Find all integer solutions to the equation $$\frac{xyz}{w}+\frac{yzw}{x}+\frac{zwx}{y}+\frac{wxy}{z}=4$$

It is known that $\odot O$ is the circumcircle of $\vartriangle ABC$ wwith diameter $AB$. The tangents of $\odot O$ at points $B$ and $C$ intersect at $P$ . The line perpendicular to $PA$ at point $A$ intersects the extension of $BC$ at point $D$. Extend $DP$ at length $PE = PB$. If $\angle ADP = 40^o$ , find the measure of $\angle E$.

If $n=p_1^{a_1},p_2^{a_2}...p_s^{a_s}$ then $\phi (n)=n \left(1- \frac{1}{p_1}\right)\left(1 - \frac{1}{p_2}\right)...\left(1- \frac{1}{p_s}\right)$. Find the smallest positive integer $n$ such that $\phi (n)=\frac{2^5}{47}n.$

If the set $S = \{1,2,3,…,16\}$ is partitioned into $n$ subsets, there must be a subset in which elements $a, b, c$ (can be the same) exist, satisfying $a+ b=c$. Find the maximum value of $n$.

As shown in figure , points $D,E,F$ lies the sides $AB$, $BC$ , $CA$ of the acute angle $\vartriangle ABC$ respectively. If $\angle EDC = \angle CDF$, $\angle FEA=\angle AED$, $\angle DFB =\angle BFE$, prove that the $CD$, $AE$, $BF$ are the altitudes of $\vartriangle ABC$.

The figure obtained by removing one small unit square from the $2\times 2$ grid table is called an $L$ ''shape". .Put $k$ L-shapes in an $8\times 8$ grid table. Each $L$-shape can be rotated, but each $L$ shape is required to cover exactly three small unit squares in the grid table, and the common area covered by any two $L$ shapes is $0$, and except for these $k$ $L$ shapes, no other $L$ shapes can be placed. Find the minimum value of $k$.

It is known that odd prime numbers $x, y z$ satisfy $$x|(y^5+1),y|(z^5+1),z|(x^5+1).$$Find the minimum value of the product $xyz$.

Given a positive integer $n \ge 3$. Find the smallest real number $k$ such that for any positive real number except $a_1, a_2,..,a_n$, $$\sum_{i=1}^{n-1}\frac{a_i}{ s-a_i}+\frac{ka_n}{s-a_n} \ge \frac{n-1}{n-2}$$where, $s=a_1+a_2+..+a_n$

As shown in figure , a circle of radius $1$ passes through vertex $A$ of $\vartriangle ABC$ and is tangent to the side $BC$ at the point $D$ , intersect sides $AB$ and $AC$ at points $E$ and $F$ respectively . Also$ EF$ bisects $\angle AFD$, and $\angle ADC = 80^o$ , Is there a triangle that satisfies the condition, so that $\frac{AB+BC+CA}{AD^2}$ is an irrational number, and the irrational number is the root of a quadratic equation with integral coefficients? If it does not exist, please prove it; if it exists, find the quadratic equation that satisfies the condition.

It is known that $a_1, a_2,...a_{108}$ are $108$ different positive integers not exceeding $2015$. Prove that there is a positive integer $k$ such that there are at least four different pairs $(i, j) $satisfying $a_i-a_j =k$.

Use $[x]$ to represent the greatest integer no more than a real number $x$. Let $$S_n=\left[1+\frac12 +\frac13+...+\frac{1}{n}\right], \,\, (n =1,2,..,)$$Prove that there are infinitely many $n$ such that $C_n^{S_n}$ is an even number. PS. Attached is the original wording which forgets left [ . I hope it is ok where I put it.

The sequence $\{a_n\}$ is defined as follows: $a_1$ is a positive rational number, $a_n= \frac{p_n}{q_n}$, ($n= 1,2,…$) is a positive integer, where $p_n$ and $q_n$ are positive integers that are relatively prime, then $a_{n+1} = \frac{p_n^2+2015}{p_nq_n}$ Is there a$_1>2015$, making the sequence $\{a_n\}$ a bounded sequence? Justify your conclusion.