Prove that \(A_n=3^{\,n+3}-4^{\,4n+2}\) is divisible by \(11\) for every positive integer \(n\).

We will prove by Mathematical Induction.

For \(n=0\), \(A_0=3^3-4^2=27-16=11\), which is divisible by \(11\).

Assume that \(A_k\) is divisible by \(11\), i.e., \(11\mid A_k\).

We shall prove that \(A_{k+1}\) is also divisible by \(11\).

\[ \begin{aligned} A_{k+1} &=3^{k+4}-4^{4k+6}\\ &=3\cdot3^{k+3}-4^4\cdot4^{4k+2}\\ &=3\cdot3^{k+3}-256\cdot4^{4k+2}\\ &=3\cdot3^{k+3}-3\cdot4^{4k+2}-253\cdot4^{4k+2}\\ &=3\left(3^{k+3}-4^{4k+2}\right)-253\cdot4^{4k+2}\\ &=3A_k-253\cdot4^{4k+2}. \end{aligned} \]

Since \(11\mid A_k\) by the induction hypothesis and \(253=11\times23\), both terms on the right-hand side are divisible by \(11\). Hence \(11\mid A_{k+1}\).

Therefore, by the Principle of Mathematical Induction, \(11\mid\left(3^{n+3}-4^{4n+2}\right)\) for every nonnegative integer \(n\).

Solution by: Thin Sokkean

Prove that \(\forall x\in\mathbb{R},\ |a\cos x+b\sin x|\le\sqrt{a^2+b^2}.\)

a. Prove that \(\forall x\in\mathbb{R}\), \(|a\cos x+b\sin x|\leq\sqrt{a^2+b^2}\).

b. Find the maximum and minimum of $$ f(x)=20\cos x+21\sin x+27. $$

Solution

a. Prove that \(\forall x\in\mathbb{R}\), \(|a\cos x+b\sin x|\leq\sqrt{a^2+b^2}\).

We have:

$$ a\cos x+b\sin x =\sqrt{a^2+b^2} \left(\frac{a}{\sqrt{a^2+b^2}}\cos x+ \frac{b}{\sqrt{a^2+b^2}}\sin x\right) $$

Let

$$ \sin\alpha=\frac{a}{\sqrt{a^2+b^2}} $$

\(X_1,X_2\) are the roots of \(X^2-(2\cos t+3\sin t)X-11\sin^2t=0\). Find the minimum value of \(A=X_1^2+X_1X_2+X_2^2\).

Solution

\(X_1,X_2\) are the roots of \(X^2-(2\cos t+3\sin t)X-11\sin^2t=0\). By Vieta's formulas, we have \(X_1+X_2=2\cos t+3\sin t\) and \(X_1X_2=-11\sin^2t\).

We have \(A=X_1^2+X_1X_2+X_2^2=(X_1+X_2)^2-X_1X_2\).

Find all polynomials \(P(x)\) which satisfy: \((x-2010)P(x+67)=xP(x)\) (2010 Baltic Way).

(2010 Baltic Way) 

Prove that \(\frac{\tan^3x}{1-3\tan^2x}=\frac18(\tan3x-3\tan x)\).

Solution

a. We know that \(\tan3x=\frac{3\tan x-\tan^3x}{1-3\tan^2x}\).

\[ \begin{aligned} \tan3x&=\frac{3\tan x-\tan^3x}{1-3\tan^2x}\\ &=\frac{3\tan x-9\tan^3x+8\tan^3x}{1-3\tan^2x}\\ &=\frac{8\tan^3x+3\tan x(1-3\tan^2x)}{1-3\tan^2x}\\ &=\frac{8\tan^3x}{1-3\tan^2x}+3\tan x. \end{aligned} \]

Find the limit of \(S_n=\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\cdots+\frac{2}{(2n+1)(2n+3)}\).

Solution

The general term of this sequence is:

$$ \frac{2}{(2k+1)(2k+3)}=\frac{1}{2k+1}-\frac{1}{2k+3}. $$

Therefore,

$$ \begin{aligned} S_n&=\left(1-\frac13\right)+\left(\frac13-\frac15\right)+\left(\frac15-\frac17\right)+\cdots+\left(\frac{1}{2n+1}-\frac{1}{2n+3}\right)\\ &=1-\frac{1}{2n+3}. \end{aligned} $$

When \(n\rightarrow\infty\), we have \(\frac{1}{2n+3}\rightarrow0\).

