MathDB

2

Part of 2001 India IMO Training Camp

Problems(5)

Functional equation !

Source: India TST 2001 Day 3 Problem 2

1/31/2015
Find all functions f ⁣:R+→R+f \colon \mathbb{R_{+}}\to \mathbb{R_{+}} satisfying : f(f(x)−x)=2xf ( f (x)-x) = 2x for all x>0x > 0.
functionalgebrapolynomialalgebra proposed
Unique word !

Source: India TST 2001 Day 1 Problem 2

1/31/2015
Two symbols AA and BB obey the rule ABBB=BABBB = B. Given a word x1x2…x3n+1x_1x_2\ldots x_{3n+1} consisting of nn letters AA and 2n+12n+1 letters BB, show that there is a unique cyclic permutation of this word which reduces to BB.
inequalitiescombinatorics proposedcombinatorics
Prime dividing values of a polynomial!

Source: India TST 2001 Day 2 Problem 2

1/31/2015
Let Q(x)Q(x) be a cubic polynomial with integer coefficients. Suppose that a prime pp divides Q(xj)Q(x_j) for j=1j = 1 ,22 ,33 ,44 , where x1,x2,x3,x4x_1 , x_2 , x_3 , x_4 are distinct integers from the set {0,1,⋯ ,p−1}\{0,1,\cdots, p-1\}. Prove that pp divides all the coefficients of Q(x)Q(x).
algebrapolynomialnumber theory unsolvednumber theory
Inverse modulo p!

Source: India TST 2001 Day 4 Problem 2

1/31/2015
Let p>3p > 3 be a prime. For each k∈{1,2,…,p−1}k\in \{1,2, \ldots , p-1\}, define xkx_k to be the unique integer in {1,…,p−1}\{1, \ldots, p-1\} such that kxk≡1(modp)kx_k\equiv 1 \pmod{p} and set kxk=1+pnkkx_k = 1+ pn_k. Prove that : ∑k=1p−1knk≡p−12(modp)\sum_{k=1}^{p-1}kn_k \equiv \frac{p-1}{2} \pmod{p}
modular arithmeticnumber theory proposednumber theory
Increasing sequence with gcd property!

Source: India TST 2001 Day 5 Problem 2

1/31/2015
A strictly increasing sequence (an)(a_n) has the property that gcd⁡(am,an)=agcd⁡(m,n)\gcd(a_m,a_n) = a_{\gcd(m,n)} for all m,n∈Nm,n\in \mathbb{N}. Suppose kk is the least positive integer for which there exist positive integers r<k<sr < k < s such that ak2=arasa_k^2 = a_ra_s. Prove that r∣kr | k and k∣sk | s.
number theorygreatest common divisornumber theory unsolved