ThEdArKlOrD

Famous example of Lagrange Interpolation Polynomials

See here https://www.math.ust...ibur/v15_n2.pdf (Example $6$)

Suppose to the contradiction that $x_{i+1}-x_i-1>3\sqrt[3]{x_ix_{i+1}}$ for all $i\in \mathbb{1,2,...,10}$

This give us $(\sqrt[3]{x_{i+1}}-\sqrt[3]{x_i}-1)(\sqrt[3]{x_{i+1}}^2+\sqrt[3]{x_i}\sqrt[3]{x_{i+1}}+\sqrt[3]{x_i}^2-\sqrt[3]{x_{i}}+\sqrt[3]{x_{i+1}}+1)>0$, so $\sqrt[3]{x_{i+1}} >\sqrt[3]{x_i}+1$, for all $i\in \mathbb{1,2,...,10}$

So $\sqrt[3]{x_{11}} >\sqrt[3]{x_1} +10\geq 11$, contradict with $x_{11}\leq 1000$

This is equivalent to APMO $1989$

See here https://www.artofpro...c6h78608p450334

If you mean $f^2(x)=f(x)^2$

Let $P(x,y)$ denote $f(x)^2=f(x-y)f(y-x)+f(3y+x)f(3y-x)$ for all $x,y \in \mathbb{R}$

$P(x,0)$ give us $f(x)^2=2f(x)f(-x)$,so $f(-x)^2=2f(x)f(-x)$ give us $f(x)^2=f(-x)^2$ for all $x\in \mathbb{R}$,

From $f(x)^2=2f(x)f(-x)$, suppose there exist $t$ such that $f(t)\neq 0$, we get $f(t)=2f(-t)$, but since $f(x)^2=f(-x)^2$ for all $x\in \mathbb{R}$, give $f(t)=f(-t)=0$, contradiction, so $f(x)=0$ for all $x\in \mathbb{R}$

If you mean $f^2(x)=f(f(x))$

Let $P(x,y)$ denote $f(f(x))=f(x-y)f(y-x)+f(3y+x)f(3y-x)$ for all $x,y \in \mathbb{R}$

$P(x,0)$ give us $f(f(x))=2f(x)f(-x)=f(f(-x))$ for all $x\in \mathbb{R}$ and $P(x,x)$ give us $f(f(x))=f(0)^2+f(4x)f(2x)$ for all $x\in \mathbb{R}$

Then $P(-y,y)$ give $f(f(-y))=f(-2y)f(2y)+f(2y)f(4y)$ for all $y\in \mathbb{R}$, so $f(0)^2+f(4x)f(2x)=f(f(x))=f(f(-x))=f(-2x)f(2x)+f(2x)f(4x)$ for all $x\in \mathbb{R}$

So $f(0)^2=f(-2x)f(2x)$ for all $x\in \mathbb{R}$, this give us $f(f(x))=f(f(-x))=2f(0)^2$ for all $x\in \mathbb{R}$

Then, $P(x,y)$ change to $2f(0)^2=f(0)^2+f(3y+x)f(3y-x)$ for all $x,y\in \mathbb{R}$, so $f(0)^2=f(3y)^2=f(0)f(6y)$ for all $y\in \mathbb{R}$

So $f(0)=0$ give $f(y)=0$ for all $y\in \mathbb{R}$ or $f(0)\neq 0$ give us $f(6y)=f(0)$ for all $y\in \mathbb{R}$

So $f$ is constant function and the rest is easy

Suppose there are $t$ elements in $S$

For each element $p\in S$, we write down $p$ in a line, under word $p$, we write down all word different from $p$ by exactly $1$ letter, note that there are total $15+1$ words from each $p$ (including $p$)

Note that among any two words differ at least $1$ letter

So $(15+1)\times t \leq 2^{15}$ give us $t\leq 2^{11}$

Impossible, let center of lower left corner cell have coordinate $(1,1)$, and suppose that each cell have side lengths $1$

We color cell with center $(i,j)$ such that $i+j \equiv_4 3$, note that there are $26$ colored cell

But each $1\times 4$ block cover exactly $1$ colored cell, so there are $26$ blocks, impossible

Let incircle touch $AB,AC$ at $D,E$ and $A-$ mixtillinear incircle touch $AB,AC$ at $M,N$ respectively

And let incircle intersect $A-$ mixtillinear incircle at $T_1,T_2$, it is enough to prove that $AT_1=AI=AT_2$