Hence,

$$ \boxed{\lim_{n\to\infty}S_n=1}. $$

Solution by: Thin Sokkean

If \(g(x)=f(x)+1-x\), find the value of \(g(2020)\).

It is given that \(f(x)\) is a function defined on \(\mathbb{R}\), satisfying \(f(1)=1\), \(f(x+5)\geq f(x)+5\), and \(f(x+1)\leq f(x)+1\). If \(g(x)=f(x)+1-x\), find the value of \(g(2020)\).

Solution

From the given conditions, we have \(f(x+5)\geq f(x)+5\) and \(f(x+1)\leq f(x)+1\).

Since \(f(x+5)\leq f(x+4)+1\) and \(f(x+4)\leq f(x+3)+1\leq f(x+2)+2\leq f(x+1)+3\), we get

$$ f(x+5)\leq f(x+1)+4. $$

Therefore,

$$ f(x)+5\leq f(x+5)\leq f(x+1)+4\leq f(x)+5. $$

Hence,

$$ f(x+1)+4=f(x)+5, $$

so

$$ f(x+1)-f(x)=1. $$

For \(x=1,2,\ldots,2019\), we have

$$ f(2)-f(1)=1,\quad f(3)-f(2)=1,\quad\cdots,\quad f(2020)-f(2019)=1. $$

Adding all equations,

$$ f(2020)-f(1)=2019. $$

Since \(f(1)=1\),

$$ f(2020)=2020. $$

Now,

$$ g(x)=f(x)+1-x. $$

For \(x=2020\),

$$ g(2020)=f(2020)+1-2020=2020+1-2020=1. $$

Therefore,

$$ \boxed{g(2020)=1}. $$

Solution by: Thin Sokkean

If \((1+2x+3x^2)^{10}=a_0+a_1x+a_2x^2+\cdots+a_{20}x^{20}\), find the coefficients \(a_1,\ a_2,\ a_3,\) and \(a_{20}\).

Solution

From \((1+2x+3x^2)^{10}=a_0+a_1x+a_2x^2+\cdots+a_{20}x^{20}\), we need to find the coefficients of \(x,x^2,x^3,\) and \(x^{20}\).

Rewrite: \((1+2x+3x^2)^{10}=[1+x(2+3x)]^{10}\).

Using Newton's formula \((a+b)^n=\sum_{i=0}^{n}C(n,i)a^{n-i}b^i\), we get \([1+x(2+3x)]^{10}=C(10,0)+C(10,1)x(2+3x)+C(10,2)x^2(2+3x)^2+\cdots+C(10,10)x^{10}(2+3x)^{10}\).

Coefficient of \(x\): \(a_1=2C(10,1)=20\).

Coefficient of \(x^2\): \(a_2=2^2C(10,2)=4\times45=180\).

Coefficient of \(x^3\): \(a_3=12C(10,2)+8C(10,3)=12\times45+8\times120=1500\).

Coefficient of \(x^{20}\): \(a_{20}=C(10,10)3^{10}=3^{10}\).

Therefore, \(a_1=20,\ a_2=180,\ a_3=1500,\ a_{20}=3^{10}\).

Part 2

Given \((1+x+2x^2)^{20}=a_0+a_1x+a_2x^2+\cdots+a_{40}x^{40}\).

When \(x=1\), \(a_0+a_1+a_2+\cdots+a_{40}=4^{20}\).

When \(x=-1\), \(a_0-a_1+a_2-a_3+\cdots-a_{39}+a_{40}=2^{20}\).

Adding the two equations, \(2a_0+2a_2+2a_4+\cdots+2a_{40}=4^{20}+2^{20}\).

Hence, \(a_0+a_2+a_4+\cdots+a_{40}=\frac{4^{20}+2^{20}}{2}\).

Since \(a_{40}=2^{20}\),

\[ a_0+a_2+a_4+\cdots+a_{38} =\frac{4^{20}+2^{20}}{2}-2^{20} =\frac{2^{40}+2^{20}}{2}-2^{20} =\frac{2^{40}-2^{20}}{2}. \]

Therefore, \(\boxed{a_0+a_2+a_4+\cdots+a_{38}=\frac{2^{40}-2^{20}}{2}}\).

Prove that \(\frac{21n+4}{14n+3}\) is irreducible for every natural number \(n\).

Solution

Method 01: Let \((a,b)\) denote the greatest common divisor of \(a\) and \(b\). Using the Euclidean Algorithm,

\[ (21n+4,14n+3)=(7n+1,14n+3)=(7n+1,1)=1. \]

Therefore, \(\frac{21n+4}{14n+3}\) is irreducible. Q.E.D.

