Radicals are complicated

Algebra Level 4

2016 + x 3 + 2016 − x 3 = 12 \large \sqrt[3]{2016+\sqrt{x}}+\sqrt[3]{2016-\sqrt{x}}=12

If x = 2 m + 201 6 n x=2^m+2016^n satisfies the equation above, where m m and n n are positive integers , find m + n m+n .


The answer is 20.

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Tommy Li by Tommy Li , 22, Hong Kong

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2 solutions

Tommy Li
Jun 13, 2016

2016 + x 3 + 2016 − x 3 = 12 \sqrt[3]{2016+\sqrt{x}}+\sqrt[3]{2016-\sqrt{x}}=12

( 2016 + x 3 + 2016 − x 3 ) 3 = 1 2 3 (\sqrt[3]{2016+\sqrt{x}}+\sqrt[3]{2016-\sqrt{x}})^3=12^3

( 2016 + x ) + 3 ( 2016 + x ) 2 3 ( 2016 − x ) 1 3 + 3 ( 2016 + x ) 1 3 ( 2016 − x ) 2 3 + ( 2016 − x ) = 1728 (2016+\sqrt{x})+3(2016+\sqrt{x})^{\frac{2}{3}}(2016-\sqrt{x})^{\frac{1}{3}}+3(2016+\sqrt{x})^{\frac{1}{3}}(2016-\sqrt{x})^{\frac{2}{3}}+(2016-\sqrt{x})=1728

( 4032 − 1728 ) + 3 ( ( 2016 + x ) 1 3 ( 2016 − x ) 1 3 ) ) ( ( 2016 + x ) 1 3 + ( 2016 − x ) 1 3 ) = 0 (4032-1728)+3((2016+\sqrt{x})^{\frac{1}{3}}(2016-\sqrt{x})^{\frac{1}{3}}))((2016+\sqrt{x})^{\frac{1}{3}}+(2016-\sqrt{x})^{\frac{1}{3}}) = 0

2304 + 3 ( 201 6 2 − x ) 1 3 ( 2016 + x 3 + 2016 − x 3 ) = 0 2304+3(2016^2-x)^{\frac{1}{3}}(\sqrt[3]{2016+\sqrt{x}}+\sqrt[3]{2016-\sqrt{x}}) = 0

2304 + 3 ( 12 ) ( 201 6 2 − x ) 1 3 = 0 2304+3(12)(2016^2-x)^{\frac{1}{3}}=0

64 + ( 201 6 2 − x ) 1 3 = 0 64+(2016^2-x)^{\frac{1}{3}}=0

( 201 6 2 − x ) 1 3 = − 2 6 (2016^2-x)^{\frac{1}{3}}=-2^{6}

201 6 2 − x = ( − 2 6 ) 3 2016^2-x=(-2^{6})^{3}

x = 2 18 + 201 6 2 x=2^{18}+2016^2

m + n = 18 + 2 = 20 m+n=18+2=20

6 Helpful 0 Interesting 0 Brilliant 0 Confused

Nice problem! One picky note: your question should specify that m m and n n are positive integers.

Even with that said, how do we know that 2 18 + 201 6 2 2^{18} + 2016^2 is the only way to write this value? (This is just a fun little exercise.)

Eli Ross Staff - 5 years ago

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2 18 + 201 6 2 = 2 10 ( 2 8 + 6 3 2 ) 2^{18} + 2016^2 = 2^{10}(2^8+63^2) . Therefore, the largest power of 2 2 which divides this number is 2 10 2^{10} .

2 m + 201 6 n 2^m + 2016^n must be divisible by 2 10 2^{10} and not by 2 11 2^{11} . But, the largest power which divides 2 m + 201 6 n 2^m + 2016^n is m i n ( m , 5 n ) min(m,5n) if m ≠ n m \neq n or m + k m+k if m = 5 n m=5n for some k k . It is easy to see that m = 10 m = 10 does not satisfy, n = 2 n=2 is the given and m = 5 n = 5 m = 5n = 5 or m = n = 0 m = n = 0 does not satisfy. Therefore, it is a unique representation.

Manuel Kahayon - 4 years, 12 months ago
Son Nguyen
Jun 15, 2016

Let a = 2016 + x 3 ; b = 2016 − x 3 a=\sqrt[3]{2016+\sqrt{x}};b=\sqrt[3]{2016-\sqrt{x}}

{ a 3 + b 3 = 4032 a + b = 12 \left\{\begin{matrix} a^3+b^3=4032 & & \\ a+b=12 & & \end{matrix}\right.

{ a 2 − a b + b 2 = 336 a + b = 12 → a = 12 − b \left\{\begin{matrix} a^2-ab+b^2=336 & & \\ a+b=12\rightarrow a=12-b & & \end{matrix}\right.

So that we have 3 b 2 − 36 b − 192 = 0 3b^2-36b-192=0

And b = − 4 ; b = 16 b=-4;b=16

2016 − x = 16 2016-\sqrt{x}=16 No solution in this case.

2016 − x = − 4 2016-\sqrt{x}=-4 so we have x = 4326400 = 201 6 2 + 2 18 x=4326400= 2016^2+2^{18}

So that m + n = 20 m+n=20

4 Helpful 0 Interesting 0 Brilliant 0 Confused

Thanks you very much! Substitution is quite helpful and efficient.

Tommy Li - 5 years ago

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No welcome,thanks.You too

Son Nguyen - 5 years ago

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