Answer to Question #145876 in Combinatorics | Number Theory for Asfandyar

Question

Let us denote Sn = (a^n) + (b^n) +( c^n) for arbitrary numbers a, b, c. It is known that S1 = 8,5, S2 = 74, 25, S3 = 639, 625 for some values of a, b, c. What is the largest possible value of (S811)^2 - S810.S812?

Answer 1

$S_{811}^2 - S_{810}*S_{812} = (a^{811} + b^{811} +c^{811})^2 -\\ (a^{810} + b^{810} +c^{810})*(a^{812} + b^{812} +c^{812}) = \\ a^{1622} + b^{1622} +c^{1622} +2a^{811}b^{811} + 2a^{811}c^{811} +\\ +2b^{811}c^{811} -a^{1622} - b^{1622} -c^{1622} - a^{810}(b^{812} +c^{812})-\\ -b^{810}(a^{812} +c^{812})-c^{810}(a^{812} +b^{812}) = \\ =2a^{811}b^{811} + 2a^{811}c^{811} +2b^{811}c^{811} - a^{810}(b^{812} +c^{812})-\\ -b^{810}(a^{812} +c^{812})-c^{810}(a^{812} +b^{812}) \\ \begin{cases} a+b+c = 8.5\\ a^2+b^2+c^2 = 74.25\\ a^3+b^3+c^3 = 639.625 \end{cases}\\ \text{since the system is symmetric there will be only one solution}:\\ a = 0 , b= \dfrac{17}{4} -\dfrac{\sqrt{305}}{4} , c= \dfrac{17}{4} +\dfrac{\sqrt{305}}{4}\\ 2(\dfrac{17}{4} -\dfrac{\sqrt{305}}{4})^{811}(\dfrac{17}{4} +\dfrac{\sqrt{305}}{4})^{811} - \\ -(\dfrac{17}{4} -\dfrac{\sqrt{305}}{4})^{810}(\dfrac{17}{4} +\dfrac{\sqrt{305}}{4})^{812} - \\ -(\dfrac{17}{4} +\dfrac{\sqrt{305}}{4})^{810}(\dfrac{17}{4} -\dfrac{\sqrt{305}}{4})^{812} = \\ = -(\dfrac{17}{4} -\dfrac{\sqrt{305}}{4})^{810}(\dfrac{17}{4} +\dfrac{\sqrt{305}}{4})^{810}*\\ *((\dfrac{17}{4} -\dfrac{\sqrt{305}}{4})^{2} +2(\dfrac{17}{4} -\dfrac{\sqrt{305}}{4})(\dfrac{17}{4} +\dfrac{\sqrt{305}}{4}) + (\dfrac{17}{4} +\dfrac{\sqrt{305}}{4})^{2}) = \\ = - (\dfrac{289-305}{16})^{811} * (\dfrac{289}{4}) = -(-1)^{811} *(\dfrac{289}{4}) = 72.25\\ answer: 72.25$