Inspired by Brian Charlesworth

Algebra Level 4

1416317954 1 1416317954 1 1416317954 1 . . . . . . 16 = a + b \large{\sqrt [ 16 ]{ 1416317954-\frac { 1 }{ 1416317954-\frac { 1 }{ 1416317954-\frac { 1 }{ ...... } } } } =a+\sqrt { b } }

If above given question can be expressed in the form of two integers a a and b b such that b b is square free. Find a + b a+b .


The answer is 5.

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1 solution

Samarth Agarwal
Sep 18, 2015

l e t 1416317954 = k l e t x = T h e g i v e n e x p r e s i o n t h e n x 16 = k x 16 x 16 + 1 x 16 = k a d d i n g 2 b o t h s i d e s ( x 8 + 1 x 8 ) 2 = 37634 2 ( x 4 + 1 x 4 ) 2 = 194 2 ( x 2 + 1 x 2 ) 2 = 14 2 ( x + 1 x ) 2 = 16 x + 1 x = 4 x 2 4 x + 1 = 0 x = 4 ± 12 2 x = 2 ± 3 x > 0 x = 2 + 3 a = 2 b = 3 a + b = 5 let\quad 1416317954=k\\ let\quad x=The\quad given\quad expresion\\ then\quad { x }^{ 16 }=k-{ x }^{ -16 }\\ { x }^{ 16 }+\frac { 1 }{ { x }^{ 16 } } =k\\ adding\quad 2\quad both\quad sides\\ { ({ x }^{ 8 }+\frac { 1 }{ { x }^{ 8 } } ) }^{ 2 }={ 37634 }^{ 2 }\\ { ({ x }^{ 4 }+\frac { 1 }{ { x }^{ 4 } } ) }^{ 2 }={ 194 }^{ 2 }\\ { ({ x }^{ 2 }+\frac { 1 }{ { x }^{ 2 } } ) }^{ 2 }={ 14 }^{ 2 }\\ { ({ x }+\frac { 1 }{ { x } } ) }^{ 2 }=16\\ \therefore \\ x+\frac { 1 }{ x } =4\\ { x }^{ 2 }-4x+1=0\\ x=\frac { 4\pm \sqrt { 12 } }{ 2 } \\ x=2\pm \sqrt { 3 } \\ x>0\\ \therefore \quad x=2+\sqrt { 3 } \\ a=2\\ b=3\\ a+b=\boxed { 5 }

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