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Inequality
lgx57   0
38 minutes ago
Source: Own
$a,b,c \in \mathbb{R}^{+}$,$\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1$. Prove that
$$a^abc+b^bac+c^cab \ge 27(ab+bc+ca)$$
0 replies
lgx57
38 minutes ago
0 replies
p divides x^x-c
mistakesinsolutions   6
N 41 minutes ago by reni_wee
Show that for integer c and a prime p, $ p |x^x-c $ has a solution
6 replies
mistakesinsolutions
Jun 13, 2023
reni_wee
41 minutes ago
exponential diophantine in integers
skellyrah   1
N 43 minutes ago by skellyrah
find all integers x,y,z such that $$ 45^x = 5^y + 2000^z $$
1 reply
skellyrah
Yesterday at 7:04 PM
skellyrah
43 minutes ago
IMO 2017 Problem 4
Amir Hossein   117
N an hour ago by ezpotd
Source: IMO 2017, Day 2, P4
Let $R$ and $S$ be different points on a circle $\Omega$ such that $RS$ is not a diameter. Let $\ell$ be the tangent line to $\Omega$ at $R$. Point $T$ is such that $S$ is the midpoint of the line segment $RT$. Point $J$ is chosen on the shorter arc $RS$ of $\Omega$ so that the circumcircle $\Gamma$ of triangle $JST$ intersects $\ell$ at two distinct points. Let $A$ be the common point of $\Gamma$ and $\ell$ that is closer to $R$. Line $AJ$ meets $\Omega$ again at $K$. Prove that the line $KT$ is tangent to $\Gamma$.

Proposed by Charles Leytem, Luxembourg
117 replies
Amir Hossein
Jul 19, 2017
ezpotd
an hour ago
x^2+y^2+z^2+xy+yz+zx=6xyz diophantine
parmenides51   7
N an hour ago by Assassino9931
Source: Greece Junior Math Olympiad 2024 p4
Prove that there are infinite triples of positive integers $(x,y,z)$ such that
$$x^2+y^2+z^2+xy+yz+zx=6xyz.$$
7 replies
parmenides51
Mar 2, 2024
Assassino9931
an hour ago
Turkish JMO 2025?
bitrak   1
N 2 hours ago by blug
Let p and q be prime numbers. Prove that if pq(p+ 1)(q + 1)+ 1 is a perfect square, then pq + 1 is also a perfect square.
1 reply
bitrak
Yesterday at 2:04 PM
blug
2 hours ago
Combi Algorithm/PHP/..
CatalanThinker   0
2 hours ago
Source: Olympiad_Combinatorics_by_Pranav_A_Sriram
5. [Czech and Slovak Republics 1997]
Each side and diagonal of a regular n-gon (n ≥ 3) is colored blue or green. A move consists of choosing a vertex and
switching the color of each segment incident to that vertex (from blue to green or vice versa). Prove that regardless of the initial coloring, it is possible to make the number of blue segments incident to each vertex even by following a sequence of moves. Also show that the final configuration obtained is uniquely determined by the initial coloring.
0 replies
CatalanThinker
2 hours ago
0 replies
Combi Proof Math Algorithm
CatalanThinker   0
2 hours ago
Source: Olympiad_Combinatorics_by_Pranav_A_Sriram
3. [Russia 1961]
Real numbers are written in an $m \times n$ table. It is permissible to reverse the signs of all the numbers in any row or column. Prove that after a number of these operations, we can make the sum of the numbers along each line (row or column) nonnegative.
0 replies
CatalanThinker
2 hours ago
0 replies
Easy Taiwanese Geometry
USJL   14
N 3 hours ago by Want-to-study-in-NTU-MATH
Source: 2024 Taiwan Mathematics Olympiad
Suppose $O$ is the circumcenter of $\Delta ABC$, and $E, F$ are points on segments $CA$ and $AB$ respectively with $E, F \neq A$. Let $P$ be a point such that $PB = PF$ and $PC = PE$.
Let $OP$ intersect $CA$ and $AB$ at points $Q$ and $R$ respectively. Let the line passing through $P$ and perpendicular to $EF$ intersect $CA$ and $AB$ at points $S$ and $T$ respectively. Prove that points $Q, R, S$, and $T$ are concyclic.

