Question #146606
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}?
1
Expert's answer
2020-12-03T07:51:06-0500

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=14(12a2+36a+972a+9)\frac{1}{4}(-\sqrt{-12*a^2+36*a+97}-2a+9)

104.625=a3+(14(12a2+36a+972a+9))3(\frac{1}{4}(-\sqrt{-12*a^2+36*a+97}-2a+9))^3+(4.5a14(12a2+36a+972a+9))3(4.5-a-\frac{1}{4}(-\sqrt{-12*a^2+36*a+97}-2a+9))^3

a=0

b=94974\frac{9}{4}-\frac{\sqrt{97}}{4}

c=94+974\frac{9}{4}+\frac{\sqrt{97}}{4}


S7372-S736*S738=2b737c737-b736(c738)-c736(b738)=2(94974)737(\frac{9}{4}-\frac{\sqrt{97}}{4})^{737} * (94+974)737(\frac{9}{4}+\frac{\sqrt{97}}{4})^{737} -(94974)736(\frac{9}{4}-\frac{\sqrt{97}}{4})^{736} (94+974)738(\frac{9}{4}+\frac{\sqrt{97}}{4})^{738} -(94974)738(\frac{9}{4}-\frac{\sqrt{97}}{4})^{738} (94+974)736(\frac{9}{4}+\frac{\sqrt{97}}{4})^{736} =-2-(94+974)2(\frac{9}{4}+\frac{\sqrt{97}}{4})^{2}-(94974)2(\frac{9}{4}-\frac{\sqrt{97}}{4})^{2}=

-2-18(89+997)\frac{1}{8}(89+9\sqrt{97}) -18(89997)\frac{1}{8}(89-9\sqrt{97}) =-2-89/8-89/8=2-89/4=-814\frac{81}{4}


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