MathDB

4

Part of 2012 Indonesia TST

Problems(8)

Quadratic residues mod n under 1+floor(n)

Source: 2012 Indonesia Round 2 TST 1 Problem 4

2/26/2012
Determine all natural numbers nn such that for each natural number aa relatively prime with nn and a≤1+⌊n⌋a \le 1 + \left\lfloor \sqrt{n} \right\rfloor there exists some integer xx with a≡x2mod  na \equiv x^2 \mod n.
Remark: "Natural numbers" is the set of positive integers.
quadraticsfloor functionsearchmodular arithmeticinductionpigeonhole principlenumber theory
Functions where d(f(x)) = x

Source: 2012 Indonesia Round 2 TST 2 Problem 4

3/4/2012
Let N\mathbb{N} be the set of positive integers. For every n∈Nn \in \mathbb{N}, define d(n)d(n) as the number of positive divisors of nn. Find all functions f:N→Nf : \mathbb{N} \rightarrow \mathbb{N} such that: a) d(f(x))=xd(f(x)) = x for all x∈Nx \in \mathbb{N} b) f(xy)f(xy) divides (x−1)yxy−1f(x)(x-1)y^{xy-1}f(x) for all x,y∈Nx,y \in \mathbb{N}
functionnumber theory unsolvednumber theory
Fibonacci powers

Source: 2012 Indonesia Round 2 TST 3 Problem 4

3/18/2012
The Fibonacci sequence {Fn}\{F_n\} is defined by F1=F2=1F_1 = F_2 = 1 and Fn+2=Fn+1+FnF_{n+2} = F_{n+1} + F_n for all positive integers nn. Determine all triplets of positive integers (k,m,n)(k,m,n) such that Fn=FmkF_n = F_m^k.
number theory unsolvednumber theory
3^m = 2^k + 7^n and geometric series

Source: 2012 Indonesia Round 2 TST 4 Problem 4

3/18/2012
Find all quadruplets of positive integers (m,n,k,l)(m,n,k,l) such that 3m=2k+7n3^m = 2^k + 7^n and mk=1+k+k2+k3+…+klm^k = 1 + k + k^2 + k^3 + \ldots + k^l.
number theory unsolvednumber theory
gcd(n, (n-m)/gcd(n,m)) = 1 for all m < n

Source: 2012 Indonesia Round 2.5 TST 1 Problem 4

5/10/2012
Determine all integer n>1n > 1 such that gcd⁡(n,n−mgcd⁡(n,m))=1\gcd \left( n, \dfrac{n-m}{\gcd(n,m)} \right) = 1 for all integer 1≤m<n1 \le m < n.
number theorygreatest common divisormodular arithmeticnumber theory proposed
a_n = sum of a_floor(n/k) + 1

Source: 2012 Indonesia Round 2.5 TST 2 Problem 4

5/21/2012
The sequence aia_i is defined as a1=1a_1 = 1 and an=a⌊n2⌋+a⌊n3⌋+a⌊n4⌋+⋯+a⌊nn⌋+1a_n = a_{\left\lfloor \dfrac{n}{2} \right\rfloor} + a_{\left\lfloor \dfrac{n}{3} \right\rfloor} + a_{\left\lfloor \dfrac{n}{4} \right\rfloor} + \cdots + a_{\left\lfloor \dfrac{n}{n} \right\rfloor} + 1 for every positive integer n>1n > 1. Prove that there are infinitely many values of nn such that an≡nmod  2012a_n \equiv n \mod 2012.
floor functionnumber theory unsolvednumber theory
1+k(p-1) is prime for many k

Source: 2012 Indonesia Round 2.5 TST 4 Problem 4

5/31/2012
Find all odd prime pp such that 1+k(p−1)1+k(p-1) is prime for all integer kk where 1≤k≤p−121 \le k \le \dfrac{p-1}{2}.
number theory proposednumber theory
Prime in {x_k, z}

Source: 2012 Indonesia Round 2.5 TST 3 Problem 4

5/21/2012
Given a non-zero integer yy and a positive integer nn. If x1,x2,…,xn∈Z−{0,1}x_1, x_2, \ldots, x_n \in \mathbb{Z} - \{0, 1\} and z∈Z+z \in \mathbb{Z}^+ satisfy (x1x2…xn)2y≤22(n+1)(x_1x_2 \ldots x_n)^2y \le 2^{2(n+1)} and x1x2…xny=z+1x_1x_2 \ldots x_ny = z + 1, prove that there is a prime among x1,x2,…,xn,zx_1, x_2, \ldots, x_n, z.
It appears that the problem statement is incorrect; suppose y=5,n=2y = 5, n = 2, then x1=x2=−1x_1 = x_2 = -1 and z=4z = 4. They all satisfy the problem's conditions, but none of x1,x2,zx_1, x_2, z is a prime. What should the problem be, or did I misinterpret the problem badly?
number theory unsolvednumber theory