Proposed by Li4 and usjl
14 replies
USJL
Jan 31, 2024
Want-to-study-in-NTU-MATH
3 hours ago
Problem 7
SlovEcience   6
N 3 hours ago by Li0nking
Consider the sequence \((u_n)\) defined by \(u_0 = 5\) and
\[
u_{n+1} = \frac{1}{2}u_n^2 - 4 \quad \text{for all } n \in \mathbb{N}.
\]a) Prove that there exist infinitely many positive integers \(n\) such that \(u_n > 2020n\).

b) Compute
\[
\lim_{n \to \infty} \frac{2u_{n+1}}{u_0u_1\cdots u_n}.
\]
6 replies
SlovEcience
May 14, 2025
Li0nking
3 hours ago
easy functional
B1t   13
N Apr 30, 2025 by AshAuktober
Source: Mongolian TST 2025 P1.
Denote the set of real numbers by $\mathbb{R}$. Find all functions $f: \mathbb{R} \to \mathbb{R}$ such that for all $x, y, z \in \mathbb{R}$,
\[
f(xf(x+y)+z) = f(z) + f(x)y + f(xf(x)).
\]
13 replies
B1t
Apr 26, 2025
AshAuktober
Apr 30, 2025
easy functional
G H J
G H BBookmark kLocked kLocked NReply
Source: Mongolian TST 2025 P1.
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B1t
24 posts
#1 • 1 Y
Y by farhad.fritl
Denote the set of real numbers by $\mathbb{R}$. Find all functions $f: \mathbb{R} \to \mathbb{R}$ such that for all $x, y, z \in \mathbb{R}$,
\[
f(xf(x+y)+z) = f(z) + f(x)y + f(xf(x)).
\]
This post has been edited 2 times. Last edited by B1t, Apr 26, 2025, 7:01 AM
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NicoN9
158 posts
#2
Y by
$f(x)f(x)$ means $f(x)^2$? or is it a typo?
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Ilikeminecraft
664 posts
#3
Y by
$x = y = z = 0$ implies $f(0) = 0$
$x = -y$ implies $f(z) = f(z) - xf(x) + f(x)^2,$ so $f\in\{0, x\}.$
assume $f(a ) = 0, f(b ) = b.$
take $y = 0,$ and we get $f(xf(x) + z) = f(z) + f(x)^2.$
if we take $x = b, y = a - b, z = a,$ we get $0 = ab.$ this implies $f \equiv 0, x.$
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B1t
24 posts
#5
Y by
NicoN9 wrote:
$f(x)f(x)$ means $f(x)^2$? or is it a typo?

I wrote it wrong. im sorry
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Ilikeminecraft
664 posts
#6
Y by
suppose $f$ is not identically $0$
suppose $x_0$ satisfies $f(x_0)\neq 0,$ and if we pick $x = x_0$ and range the values of $y,$ we obtain that $f$ is surjective. ignore the definition of $x_0$
take $y = 0$ and we get $f(xf(x) + z) = f(xf(x)) + f(z)$
take $x = - y$ yields $f(xf(x))=xf(x).$
thus, $f(f(xf(x)) + z) = f(z) + f(f(xf(x)))$ which implies that the function is cauchy
plugging back into original equation, one gets $f(xf(y)) = f(x)y$
pick $x = 1$ to get $f(f(y)) = y,$ which is well-known to imply that $|f(y)| = |y|.$

assume $A$ is the set of $x$ such that $f(x) = x$ and $B$ is the set of $x$ such that $f(x) = -x.$
if $x, y \in A,$ then $f(xy) = xy,$ so $xy \in A$
if $x\in A, y\in B,$ then $f(xy) = -xy,$ so $xy\in B$
if $x \in B, y \in A,$ then $f(xy) = -xy,$ so $xy \in B$
if $x \in B, y\in B,$ then $f(xy) = xy,$ so $xy\in A$

