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

Source: IMO Shortlist 1993, Georgia 1

October 24, 2005
functionalgebraequationIMO Shortlist

Problem Statement

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.
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