Mathematical Olympiads 2000–2001
... equality indeed holds for arbitrarily large n. Define (m0 , n0 ) = (1, 1) and (mk+1 , nk+1 ) = (2mk + 3nk , mk + 2nk ) for k ≥ 1. It is easily verified that m2k+1 − 3n2k+1 = m2k − 3n2k . Thus, because the equation 3n2k −2 = m2k holds for k = 0, it holds for all k ≥ 1. Because n1 , n2 , . . . ...
... equality indeed holds for arbitrarily large n. Define (m0 , n0 ) = (1, 1) and (mk+1 , nk+1 ) = (2mk + 3nk , mk + 2nk ) for k ≥ 1. It is easily verified that m2k+1 − 3n2k+1 = m2k − 3n2k . Thus, because the equation 3n2k −2 = m2k holds for k = 0, it holds for all k ≥ 1. Because n1 , n2 , . . . ...
Wilson`s Theorem and Fermat`s Theorem
... • If n is a positive integer, φ(n) is the number of integers in the range {1, . . . , n} which are relatively prime to n. φ is called the Euler phi-function. • Euler’s theorem generalizes Fermat’s theorem to the case where the modulus is not prime. It says that if n is a positive integer and (a, n) ...
... • If n is a positive integer, φ(n) is the number of integers in the range {1, . . . , n} which are relatively prime to n. φ is called the Euler phi-function. • Euler’s theorem generalizes Fermat’s theorem to the case where the modulus is not prime. It says that if n is a positive integer and (a, n) ...
Mathematical Olympiads 2000–2001
... equality indeed holds for arbitrarily large n. Define (m0 , n0 ) = (1, 1) and (mk+1 , nk+1 ) = (2mk + 3nk , mk + 2nk ) for k ≥ 1. It is easily verified that m2k+1 − 3n2k+1 = m2k − 3n2k . Thus, because the equation 3n2k −2 = m2k holds for k = 0, it holds for all k ≥ 1. Because n1 , n2 , . . . ...
... equality indeed holds for arbitrarily large n. Define (m0 , n0 ) = (1, 1) and (mk+1 , nk+1 ) = (2mk + 3nk , mk + 2nk ) for k ≥ 1. It is easily verified that m2k+1 − 3n2k+1 = m2k − 3n2k . Thus, because the equation 3n2k −2 = m2k holds for k = 0, it holds for all k ≥ 1. Because n1 , n2 , . . . ...