this implies $f$ is multiplicative
hence, since $f$ is both multiplicative and additive, $f$ is identity
This post has been edited 1 time. Last edited by Ilikeminecraft, Apr 26, 2025, 3:22 PM
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Haris1
77 posts
#7
Y by
I wont write the solution , i will just write the steps.
$1.$ Prove that the function is additive
$2.$ Prove that its either constant or bijective
$3.$ Prove that its multiplicative and finish
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cazanova19921
555 posts
#8 • 1 Y
Y by farhad.fritl
B1t wrote:
Denote the set of real numbers by $\mathbb{R}$. Find all functions $f: \mathbb{R} \to \mathbb{R}$ such that for all $x, y, z \in \mathbb{R}$,
\[
P(x,y,z):\, f(xf(x+y)+z) = f(z) + f(x)y + f(xf(x)).
\]
$P(0,0,0)$ $\implies$ $f(0)=0$.
$P(x,-x,0)$ $\implies$ $f(xf(x))=xf(x)$
So $P(x,y-x,z): \, f(xf(y)+z)=f(z)+yf(x)$ (new $P$)
- If $f(t)=0$ for some $t \neq 0$, then $P(x,t,0)$ $\implies$ $\boxed{f=0}$ which is a solution.
- Suppose $f(t)=0$ $\iff$ $t=0$.
$P(1,x,0)$ $\implies$ $f(f(x))=xf(1)$. hence $f$ is bijective, replace $x=1$ in this equation, we get $f(f(1))=f(1)$ so $f(1)=1$.
Therefore $f(f(x))=x$ for all $x$.
$P(x, f(y), 0)$ $\implies$ $f(xy)=f(x)f(y)$ for all $x, y$
$P(x, 1, y)$ $\implies$ $f(x+y)=f(x)+f(y)$ for all $x, y$
So $f$ is additive and multiplicative, hence $\boxed{f=\mathrm{Id}}$ which is also a valid solution.
This post has been edited 1 time. Last edited by cazanova19921, Apr 26, 2025, 1:55 PM
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MathLuis
1556 posts
#9
Y by
Denote $P(x,y,z)$ the assertion of the given F.E.
Notice from $P(0,y,z)$ we can inmediately get $f(0)=0$ and also $P(x,-x,z)$ gives that $f(xf(x))=xf(x)$ but also $P(x,y,0)$ now gives that $f(xf(x+y))=(x+y)f(x)$ so in fact shifting gives $Q(x,y)$ which is $f(xf(y))=yf(x)$ now notice if ther existed some $c \ne 0$ for which $f(c)=0$ then $Q(x,c)$ gives that $f(x)=0$ for all reals $x$ so its either that or $f$ is injective at zero, but now basically note that our F.E. may now by re-written as $R(x,y,z)$ to be $f(xf(y)+z)=f(z)+yf(x)$ for all reals $x,y,z$ but also take $y \ne 0$ and shift $x$ to get that $f$ is additive, but also from $Q(x, f(x))$ we get that $f(f(1)x^2)=f(x)^2$ and therefore shifting $x \to f(x)$ gives $f(f(1)f(x)^2)=f(1)^2 \cdot x^2$ so for example $f$ is surjective on all non-negative reals also from $Q(x,y)$ we have $f$ injective trivially when setting $x \ne 0$ and this take $f(d)=1$ and $Q(x,d)$ to get that $d=1$ and thus $f(1)=1$ so $Q(1,x)$ gives $f$ is an involution so by $Q(x,f(y))$ we get $f$ multiplicative so addivitive+multiplicative means $f$ is the identity function or constant (later case gives $f$ is zero everywhere so doesn't count), so $f(x)=0,x$ for all reals $x$ are the only two solutions that work thus we are done :cool:.
This post has been edited 1 time. Last edited by MathLuis, Apr 27, 2025, 3:32 PM
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jasperE3
11385 posts
#10
Y by
B1t wrote:
Denote the set of real numbers by $\mathbb{R}$. Find all functions $f: \mathbb{R} \to \mathbb{R}$ such that for all $x, y, z \in \mathbb{R}$,
\[
f(xf(x+y)+z) = f(z) + f(x)y + f(xf(x)).
\]

Let $P(x,y)$ be the assertion $f(xf(x+y)+z)=f(z)+yf(x)+f(xf(x))$.
Note that $\boxed{f(x)=0}$ is a solution, else there is some $j$ with $f(j)\ne0$.
$P(0,0,0)\Rightarrow f(0)=0\Rightarrow j\ne0$
$P(j,-j,0)\Rightarrow f(jf(j))=jf(j)$
$P(j,0,x)\Rightarrow f(x+jf(j))=f(x)+jf(j)$
$P(x,y+jf(j),z)\Rightarrow f(xf(x+y)+jf(j)x+z)=f(z)+yf(x)+f(xf(x))+jf(j)f(x)$
Comparing this last line with $P(x,y)$ we get:
$$f(xf(x+y)+jf(j)x+z)=f(xf(x+y)+z)+jf(j)f(x)$$and setting $z=-xf(x+y)$ and $x=1$ this is $jf(j)(f(1)-1)=0$, so $f(1)=1$.
$P(1,x-1,y)\Rightarrow f(f(x)+y)=x+f(y)$
Call this assertion $Q(x,y)$.
$Q(x,0)\Rightarrow f(f(x))=x$
$Q(f(x),y)\Rightarrow f(x+y)=f(x)+f(y)$
$P(x,0,y)\Rightarrow f(xf(x)+y)=f(xf(x))+f(y)$
$P(x,-x,0)\Rightarrow f(xf(x))=xf(x)$
Now $P(x,y)$ becomes:
\begin{align*}
xf(x)+f(z)+f(xf(y))&=f(xf(x))+f(z)+f(xf(y))\\
&=f(xf(x)+xf(y)+z)\\
&=f(xf(x+y)+z)\\
&=f(z)+yf(x)+f(xf(x))\\
&=f(z)+yf(x)+xf(x)
\end{align*}and so $f(xf(y))=yf(x)$. Taking $y\mapsto f(y)$ we have $f(xy)=f(x)f(y)$, well-known that the only additive and multiplicative functions are $\boxed{f(x)=x}$ and $\boxed{f(x)=-x}$ which both work.
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GreekIdiot
269 posts
#11
Y by
why is mongolian tst so easy?
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B1t
24 posts
#12
Y by
GreekIdiot wrote:
why is mongolian tst so easy?

true
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MuradSafarli
112 posts
#13 • 2 Y
Y by B1t, Nuran2010
Interesting problem!