Method 02: We prove by contradiction. Assume that \(\frac{21n+4}{14n+3}\) is reducible. Then there exists a prime number \(p\) such that \(21n+4\equiv0\pmod p\) and \(14n+3\equiv0\pmod p\).

Multiplying the first equation by \(2\), we get \(42n+8\equiv0\pmod p\). Subtracting \(3(14n+3)\equiv0\pmod p\),

\[ (42n+8)-(42n+9)\equiv0\pmod p \]

so \(-1\equiv0\pmod p\), which is impossible. Hence, \(\frac{21n+4}{14n+3}\) is irreducible. Q.E.D.

Solution by: Thin Sokkean

Vietnam Math Out Standing Student 2012-13

Vietnam Math Outstanding Student 2012-2013

Problem 01: Solve the equation \(x^{4n}+\sqrt{x^{2n}+2012}=2012\), for all \(n\in\mathbb{N}\).

Problem 02: It is given the sequence \((U_n)\) determined by \(U_1=3\) and \(U_{n+1}=\frac13\left(2U_n+\frac3{U_n^2}\right)\), for all \(n\in\mathbb{N}\). Find the limit \(\lim_{n\rightarrow\infty}U_n\).

Problem 03: Given three non-negative real numbers \(x,y,z\), prove that \(\frac1x+\frac1y+\frac1z\geq\frac{36}{9+x^2y^2+y^2z^2+z^2x^2}\).

Problem 04: Find the positive roots of the equation \(\sqrt{x+2\sqrt3}=\sqrt y+\sqrt z\).

Vietnam Math Outstanding Student 2012-2013

Problem 01: Solve the equation \(x^{4n}+\sqrt{x^{2n}+2012}=2012\), for all \(n\in\mathbb{N}\).

Solution: Let \(t=x^{2n}\). Then the equation becomes \(t^2+\sqrt{t+2012}=2012\).

We have \(t^2=2012-\sqrt{t+2012}\), so adding \(t+\frac14\) to both sides gives \((t+\frac12)^2=t+2012-\sqrt{t+2012}+\frac14=(\sqrt{t+2012}-\frac12)^2\).

Since \(t=x^{2n}\geq0\), we get \(t+\frac12=\sqrt{t+2012}-\frac12\), hence \(t+1=\sqrt{t+2012}\).

Squaring both sides, \(t^2+t+1=t+2012\), so \(t^2-2011=0\).

Therefore, \(t=\sqrt{2011}\), and since \(t=x^{2n}\), we have \(x^{2n}=\sqrt{2011}\).

 

Find all functions \(f:\mathbb{R}\rightarrow\mathbb{R}\) such that \(f(x)f(y)=f(xy-1)+xf(y)+yf(x)\), \(\forall x,y\in\mathbb{R}\).

Solution

Replace \(y=0\), we get \(f(x)f(0)=f(-1)+xf(0)\).

If \(f(0)\neq0\), then \(f(x)=x+c\), where \(c\) is a constant. Substituting into the original equation gives a contradiction. Therefore, \(f(0)=0\), and hence \(f(-1)=0\).

Putting \(x=y=1\), we have \(f(1)^2=2f(1)\), so \(f(1)=0\) or \(f(1)=2\).

Putting \(y=-1\), we get \(f(-x-1)=f(x)\).

Replacing \(y\) by \(-y-1\) in the original equation, using \(f(-y-1)=f(y)\) and \(f(xy+x)=f(x(-y-1)-1)\), we obtain

\[ f(xy-1)+yf(x)=f(xy+x)+xf(y)-(y+1)f(x). \]

For \(x\neq0\), let \(y=\frac1x\), then

\[ f(x+1)=\frac{x+2}{x}f(x), \]

or equivalently,

\[ f(x-1)=\frac{x-1}{x+1}f(x). \]

Now replace \(y=1\) in the original equation:

\[ f(x)f(1)=f(x-1)+xf(1)+f(x) \]

Using \(f(x-1)=\frac{x-1}{x+1}f(x)\), we have:

Case 1: \(f(1)=0\), then \(f(x)=0\) for all \(x\in\mathbb{R}\).

Case 2: \(f(1)=2\), then \(f(x)=x(x+1)\).

Both functions satisfy the original equation.

Therefore, the solutions are:

\[ \boxed{f(x)\equiv0\quad\text{or}\quad f(x)=x(x+1),\ \forall x\in\mathbb{R}.} \]

Solution by: Thin Sokkean