Note that $\angle{IMA}=90^{\circ}$ and $\angle{ADM}=90^{\circ}$

Then we have $AD\cdot AM=AI^2$, similarly $AE\cdot AN=AI^2$

Then we use inversion center at $A$ with $r=\sqrt{AD\cdot AM}$

Note that incircle "swap" with mixtillinear incircle, but intersection point $T_1,T_2$ go to itself

So $AT_1^2=AD\cdot AM=AI^2$, this complete the prove

We can use the result that if $P(x)=a_nx^n+a_{n-1}x^{n-1}+...+a_1x+a_0\in \mathbb{Z}[x]$ with $|a_0|$ is prime and $|a_0|>|a_1|+|a_2|+...+|a_n|$ then $P(x)$ is irreducible over $\mathbb{Z}[x]$

 

 

Chúng ta có thể dùng kết quả là nếu $P(x) = a_{n}x^{n} + a_{n - 1}x^{n - 1} + \cdots + a_{1}x + a_{0} \in \mathbb{Z}[x]$ với $|a_{0}|$ là số nguyên tố và $|a_{0}| > |a_{1}| + |a_{2}| + \cdots + |a_{n}|$ thì $P(x)$ bất khả quy trên $\mathbb{Z}[x]$

Let $P(x,y)$ be the assertion of $f(x+y)+f(x)f(y)=f(xy)+f(x)+f(y)$ for all $x,y\in \mathbb{R}^+$

$P(x,y+z)$ give us $f(x+y+z)=f(x) + f(y + z) + f(xy + xz)-f(x)f(y + z)=f(x)+f(y)+f(z)+f(xy)+f(yz)+f(zx) + f(x)f(y)f(z)-f(x)f(y)-f(y)f(z)-f(z)f(x)+f(x^2yz)-f(xy)f(xz)-f(x)f(yz)$ for all $x,y,z\in \mathbb{R}^+$

Similarly, use this with $P(y,z+x)$ give us $f(x^2yz)-f(xy)f(xz)-f(x)f(yz) = f(xy^2z)-f(xy)f(yz)-f(y)f(xz)$ for all $x,y,z\in \mathbb{R}^+$

Set $y=1$ in that equation give us $f(x^2z)-f(x)f(xz)-f(x)f(z)=f(xz)-f(x)f(z)-f(1)f(xz)$ for all $x,z\in \mathbb{R}^+$

Which is $f(x^2z)=(1-f(1))f(xz)+f(x)f(xz)$ for all $x,z\in \mathbb{R}^+$, so $f(xy)=(1-f(1))f(y)+f(x)f(y)$ for all $x,y\in \mathbb{R}^+$

Since $f(xy)=(1-f(1))f(y)+f(x)f(y)=(1-f(1))f(x)+f(x)f(y)$ for all $x,y\in \mathbb{R}^+$ give us $f(x)=c$ constant or $f(1)=1$

If $f(x)=c$, we get $c+c^2=3c$, so $f(x)=2$ for all $x\in \mathbb{R}^+$

Otherwise, $f(1)=1$, so $f(xy)=f(x)f(y)$ for all $x,y\in \mathbb{R}^+$, so $f(x+y)=f(x)+f(y)$ for all $x,y\in \mathbb{R}^+$

Since $f:\mathbb{R}^+ \rightarrow \mathbb{R}^+$, so we get $f(x)=cx$ for constant $c$, check in $P(x,y)$ give $c=1$

So the answer is $f(x)=2$ for all $x\in \mathbb{R}^+$ and $f(x)=x$ for all $x\in \mathbb{R}^+$

 

Gọi $P(x,y)$ một phép thế tương ứng của $f(x+y)+f(x)f(y)=f(xy)+f(x)+f(y)$ với mọi $x,y\in \mathbb{R}^+$

$P(x,y+z)$ cho ta $f(x+y+z)=f(x) + f(y + z) + f(xy + xz)-f(x)f(y + z)=f(x)+f(y)+f(z)+f(xy)+f(yz)+f(zx) + f(x)f(y)f(z)-f(x)f(y)-f(y)f(z)-f(z)f(x)+f(x^2yz)-f(xy)f(xz)-f(x)f(yz)$ với mọi $x,y,z\in \mathbb{R}^+$

