MathDB

5

Part of 1993 IMO Shortlist

Problems(3)

Pairs of integers for ad^2 + 2bde + ce^2= n

Source: IMO Shortlist 1993, Georgia 1

10/24/2005
a>0a > 0 and bb, cc are integers such that acac – b2b^2 is a square-free positive integer P. [hide="For example"] P could be 3∗53*5, but not 32∗53^2*5. Let f(n)f(n) be the number of pairs of integers d,ed, e such that ad2+2bde+ce2=nad^2 + 2bde + ce^2= n. Show thatf(n)f(n) is finite and that f(n)=f(Pkn)f(n) = f(P^{k}n) for every positive integer kk.
Original Statement:
Let a,b,ca,b,c be given integers a>0,a > 0, ac−b2=P=P1⋯Pnac-b^2 = P = P_1 \cdots P_n where P1⋯PnP_1 \cdots P_n are (distinct) prime numbers. Let M(n)M(n) denote the number of pairs of integers (x,y)(x,y) for which ax2+2bxy+cy2=n. ax^2 + 2bxy + cy^2 = n. Prove that M(n)M(n) is finite and M(n)=M(Pk⋅n)M(n) = M(P_k \cdot n) for every integer k≥0.k \geq 0. Note that the "nn" in PNP_N and the "nn" in M(n)M(n) do not have to be the same.
functionalgebraequationIMO Shortlist
Poland goes Combinatorics

Source: IMO Shortlist 1993, Poland 1

3/25/2006
Let SnS_n be the number of sequences (a1,a2,…,an),(a_1, a_2, \ldots, a_n), where ai∈{0,1},a_i \in \{0,1\}, in which no six consecutive blocks are equal. Prove that Sn→∞S_n \rightarrow \infty when n→∞.n \rightarrow \infty.
inductioncombinatoricsRamsey TheoryPermutation patternspatternIMO ShortlistPoland
Composited function and relatively prime positive integers

Source: IMO Shortlist 1993, Ireland 3

3/16/2006
Let SS be the set of all pairs (m,n)(m,n) of relatively prime positive integers m,nm,n with nn even and m<n.m < n. For s=(m,n)∈Ss = (m,n) \in S write n=2k⋅non = 2^k \cdot n_o where k,n0k, n_0 are positive integers with n0n_0 odd and define f(s)=(n0,m+n−n0). f(s) = (n_0, m + n - n_0). Prove that ff is a function from SS to SS and that for each s=(m,n)∈S,s = (m,n) \in S, there exists a positive integer t≤m+n+14t \leq \frac{m+n+1}{4} such that ft(s)=s, f^t(s) = s, where ft(s)=(f∘f∘⋯∘f)⏟t times(s). f^t(s) = \underbrace{ (f \circ f \circ \cdots \circ f) }_{t \text{ times}}(s). If m+nm+n is a prime number which does not divide 2k−12^k - 1 for k=1,2,…,m+n−2,k = 1,2, \ldots, m+n-2, prove that the smallest value tt which satisfies the above conditions is [m+n+14]\left [\frac{m+n+1}{4} \right ] where [x]\left[ x \right] denotes the greatest integer ≤x.\leq x.
functionmodular arithmeticnumber theoryIterationfunctional equationIMO Shortlist