Let $a, b, c, d, e$ be distinct positive integers such that $a^4 + b^4 = c^4 + d^4 = e^5$. Show that $ac + bd$ is a composite number.

Solve in positive integers $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{4}{5}$

Prove there exist infinite number of positive integer $a$ such that for any positive integer $n$, $n^4 + a$ is not a prime number.

Prove for any positive integer $n$, the fraction $\frac{21n+4}{14n+3}$ cannot be further simplified.

If integer d is not equal to 2,5 or 13. Prove: there must exist two different elements $a$ and $b$ in set {2, 5, 3, d} such that $ab -

Let $f(n)$ denote the sum of the digits of $n$. Find $f(f(f(4444^{4444})))$.

Prove that $\frac{5^{125}-1}{5^{25}-1}$ is composite.

Find all integers $a$, $b$, $c$ with $1 < a < b < c$ such that the number $(a-1)(b-1)(c-1)$ is a divisor of $abc-1$.

Prove that if positive integer $a$ and $b$ are such that $ab+1$ divides $a^2 + b^2$. then $$\frac{a^2+b^2}{ab+1}$$ is a square number.

Let sequence $g(n)$ satisfy $g(1)=0, g(2)=1, g(n+2)=g(n+1)+g(n)+1$ where $n\ge 1$. Show that if $n$ is a prime greater than 5, then $n\mid g(n)[g(n)+1]$.

Find all postive integers $(a,b,c)$ such that
$$ab-c,\quad bc-a,\quad ca-b$$are all powers of $2$.
Proposed by Serbia

Show that every integer $k > 1$ has a multiple which is less than $k^4$ and can be written in base 10 using at most 4 different digits.

In a sports contest, there were $m$ medals awarded on $n$ successive days ($n > 1$). On the first day, one medal and $1/7$ of the remaining $m − 1$ medals were awarded. On the second day, two medals and $1/7$ of the now remaining
medals were awarded; and so on. On the $n^{th}$ and last day, the remaining $n$ medals were awarded. How many days did the contest last, and how many medals were awarded altogether?

Determine all three-digit numbers $N$ having the property that $N$ is divisible by 11, and $\dfrac{N}{11}$ is equal to the sum of the squares of the digits of $N$.

Show that for any positive integer $n$, the value of $\displaystyle\sum_{k=0}^{n}2^{3k}\binom{2n+1}{2k+1}$ is not a multiple of $5$.