Tương tự, $P(y,z+x)$ cho $f(x^2yz)-f(xy)f(xz)-f(x)f(yz) = f(xy^2z)-f(xy)f(yz)-f(y)f(xz)$ với mọi $x,y,z\in \mathbb{R}^+$

Thay $y:=1$ vào PT trên $f(x^2z)-f(x)f(xz)-f(x)f(z)=f(xz)-f(x)f(z)-f(1)f(xz)$ với mọi $x,z\in \mathbb{R}^+$

Có nghĩa là $f(x^2z)=(1-f(1))f(xz)+f(x)f(xz)$ với mọi $x,z\in \mathbb{R}^+$, vậy $f(xy)=(1-f(1))f(y)+f(x)f(y)$ với mọi $x,y\in \mathbb{R}^+$

Từ $f(xy)=(1-f(1))f(y)+f(x)f(y)=(1-f(1))f(x)+f(x)f(y)$ với mọi $x,y\in \mathbb{R}^+$ thu được $f(x)=c$ hằng số hoặc $f(1)=1$

Nếu $f(x)=c$, ta thu được $c+c^2=3c$, vậy $f(x)=2$ với mọi $x\in \mathbb{R}^+$

Ngược lại, nếu $f(1)=1$, vậy $f(xy)=f(x)f(y)$ với mọi $x,y\in \mathbb{R}^+$, suy ra $f(x+y)=f(x)+f(y)$ với mọi $x,y\in \mathbb{R}^+$

Vì $f:\mathbb{R}^+ \rightarrow \mathbb{R}^+$, nên ta suy ra $f(x)=cx$ với hằng số $c$, kiểm tra với $P(x,y)$ cho ra $c=1$

Vậy hàm cần tìm là $f(x)=2$ với mọi $x\in \mathbb{R}^+$ và $f(x)=x$ với mọi $x\in \mathbb{R}^+$

$x=1$ give $f(1)f(f(1))=1$, so $f(1)=1$

It is easy to see that $f$ is injective

$x=p$ where $p$ is prime number give us $f(p)f(f(p))=p^2$, since $f(1)=1$ so $f(p)\neq 1$

Then $f(p)=p^2$ or $f(p)=p$, if $f(p)=p^2$, we get $f(p^2)=1$, contradiction

So $f(p)=p$ for all prime number $p$

Then, we will induction that $f(n)=n$

Note that $f(1)=1,f(2)=2,f(3)=3$

Suppose it's true until $f(m)=m$ for some $m\in \mathbb{Z}^+$

Since $f$ is injective, $f(m+1)\geq m+1$

So $f(\text{ something greater than }m)\geq m+1$

So $f(m+1)f(f(m+1))\geq (m+1)^2$, equality must hold, so $f(m+1)=m+1$, complete induction step

 

$x=1$ cho ta $f(1)f(f(1))=1$ nên $f(1)=1$

Dễ thấy rằng $f$ là đơn ánh

$x=p$ với $p$ là số nguyên tố cho ta $f(p)f(f(p))=p^2$,do $f(1)=1$ nên $f(p)\neq 1$

Do đó $f(p)=p^2$ hoặc $f(p)-p$, Nếu $f(p)=p^2$ ta suy ra $f(p^2)=1$ , mâu thuẫn.

Vậy $f(p)=p$ với mọi $p$ nguyên tố

Ta dự đoán $f(n)=n$

Để ý rằng $f(1)=1,f(2)=2,f(3)=3$

Giả sử đúng đến $f(m)=m$ với $m\in \mathbb{Z}^+$

Do $f$ là đơn ánh nên $f(m+1)\geq m+1$

Vì vậy $f$ của một số lớn hơn $m\geq m+1$

Do đó $f(m+1)f(f(m+1))\geq (m+1)^2$, đẳng thức phải xảy ra nên $f(m+1)=m+1$.

Theo nguyên lí quy nạp toán học ta suy ra $f(n)=n$.

Define sequence $\{ a_n\}$ by $a_{n+3}=a_{n+2}+a_{n+1}-a_{n}$ for all $n\in \mathbb{Z}^+$
Find all $a_1,a_2,a_3$ such that $a_n$ is square of integer for all $n\in \mathbb{Z}^+$

Find all two variables polynomials $P(x,y)$ such that for any real numbers $a,b,c$, we have 

$P(ab,c^2-2)+P(ac,b^2-2)+P(bc,a^2-2)=0$

Here is Epsilon No. 8

Link; https://drive.google...EpTS0dka1E/view