MathDB

4

Part of 2005 Moldova Team Selection Test

Problems(3)

function over set of polynoms

Source: Moldova TST 2005

3/21/2005
nn is a positive integer, KK the set of polynoms of real variables x1,x2,...,xn+1x_1,x_2,...,x_{n+1} and y1,y2,...,yn+1y_1,y_2,...,y_{n+1}, function f:K→Kf:K\rightarrow K satisfies f(p+q)=f(p)+f(q),  f(pq)=f(p)q+pf(q),  (\forall)p,q\in K. If f(x_i)=(n-1)x_i+y_i,  f(y_i)=2ny_i for all i=1,2,...,n+1i=1,2,...,n+1 and ∏i=1n+1(txi+yi)=∑i=0n+1pitn+1−i\prod_{i=1}^{n+1}(tx_i+y_i)=\sum_{i=0}^{n+1}p_it^{n+1-i} for any real tt, prove, that for all k=1,...,n+1k=1,...,n+1 f(pk−1)=kpk+(n+1)(n+k−2)pk−1f(p_{k-1})=kp_k+(n+1)(n+k-2)p_{k-1}
functionalgebra proposedalgebra
largest p

Source: Moldova TST 2005

4/9/2005
Find the largest positive pp (p>1p>1) such, that ∀a,b,c∈[1p,p]\forall a,b,c\in[\frac1p,p] the following inequality takes place 9(ab+bc+ca)(a2+b2+c2)≥(a+b+c)49(ab+bc+ca)(a^2+b^2+c^2)\geq(a+b+c)^4
inequalitiesinequalities unsolved
f(n)=n

Source: Moldova TST 2005

4/10/2005
Given functions f,g:N∗→N∗f,g:N^*\rightarrow N^*, gg is surjective and 2f(n)2=n2+g(n)22f(n)^2=n^2+g(n)^2, ∀n>0\forall n>0. Prove that if ∣f(n)−n∣≤2005n|f(n)-n|\le2005\sqrt n, ∀n>0\forall n>0, then f(n)=nf(n)=n for infinitely many nn.
functionnumber theory unsolvednumber theory