S
This post has been edited 1 time. Last edited by RopuToran, Aug 8, 2021, 4:38 PM
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MathLuis
1537 posts
#15 • 1 Y
Y by jhu08
WakeUp wrote:
Let $\mathbb{R}$ denote the set of real numbers. Find all functions $f:\mathbb{R}\rightarrow\mathbb{R}$ such that
\[f(x^2)+f(xy)=f(x)f(y)+yf(x)+xf(x+y)\]for all $x,y\in\mathbb{R}$.

Case 1: $f$ is constant.
We will set $f(x)=c$ where $c$ is some real constant. Plugging this on the F.E. we get:
$$2c=c^2+cy+cx \implies c=0 \implies f(x)=0$$Cade 2: $f$ is non-constant.
Let $P(x,y)$ the assertion of the given F.E.
$P(0,0)$
$$f(0)^2=2f(0) \implies f(0)=0 \; \text{or} \; f(0)=2$$Case 2.1: $f(0)=2$
$P(0,x)$
$$4=2f(x)+2x \implies f(x)=2-x$$Case 2.2: $f(0)=0$
$P(x,0)$
$$f(x^2)=xf(x) \implies f \; \text{odd}$$$P(x,-x)$ where $x$ is any non-cero real
$$f(x)^2+xf(x)=0 \implies f(x)=-x$$Since $f(0)=0$ we have that $f(x)=-x \; \forall x \in \mathbb R$
Thus the solutions are:

$\boxed{f(x)=0 \; \forall x \in \mathbb R}$

$\boxed{f(x)=2-x \; \forall x \in \mathbb R}$

$\boxed{f(x)=-x \; \forall x \in \mathbb R}$

Thus we are done :blush:
This post has been edited 1 time. Last edited by MathLuis, Aug 8, 2021, 6:26 PM
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rama1728
800 posts
#16 • 1 Y
Y by jhu08
WakeUp wrote:
Let $\mathbb{R}$ denote the set of real numbers. Find all functions $f:\mathbb{R}\rightarrow\mathbb{R}$ such that
\[f(x^2)+f(xy)=f(x)f(y)+yf(x)+xf(x+y)\]for all $x,y\in\mathbb{R}$.

A good problem for oddness of a function and how to tackle pointwise traps. Other steps are natural.

Solution
This post has been edited 3 times. Last edited by rama1728, Aug 8, 2021, 7:14 PM
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JustKeepRunning
2958 posts
#17 • 1 Y
Y by jhu08
A nice exercise for pointwise trap!

The answers are $f\equiv 0, 2-x, -x$. These work.

Denote the assertion by $P(x,y)$. $P(0,0)$ gives that $f(0)=0,2$.

Case 1: $f(0)=2$.

$P(0,y)$ gives that $f(y)=2-y.$

Case 2: $f(0)=0$

$P(x,0)$ gives that $f(x^2)=xf(x),$ so we have that $f$ is odd. Then $P(x,-x)$ gives that $0=f(x)(f(x)+x),$ so $f(x)=0,-x$. To avoid pointwise trap, suppose that $f(x)=0$ and $f(y)=-y$ for some $x,y\neq 0$. Then from $P(x,y)$ and $P(y,x)$ and subtracting, we get that $xf(x)-yf(y)=yf(x)-xf(y)+(x-y)f(x+y)$. Obviously, $x\neq y,$ and simplifying gives that $f(x+y)=f(y)=-y$. If $y=0,$ we are done, and if $x+y=y,$ then $x=0,$ and we are done as well.
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RopuToran
609 posts
#18 • 1 Y
Y by jhu08
JustKeepRunning wrote:
Obviously, $x\neq y,$ and simplifying gives that $f(x+y)=f(y)=-y$. If $y=0,$ we are done, and if $x+y=y,$ then $x=0,$ and we are done as well.

How did you simplify the equation into this?
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jasperE3
11352 posts
#19 • 1 Y
Y by jhu08
They used the properties $f(x)=0$ and $f(y)=-y$.
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ZETA_in_olympiad
2211 posts
#20
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Let $P(x,y)$ denote the assertion. Quickly $P(0,x)$ gives $f(x)\equiv 2-x$ or $f(0)=0.$ The former works, so we explore the latter.

$P(x,0)$ gives $f(x^2)=xf(x).$ And so comparing $P(x,y)$ with $P(y,x)$ shows that $f(x+y)=f(x)+f(y)$ for all $x\neq y.$ To conclude $P(x,-x)$ implies $f(x)\in \{0,-x\}$ but since additive $f\equiv 0$ or $f\equiv -x$ and both satisfy.
This post has been edited 1 time. Last edited by ZETA_in_olympiad, Aug 1, 2022, 11:08 AM
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HamstPan38825
8867 posts
#22
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This dies to basically anything.

Setting $x=y=0$ yields $2f(0) = f(0)^2$. If $f(0) = 2$, setting $x=0$ yields $f(x) = 2-x$ immediately.

If $f(0) = 0$, setting $y=0$ yields $f(x^2) = xf(x)$, implying $f$ is odd. Setting $y=-x$ in the original, \[0=f\left(x^2\right) + f\left(-x^2\right) = f(x)f(-x)-xf(x) = -f(x)^2 - xf(x).\]So for each $x$, either $f(x) = 0$ or $f(x) = -x$. There are many ways to resolve the pointwise trap, but here is a really stupid way. By setting $y=x$ we get $f(x)^2 + xf(2x) = xf(x)$, i.e. $xf(2x) = 2xf(x)$ or $f(2x) = f(x)$. Furthermore, by swapping $x$ and $y$, we get \[(x-y)f(x) - xf(x+y)=(y-x)f(y) - yf(x+y)\]so $f(x+y) = f(x)+f(y)$ for all $x \neq y$. Combining this with the previous equation yields that $f$ is Cauchy and bounded below on $x \geq 0$, thus $f$ is linear. We can check that only $f \equiv 0$ works here.
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math-olympiad-clown
32 posts
#23
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case 1. f is a constant function :
c+c=c^2+cy+xc x=y=1 plug in we get c=0
so f(x)=0

case 2. f is not a constant function :
P(0,y) :2f(0)=f(0)f(y)+yf(0)

2-1. if f(0) is not 0 then y=1 plug in we get f(1)=1
P(1,y): 1+f(y)=f(y)+y+f(y+1) and we know that f(y)=2-y

2-2.f(0)=0 : P(0,0): f(x^2)=xf(x) this imply f is a odd function
P(x,-x) : 0=-(f(x)^2)-xf(x) we get f(x)=-x

so the answer is f(x)=0 or 2-x or -x
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CrazyInMath
457 posts
#24
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$f(x)=0$ works, now assume $f$ is not constant
$P(0,0)$ gives $f(0)=0, 2$

If $f(0)=2$, $P(0, x)$ gives $f(x)=2-x$ which works
If $f(0)=0$, $P(x, 0)$ gives $f(x^2)=xf(x)$ so $f$ is odd
then $P(x, -x)$ gives $-f(x)^2-xf(x)=0$ so $f(x)(f(x)-x)=0$, so $f(x)=-x, 0$.
If $f(a)=0$, $f(b)=-b$, by $P(a, b)$ and $P(b, a)$ we have contradiction.
So $f(x)=-x$ or $f(x)=0$ in this case.
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