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Number Theory Level 4

Let a , b , c , d a, b, c, d be 4 distinct positive integers such that a 3 + b 3 = c 3 + d 3 a^3+b^3=c^3+d^3 . Find the minimum value of a + b + c + d a+b+c+d .


The answer is 32.

Piyush Palawat by Piyush Palawat , 22, India

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5 solutions

The smallest possible number which can be represented as sum of two distinct perfect cubes in two different ways is 1729 1729 known as the Hardy-Ramanujan number.
1729 = 1 3 + 1 2 3 = 1 0 3 + 9 3 1729 = 1^{3} + 12^{3} = 10^{3} + 9^{3}
∴ a + b + c + d = 1 + 12 + 9 + 10 = 32 \therefore a + b + c + d = 1+ 12 + 9 + 10 = 32
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Arjen Vreugdenhil
Jan 16, 2016

A famous fact of the History of Mathematics: Hardy's taxicab number. 1729 = 1000 + 729 = 1728 + 1 is the smallest number that can be written as the sum of two cubes in two different ways!

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Rohit Udaiwal
Jan 14, 2016

Hardy-Ramanujan Number

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I believe that Ramanujan was in the hospital being visited by Hardy, who told Ramanujan that the taxi-cab license number that hit him was 1729, and that must be a very bad number, to which Ramanujan replied:"No, it is the smallest number that can be written as the sum of two cubes in two different ways."Ed Gray

Edwin Gray - 2 years, 4 months ago
Iwan Sandjaja
Jan 15, 2016 
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for x in combinations(range(50),4):
     y = x[3]**3+x[0]**3-x[1]**3-x[2]**3
     if y==0:
         print(x[3], x[0], x[1], x[2], y)
         break

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Shubhang Mundra
Jan 14, 2016

Simply use the RAMANUJAN NUMBER .

a 3 + b 3 a^3+b^3 =1729

➡ 1 3 + 1 2 3 = 1729 1^3+12^3=1729

➡a=1 , b=12

c 3 + d 3 c^3+d^3 =1729

➡ 9 3 + 1 0 3 = 1729 9^3+10^3=1729

➡c=9 , d=10

Therefore, a+c+c+d=1+12+9+10

=32

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same logic guy..

Shreyansh Choudhary - 5 years, 5 months ago

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