Transcription of INMO 2021
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INMO 2021
INMO 2021 Official SolutionsProblem 2is an integer, and letm1,n1,m2,n2, ,mr,nrbe2rintegerssuch that|minj mjni|= 1for any two integersiandjsatisfying1 i < j r. Determine the maximum possiblevalue ,n1,m2,n2be integers satisfyingm1n2 m2n1= 1. By changing thesigns ofm2,n2if need be, we may assume thatm1n2 m2n1= ,n3are integers satisfyingm1n3 m3n1= 1, again we may assume (by changingtheir signs if necessary) thatm1n3 m3n1= (n2 n3) =n1(m2 m3).Asm1,n1are relatively prime,m1dividesm2 m3; say,m2 m3=m1afor some integera. Thus, we getn2 n3=n1a. In other words,m3=m2 m1a , n3=n2 , ifm2n3 n2m3= 1, we get 1 =m2(n2 n1a) n2(m2 m1a) = (m1n2 m2n1)a= ,m3=m2 m1a=m2 m1,n3=n2 n1a=n2 if we were to have another pair of integersm4,n4such thatm1n4 n1m4= 1,we may assume thatm1n4 n1m4= 1. As seen above,m4=m2 m1,n4=n2 n1.
The proof is now complete. Alternate Solution. The only such pair is (0;0), which clearly works. Let us prove this is the only one. In what follows, we use 2(n) to denote the largest integer kso that 2kjn for any non-zero n2Z. If one of the cubics has 0 as a root, say the first one, then 03 + 0 a+ b= 0, so b= 0.
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