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k a March Highlights and 2025 AoPS Online Class Information
jlacosta   0
Mar 2, 2025
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0 replies
jlacosta
Mar 2, 2025
0 replies
number theory
nguyenlehoang   0
a few seconds ago
Prove that if $k > 1$ is an integer then there are infinitely many positive
integers $n$ such that $n | k^{n} + 1$.
0 replies
nguyenlehoang
a few seconds ago
0 replies
Interesting inequality
sqing   7
N 6 minutes ago by sqing
Source: Own
Let $ a,b> 0$ and $ a+b=1 . $ Prove that
$$ \frac{1}{a}+\frac{1}{b}\geq \frac{2k}{1+k^2 a^2b^2}$$Where $ 4\leq k\in N^+.$
$$ \frac{1}{a}+\frac{1}{b}\geq \frac{4+\frac{k}{4}}{1+ ka^2b^2}$$Where $16\geq  k>0 .$
7 replies
+1 w
sqing
Today at 3:45 AM
sqing
6 minutes ago
Elegant inequality
SunnyEvan   2
N 7 minutes ago by cube4320
Source: proposed by Zhenping An
Let $a$, $b$, $c$, $d$ be non-negative real numbers such that
\[2a+2b+2c+2d+ab+bc+cd+da+3=abcd.\]prove that : \[\sqrt[4]{abc}+\sqrt[4]{bcd}+\sqrt[4]{cda}+\sqrt[4]{dab}\le\sqrt[4]{27(1+a)(1+b)(1+c)(1+d)}.\]
2 replies
SunnyEvan
3 hours ago
cube4320
7 minutes ago
Inequality
spiderman0   0
13 minutes ago
given a,b,c are positive real numbers prove that$ \frac{a^4}{a^2+ab+b^2}+ \frac{b^4}{b^2+bc+c^2}+ \frac{c^4}{c^2+ca+a^2}\ge \frac{a^3+b^3+c^3}{a+b+c}$
0 replies
spiderman0
13 minutes ago
0 replies
No more topics!
Easiest Functional Equation
NCbutAN   7
N Mar 16, 2025 by InftyByond
Source: Random book
Find all functions $f: \mathbb R \to \mathbb R$ such that $$f(yf(x)+f(xy))=(x+f(x))f(y)$$Follows for all reals $x,y$.
7 replies
NCbutAN
Mar 2, 2025
InftyByond
Mar 16, 2025
Easiest Functional Equation
G H J
G H BBookmark kLocked kLocked NReply
Source: Random book
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NCbutAN
1 post
#1 • 1 Y
Y by MathLuis
Find all functions $f: \mathbb R \to \mathbb R$ such that $$f(yf(x)+f(xy))=(x+f(x))f(y)$$Follows for all reals $x,y$.
This post has been edited 1 time. Last edited by NCbutAN, Mar 2, 2025, 4:13 PM
Reason: .
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InftyByond
196 posts
#2 • 1 Y
Y by Yihangzh
mb chat I goofed
This post has been edited 1 time. Last edited by InftyByond, Mar 16, 2025, 7:14 AM
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pco
23458 posts
#3
Y by
InftyByond wrote:
literally plugging y=0 solves the problem immediately
How ?
If $f(0)=0$ plugging $y=0$ solves nothing

In my opinion
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maromex
110 posts
#4
Y by
Plugging $y=0$ just gives $f(0) = 0$. It doesn't immediately solve the problem but it looks like it helps.

Edit: After thinking further, I don't actually think it helps.

@below

Also, consider that $f(x) = 0$ and $f(x) = x$ are answers to the equation.
This post has been edited 4 times. Last edited by maromex, Mar 2, 2025, 6:39 PM
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Indpsolver
43 posts
#5
Y by
but plugging $$ y=0 $$gives $$ f(x)= -x $$, is it right
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megarnie
5538 posts
#6 • 3 Y
Y by maromex, scannose, KevinYang2.71
We claim the only solutions are $f\equiv 0$ and $f(x) = x$. These work. Now we show they are the only ones. Suppose $f$ was not identically zero or the identity.

Claim: $f$ is nonlinear.
Proof: Suppose $f$ was linear.

Setting $y = 0$ gives $f(f(0)) = (x + f(x)) f(0)$. If $f(0) \ne 0$, then $f(x) + x$ is a constant, implying $f$ is an involution, so $f(f(0)) = 0$, after which $P(0,0)$ gives $f(0) = 0$, absurd. Hence $f(0)= 0$. Let $f(x) = kx$.

$P(1,1): f(2k) = k(k+1)$, so $2k^2 = k^2 + k$, which implies $k = k^2 \implies k \in \{0,1\}$. $\square$

Let $P(x,y)$ be the given assertion and $g(x) = x + f(x)$.

