MathDB

3

Part of 2003 Vietnam Team Selection Test

Problems(2)

Challenging functional equation

Source: Vietnam TST 2003 for the 44th IMO, problem 3

6/26/2005
Let f(0,0)=52003,f(0,n)=0f(0, 0) = 5^{2003}, f(0, n) = 0 for every integer n≠0n \neq 0 and  f(m,n)=f(m−1,n)−2⋅⌊f(m−1,n)2⌋+⌊f(m−1,n−1)2⌋+⌊f(m−1,n+1)2⌋\begin{array}{c}\ f(m, n) = f(m-1, n) - 2 \cdot \Bigg\lfloor \frac{f(m-1, n)}{2}\Bigg\rfloor + \Bigg\lfloor\frac{f(m-1, n-1)}{2}\Bigg\rfloor + \Bigg\lfloor\frac{f(m-1, n+1)}{2}\Bigg\rfloor \end{array} for every natural number m>0m > 0 and for every integer nn.
Prove that there exists a positive integer MM such that f(M,n)=1f(M, n) = 1 for all integers nn such that ∣n∣≤(52003−1)2|n| \leq \frac{(5^{2003}-1)}{2} and f(M,n)=0f(M, n) = 0 for all integers n such that ∣n∣>52003−12.|n| > \frac{5^{2003}-1}{2}.
floor functioncalculusintegrationinductionalgebra unsolvedalgebra
2n + 1 has no prime divisor of the form 8k - 1

Source: Vietnam TST 2003 for the 44th IMO, problem 6

6/26/2005
Let nn be a positive integer. Prove that the number 2n+12^n + 1 has no prime divisor of the form 8⋅k−18 \cdot k - 1, where kk is a positive integer.
modular arithmeticquadraticsnumber theory unsolvednumber theory