Answer to Question #146067 in Combinatorics | Number Theory for Asfandyar

Question

Let us denote S_n = aⁿ + bⁿ + cⁿ 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 S²₈₁₁ — S₈₁₀S₈₁₂? 

Answer 1

$S_n = a^n + b^n +c^n$

$S_{811}^2-S_{810}\cdot S_{812}=(a^{811}+b^{811}+c^{811})^2-(a^{810}+b^{810}+c^{810})(a^{812}+b^{812}+c^{812})=a^{1622}+b^{1622}+c^{1622}+2a^{811}b^{811}+2a^{811}c^{811}+2b^{811}c^{811}-a^{1622}-b^{1622}-c^{1622}-a^{810}(b^{812}+c^{812})-b^{810}(a^{812}+c^{812})-c^{810}(a^{812}+b^{812})=2a^{811}b^{811}+2a^{811}c^{811}+2b^{811}c^{811}-a^{810}(b^{812}+c^{812})-b^{810}(a^{812}+c^{812})-c^{810}(a^{812}+b^{812})$

$S_1=8.5,\ S_2=74.25, \ S_3=639.625:$

$\begin{cases} a+b+c=8.5\\a^2+b^2+c^2=74.25\\a^3+b^3+c^3=639.625 \end{cases}$

Solving the system of equations:

$a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ac)=8.5^2-2(ab+bc+ac)$

and

$a^3+b^3+c^3=(a+b+c)(a^2+b^2+c^2-(ab+bc+ac))+3abc=$

$=8.5(8.5^2-2(ab+bc+ac)-(ab+bc+ac))+3abc=$

$=8.5(8.5^2-3(ab+bc+ac))+3abc$

Then we have

$\begin{cases} a+b+c=8.5\\8.5^2-2(ab+bc+ac)=74.25\\8.5(8.5^2-3(ab+bc+ac))+3abc=639.625 \end{cases} \implies$

$ab+bc+ac=-1$

$3abc=639.625-8.5(8.5^2+3)=0$

Since $abc=0$ lets assume $a=0$

Then we have

$\begin{cases}b+c=8.5\\b^2+c^2=74.25\end{cases}\implies \begin{cases}b=8.5-c\\(8.5-c)^2+c^2=74.25\end{cases}$

$c^2-8.5c-1=0\implies c=\frac{17\pm\sqrt{305}}{4}$

Then $b=8.5-c=\frac{17}{2}-\frac{17\pm\sqrt{305}}{4}=\frac{17\mp\sqrt{305}}{4}$

Since the system is symmetric there is one solution:

$a=0, \ b = \frac{17-\sqrt{305}}{4}, \ c=\frac{17+\sqrt{305}}{4}$

Thus

$S_{811}^2-S_{810}\cdot S_{812}=2(\frac{17-\sqrt{305}}{4})^{811}(\frac{17+\sqrt{305}}{4})^{811}-(\frac{17-\sqrt{305}}{4})^{810}(\frac{17+\sqrt{305}}{4})^{812}-(\frac{17+\sqrt{305}}{4})^{810}(\frac{17-\sqrt{305}}{4})^{812}=-(\frac{17-\sqrt{305}}{4})^{810}(\frac{17+\sqrt{305}}{4})^{810}((\frac{17-\sqrt{305}}{4})^2+2\frac{17-\sqrt{305}}{4}\frac{17+\sqrt{305}}{4}+(\frac{17+\sqrt{305}}{4})^2)=-(\frac{289-305}{16})^{811}\cdot\frac{289}{4}=-(-1)^{811}\cdot\frac{289}{4}=72.25$

Answer: 72.25