Answer to Question #146063 in Discrete Mathematics for Asfandyar

Question

Let us denote S_n = aⁿ + bⁿ + cⁿ for arbitrary numbers a, b, c. 

It is known that S₁ = 8, S₂ = 66, S₃ = 536 for some values of

a, b, c. What is the smallest possible value of S²₄₂ — S₄₁ S₄₃?

Answer 1

$S_{42}^2 - S_{41}*S_{43} = (a^{42} + b^{42} +c^{42})^2 -\\ (a^{41} + b^{41} +c^{41})*(a^{43} + b^{43} +c^{43}) = \\ a^{84} + b^{84} +c^{84} +2a^{42}b^{42} + 2a^{42}c^{42} +\\ +2b^{42}c^{42} -a^{84} - b^{84} -c^{84} - a^{41}(b^{43} +c^{43})-\\ -b^{41}(a^{43} +c^{43})-c^{41}(a^{43} +b^{43}) = \\ =2a^{42}b^{42} + 2a^{42}c^{42} +2b^{42}c^{42} - a^{41}(b^{43} +c^{43})-\\ -b^{41}(a^{43} +c^{43})-c^{41}(a^{43} +b^{43}) \\ \begin{cases} a+b+c = 8\\ a^2+b^2+c^2 = 66\\ a^3+b^3+c^3 = 536 \end{cases}\\ \text{since the system is symmetric there will be only one solutions}:\\ \begin{cases} c = 8 -a- b\\ a^2+b^2+c^2 = 66\\ a^3+b^3+c^3 = 536 \end{cases}\\ \begin{cases} a^2+b^2+64+a^2+b^2-16a-16b-2ab = 66\\ a^3+b^3+c^3 = 536 \end{cases}\\ \begin{cases} 2a^2+2b^2-16a-16b-2ab = 2\\ a^3+b^3+512 - 192(a+b) + 24(a+b)^2-(a+b)^3 = 536 \end{cases}\\ a = 0 , b= 4 -\sqrt{17} , c=4 +\sqrt{17}\\ 2(4 -\sqrt{17})^{42}(4 +\sqrt{17})^{42} - \\ -(4 -\sqrt{17})^{41}(4 +\sqrt{17})^{43} - \\ -(4 +\sqrt{17})^{41}(4 -\sqrt{17})^{43} = \\ = -(4 -\sqrt{17})^{41}(4 +\sqrt{17})^{41}*\\ *((4 -\sqrt{17})^{2} +2(4 -\sqrt{17})(4 +\sqrt{17}) + (4 +\sqrt{17})^{2}) = \\ = - (16-17)^{41} * (64) = -(-1)^{41} *(64) = 64\\ answer: 64$