Answer to Question #146606 in Combinatorics | Number Theory for ankit

Question

Let us denote S_n=a^n+b^n+c^n for arbitrary numbers a,b,c. It is known that S_1=4,5, S_2=22,25, S_3=104,625 for some values of a,b,c. What is the largest possible value of S_{737}^2-S_{736}S_{738}?

Answer 1

S₇₃₇²-S₇₃₆*S₇₃₈=(a⁷³⁷+b⁷³⁷+c⁷³⁷)²-(a⁷³⁶+b⁷³⁶+c⁷³⁶)(a⁷³⁸+b⁷³⁸+c⁷³⁸)=

a¹⁴⁷⁴+a⁷³⁷b⁷³⁷+a⁷³⁷c⁷³⁷+b⁷³⁷a⁷³⁷+b¹⁴⁷⁴+b⁷³⁷c⁷³⁷+c⁷³⁷a⁷³⁷+c⁷³⁷b⁷³⁷+c¹⁴⁷⁴-

-(a¹⁴⁷⁴+a⁷³⁶b⁷³⁸+a⁷³⁶c⁷³⁸+b⁷³⁶a⁷³⁸+b¹⁴⁷⁴+b⁷³⁶c⁷³⁸+c⁷³⁶a⁷³⁸+c⁷³⁶b⁷³⁸+c¹⁴⁷⁴)=

2a⁷³⁷b⁷³⁷+2a⁷³⁷c⁷³⁷+2b⁷³⁷c⁷³⁷-a⁷³⁶(b⁷³⁸+c⁷³⁸)-b⁷³⁶(a⁷³⁸+c⁷³⁸)-c⁷³⁶(a⁷³⁸+b⁷³⁸)

It is known that:

4.5=a+b+c

22.25=a²+b²+c²

104.625=a³+b³+c³

Since the system is symmetric we can consider only one solution:

c=4.5-a-b

22.25=a²+b²+(4.5-a-b)²

b=$\frac{1}{4}(-\sqrt{-12*a^2+36*a+97}-2a+9)$

104.625=a³+$(\frac{1}{4}(-\sqrt{-12*a^2+36*a+97}-2a+9))^3$+$(4.5-a-\frac{1}{4}(-\sqrt{-12*a^2+36*a+97}-2a+9))^3$

a=0

b=$\frac{9}{4}-\frac{\sqrt{97}}{4}$

c=$\frac{9}{4}+\frac{\sqrt{97}}{4}$

S₇₃₇²-S₇₃₆*S₇₃₈=2b⁷³⁷c⁷³⁷-b⁷³⁶(c⁷³⁸)-c⁷³⁶(b⁷³⁸)=2$(\frac{9}{4}-\frac{\sqrt{97}}{4})^{737}$ * $(\frac{9}{4}+\frac{\sqrt{97}}{4})^{737}$ -$(\frac{9}{4}-\frac{\sqrt{97}}{4})^{736}$ $(\frac{9}{4}+\frac{\sqrt{97}}{4})^{738}$ -$(\frac{9}{4}-\frac{\sqrt{97}}{4})^{738}$ $(\frac{9}{4}+\frac{\sqrt{97}}{4})^{736}$ =-2-$(\frac{9}{4}+\frac{\sqrt{97}}{4})^{2}$-$(\frac{9}{4}-\frac{\sqrt{97}}{4})^{2}$=

-2-$\frac{1}{8}(89+9\sqrt{97})$ -$\frac{1}{8}(89-9\sqrt{97})$ =-2-89/8-89/8=2-89/4=-$\frac{81}{4}$