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c=a^{2014}+b^{2014} ,d=a^{2015}+b^{2015}, rel. prime (HOMC 15 Q13)

Source:

September 7, 2019

number theoryrelatively primeexponential

Problem Statement

Let

m

be given odd number, and let

a, b

denote the roots of equation

x^2 + mx - 1 = 0

and

c = a^{2014} + b^{2014}

d =a^{2015} + b^{2015}

. Prove that

c

and

d

are relatively prime numbers.