Fraction of Fraction

a + 1 b + 1 c + 1 d = 229 99 \large a + \frac{1}{b + \frac{1}{c + \frac{1}{d}}} = \dfrac{229}{99}

If a , b , c a, b, c and d d are positive integers satisfying the equation above, find the value of a 2 + b 2 + c 2 + d 2 a^{2} + b^{2} + c^{2} + d^{2} .


The answer is 74.

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1 solution

Raihan Fauzan
May 5, 2016

Make each variables have closest value to the result.

a = 198 99 = 2 a = \frac{198}{99} = 2

1 b + 1 c + 1 d = 31 99 \frac{1}{b + \frac{1}{c + \frac{1}{d}}} = \frac{31}{99}

b + 1 c + 1 d = 99 31 b + \frac{1}{c + \frac{1}{d}} = \frac{99}{31}

b = 93 31 = 3 b = \frac{93}{31} = 3

1 c + 1 d = 6 31 \frac{1}{c + \frac{1}{d}} = \frac{6}{31}

c + 1 d = 31 6 c + \frac{1}{d} = \frac{31}{6}

c = 30 6 = 5 c = \frac{30}{6} = 5

1 d = 1 6 \frac{1}{d} = \frac{1}{6}

d = 6 d = 6

So,

a 2 + b 2 + c 2 + d 2 = 2 2 + 3 2 + 5 2 + 6 2 = 4 + 9 + 25 + 36 = 74 a^{2} + b^{2} + c^{2} + d^{2} = 2^{2} + 3^{2} + 5^{2} + 6^{2} = 4 + 9 + 25 + 36 = \boxed{74}

Bonus question: can you prove that this is the unique solution?

展豪 張 - 5 years, 1 month ago

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What do you mean?

Raihan Fauzan - 5 years, 1 month ago

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Prove that ( a , b , c , d ) = ( 2 , 3 , 5 , 6 ) (a,b,c,d)=(2,3,5,6) is the only integer solution.

展豪 張 - 5 years, 1 month ago

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@展豪 張 Until now, I haven't found the other solution. Can you find it?

Raihan Fauzan - 5 years, 1 month ago

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@Raihan Fauzan This is not a proof that there is no other solution. Can you prove it?

展豪 張 - 5 years, 1 month ago

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@展豪 張 No, I cannot

Raihan Fauzan - 5 years, 1 month ago

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@Raihan Fauzan In the first line of your solution you wrote 'Make each variables have closest value to the result.'
Can you figure out why doing so?

展豪 張 - 5 years, 1 month ago

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