$P(x,0)$ gives that $f(0) = 0$.

Claim: $f(1) \ne 0$.
Proof: Suppose $f(1) = 0$.

Setting $y = 1$ gives $f(2f(x)) = 0$ for all reals $x$.

Setting $x =1$ gives $f(f(x)) = f(x)$ for all reals $x$.

$P(f(x),2): 0 = 2f(x) f(2)$. Since $f$ isn't linear, we also have $f(2) = 0$.

$P(2, f(x)): 0 = 2f(f(x)) = 2f(x)$, so $f$ is identically zero, a contradiction. $\square$

Claim: $f$ is injective.
Proof: If $f(c) = f(d)$, then $P(c,1)$ compared with $P(d,1)$ gives that $(c + f(c)) f(1) = (d + f(d)) f(1)$, so $c = d$. $\square$

Claim: $g$ is injective.
Proof: Suppose for some reals $a,b$, we have $g(a) = g(b)$.

$P(a,1)$ compared with $P(b,1)$ gives $f(2f(a)) = f(2f(b))$, so $2f(a) = 2f(b) \implies f(a) = f(b)\implies a= b$. $\square$

Now, we note that $f(y (g(x) - x) + g( xy) - xy) = g(x) (g(y) - y)$, so adding both sides by $y(g(x) - x) + g(xy) - xy$ gives that \[g(y(g(x) - x) + g(xy) - xy) = g(x) (g(y) - y) + y(g(x) - x) + g(xy) - xy = g(x) g(y) + g(xy) - 2xy \ \ \ \ \ (1)  \]
Swapping $x,y$ and comparing gives $g(y(g(x) - x) + g(xy) - xy) = g(x(g(y) - y) + g(xy) - xy))$, so $yf(x) = x f(y)$, which implies for $x,y\ne 0$ that $\frac{f(x)}{x} = \frac{f(y)}{y}$, so $\frac{f(x)}{x}$ is constant over reals $x\ne 0$. Since $f(0) = 0$, we have $f(x) = cx$ for some constant $c$ and all reals $x$, a contradiction because $f$ is nonlinear.
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KevinYang2.71
407 posts
#7 • 2 Y
Y by megarnie, NCbutAN
The only solutions are $\boxed{f(x)=0}$ and $\boxed{f(x)=x}$. These both work.

Let $P(x,\,y)$ denote the assertion. From $P(0,\,0)$ and $P(x,\,0)$ we get $f(f(0))=f(0)^2=f(0)(x+f(x))$. If $f(0)\neq 0$, we have $f(x)=f(0)-x$. Plugging this into $P$ yields $f(0)^2=2xy$, which is a contradiction for $x=y=0$. Hence $f(0)=0$.

Suppose $f(1)\neq 0$. Then $x=\frac{f(2f(x))}{f(1)}-f(x)$ so $f$ is injective. $P(x,\,1)$ gives $x=f^{-1}\left(\frac{1}{2}f^{-1}(f(1)(x+f(x)))\right)$ so $x\mapsto x+f(x)$ is also injective. Comparing $P(x,\,y)$ and $P(y,\,x)$ gives
\[
yf(x)+f(xy)+f(yf(x)+f(xy))=xf(y)+yf(x)+f(xy)+f(x)f(y)=xf(y)+f(xy)+f(xf(y)+f(xy))
\]so $xf(y)=yf(x)$ by injectivity. Thus $f(x)=cx$ for $c:=f(1)$. A simple calculation shows that $c\in\{0,\,1\}$.

Suppose now $f(1)=0$. We claim that $f(x)=0$. Assume FTSOC there exists $a$ such that $b:=f(a)\neq 0$. $P(1,\,x)$ gives $f(f(x))=f(x)$ so $f(b)=b$. $P(x,\,1)$ gives $f(2f(x))=0$ so $f(2b)=0$. Then $P(2b,\,x)$ yields $f(2bx)=2bf(x)$. From $P\left(b,\,\frac{1}{b}\right)$ we get $0=2bf\left(\frac{1}{b}\right)$ so $f\left(\frac{1}{b}\right)=0$. Then $f(2)=f\left(2b\cdot\frac{1}{b}\right)=0$ so $P(2,\,x)$ gives $f(2x)=2f(x)$. However, $0=f(2b)=2f(b)=2b$, a contradiction. $\square$
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InftyByond
196 posts
#8
Y by
may just be joever for me I misread the problem statement
this is about as easy as it is for me to not silly (impossible)
This post has been edited 1 time. Last edited by InftyByond, Mar 16, 2025, 8:15 AM
Reason: AAAAAAAAAA IM SORRY PCO IM SORRY EVERYONE
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