Olympiad problems

2009 JBMO Shortlist, 4

Determine all prime numbers $p_1, p_2,..., p_{12}, p_{13}, p_1 \le p_2 \le ... \le p_{12} \le p_{13}$, such that $p_1^2+ p_2^2+ ... + p_{12}^2 = p_{13}^2$ and one of them is equal to $2p_1 + p_9$.

1989 All Soviet Union Mathematical Olympiad, 499

Tags: irrational , sum , sum of powers , rational , algebra

Do there exist two reals whose sum is rational, but the sum of their $n$ th powers is irrational for all $n > 1$? Do there exist two reals whose sum is irrational, but the sum of whose $n$ th powers is rational for all $n > 1$?

2007 Thailand Mathematical Olympiad, 7

Tags: algebra , sum of powers

Let $a, b, c$ be complex numbers such that $a+b+c = 1$, $a^2+b^2+c^2 = 2$ and $a^3+b^3+c^3 = 3$. Find the value of $a^4 + b^4 + c^4$.

2001 Estonia National Olympiad, 4

Tags: composite , sum of powers , prime , number theory

Prove that for any integer $a > 1$ there is a prime $p$ for which $1+a+a^2+...+ a^{p-1}$ is composite.

1997 Czech and Slovak Match, 5

Tags: number theory , integer , sum of powers

The sum of several integers (not necessarily distinct) equals $1492$. Decide whether the sum of their seventh powers can equal (a) $1996$; (b) $1998$.

1998 Abels Math Contest (Norwegian MO), 3

Tags: sum , number theory , sum of powers

Let $n$ be a positive integer. (a) Prove that $1^5 +3^5 +5^5 +...+(2n-1)^5$ is divisible by $n$. (b) Prove that $1^3 +3^3 +5^3 +...+(2n-1)^3$ is divisible by $n^2$.

1995 Romania Team Selection Test, 1

Tags: sum , sum of powers , inequalities , algebra

Let $a_1, a_2,...., a_n$ be distinct positive integers. Prove that $(a_1^5 + ...+ a_n^5) + (a_1^7 + ...+ a_n^7) \ge 2(a_1^3 + ...+ a_n^3)^2$ and find the cases of equality.

1996 Estonia National Olympiad, 4

Tags: number theory , divisible , divide , sum of powers , sum

Prove that, for each odd integer $n \ge 5$, the number $1^n+2^n+...+15^n$ is divisible by $480$.

2001 Regional Competition For Advanced Students, 1

Tags: number theory , sum of powers , sum , digit , power

Let $n$ be an integer. We consider $s (n)$, the sum of the $2001$ powers of $n$ with the exponents $0$ to $2000$. So $s (n) = \sum_{k=0}^{2000}n ^k$ . What is the unit digit of $s (n)$ in the decimal system?

2004 Thailand Mathematical Olympiad, 17

Tags: number theory , remainder , sum of powers

Compute the remainder when $1^{2547} + 2^{2547} +...+ 2547^{2547}$ is divided by $25$.

1999 Tournament Of Towns, 2

Tags: number theory , prime , sum of powers , sum

Let $d = a^{1999} + b^{1999} + c^{1999}$ , where $a, b$ and $c$ are integers such that $a + b + c = 0$. (a) May it happen that $d = 2$? (b) May it happen that $d$ is prime? (V Senderov)

1996 Swedish Mathematical Competition, 5

Tags: number theory , power of 2 , sum of powers

Let $n \ge 1$. Prove that it is possible to select some of the integers $1,2,...,2^n$ so that for each $p = 0,1,...,n - 1$ the sum of the $p$-th powers of the selected numbers is equal to the sum of the $p$-th powers of the remaining numbers.

2000 Tournament Of Towns, 3

Tags: inequalities , sum of powers , algebra

Prove the inequality $$ 1^k+2^k+...+n^k \le \frac{n^{2k}-(n-1)^k}{n^k-(n-1)^k}$$ (L Emelianov)

2002 Junior Balkan Team Selection Tests - Romania, 1

Tags: number theory , divide , divisible , divisor , sum of powers

Let $n$ be an even positive integer and let $a, b$ be two relatively prime positive integers. Find $a$ and $b$ such that $a + b$ is a divisor of $a^n + b^n$.

2003 Junior Balkan Team Selection Tests - Romania, 2

Tags: number theory , consecutive , divide , sum of powers , prime

Consider the prime numbers $n_1< n_2 <...< n_{31}$. Prove that if $30$ divides $n_1^4 + n_2^4+...+n_{31}^4$, then among these numbers one can find three consecutive primes.

2008 Korea Junior Math Olympiad, 6

Tags: number theory , sum of powers , sum , divisible , divisor

If $d_1,d_2,...,d_k$ are all distinct positive divisors of $n$, we define $f_s(n) = d_1^s+d_2^s+..+d_k^s$. For example, we have $f_1(3) = 1 + 3 = 4, f_2(4) = 1 + 2^2 + 4^2 = 21$. Prove that for all positive integers $n$, $n^3f_1(n) - 2nf_9(n) + n^2f_3(n)$ is divisible by $8$.