Let \( P(x, -x, y) \) denote the assertion of the functional equation:
\[
xf(x) = f(xf(x)) \tag{1}
\]
Now, consider the following:

- \( P(1, -1, x) \) gives:
\[
f(1) = f(f(1)).
\]- \( P(1, x-1, 0) \) gives:
\[
f(f(x)) = x \cdot f(1). \tag{2}
\]
Now, we consider two cases:

---

**Case 1:** \( f(1) = 0 \)

From equation (2), we have:
\[
f(f(x)) = 0 \quad \text{for all } x.
\]
Applying \( f \) to both sides of equation (1):
\[
xf(x) = f(xf(x)) = f(f(xf(x))) = 0,
\]thus implying:
\[
f(x) = 0 \quad \text{for all } x.
\]
---

**Case 2:** \( f(1) \neq 0 \)

From equation (2), we can conclude that \( f \) is bijective.

Suppose there exists some \( k \) such that \( f(k) = 1 \).
Applying \( P(k, -k, x) \) gives:
\[
k = 1,
\]thus \( k = 1 \).

Moreover, from (2), we obtain:
\[
f(f(x)) = x.
\]
Now, consider \( P(1, f(x) - 1, y) \). We get:
\[
f(x) + f(y) = f(x+y),
\]meaning \( f \) is **additive**.

Since \( f \) is additive and satisfies \( f(f(x)) = x \), we deduce that \( f(x) = cx \) for some constant \( c \). Substituting back into the original functional equation shows that \( c = 1 \), and thus:
\[
f(x) = x.
\]
---

**Final answer:**
1. \( f(x) = 0 \) for all real numbers \( x \), or
2. \( f(x) = x \) for all real numbers \( x \).
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complex2math
7 posts
#14
Y by
Denote by $P(x, y, z)$ the assertion of the given functional equation. Then $P(0, 0, 0)$ and $P(x, -x, 0)$ gives the following

Claim 1. $f(0) = 0$ and $f(xf(x)) = xf(x)$.

Now we can rewrite the given functional equation as
\[
f(xf(x + y) + z) = f(z) + f(x)(x + y) \qquad (\heartsuit)
\]and let $P(x, y, z)$ re-denote the assertion of equation $(\heartsuit)$ and set $f(1) = a$.

Claim 2. If $a = 0$, then $f(x) \equiv 0$.

Proof. $P(x, 1 - x, z)$ implies $f(z) = f(ax + z) = f(z) + f(x)$, so $f(x) = 0$ for all $x \in \mathbb{R}$.

In what follows, we always assume $a \ne 0$.

Claim 3. $f(x)$ is bijective when $a \ne 0$.

Proof. From $P(1, y, 0)$ we get $f(f(1 + y)) = a(y + 1) \implies f(f(y)) = ay$. This is a bijection whenever $a \ne 0$.

Claim 4. $a = f(1) = 1$ and $f(x)$ is additive, i.e. $f(x + y) = f(x) + f(y)$ holds.

Proof. Since $xf(x) = f(xf(x))$, substituting $x = 1$ we obtain $f(1) = f(f(1)) \implies f(1) = 1$ as $f$ is injective. Then $P(x, 1 - x, z)$ gives $f(x + z) = f(z) + f(x)$.

Claim 5. $f(x)$ is multiplicative, i.e. $f(xy) = f(x)f(y)$ holds.

Proof. We have $f(f(y)) = ay = y$ so $P(x, y, 0)$ gives $f(xf(x + y)) = f(x)(x + y) = f(x)\cdot f(f(x + y))$. Then note that
\[
S_x := \{f(x + y): y \in \mathbb{R}\} = \mathbb{R}
\]for fixed $x \in \mathbb{R}$ since $f$ is surjective.

It's well-known that if $f(x)$ is both additive and multiplicative, then either $f(x) \equiv 0$ or $f(x) = x$.
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AshAuktober
1009 posts
#15
Y by
The claims I made in order to solve:
1) $f(0) = 0$.
2) $f(xf(x)) = xf(x)$.
3) $f(xf(y)) = yf(x)$.
4) Either $f \equiv 0$ or $f(x) = 0 \implies x = 0$.
(Now onwards taking second case...)
5) $f$ is an involution
6) $f$ is additive
7) $f$ is multiplicative
And done!
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