What is the smallest positive number which can be expressed as sum of 2 different cubes in 2 different ways?
Note : a = b 3 + c 3 = c 3 + b 3 is a single way.
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Thank you. I was looking for a good proof. Is there any way so that we don't need to check so many numbers before getting 91?
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We can improve the above argument a bit. We now that a , b must be the roots of a quadratic, and those roots are 2 1 [ d ± 3 1 ( d 4 n − d 2 ) ] Since these roots have to be integers we must have that 3 1 ( d 4 n − d 2 ) = u 2 where u is a non-negative integer which has the same parity as d . Indeed, since we want a = b , we know that u must be a positive integer. All this tells us that n = 4 1 d ( d 2 + 3 u 2 ) where d , u are positive integers with the same parity (note that, since d and u have the same parity, d 2 + 3 u 2 is divisible by 4 , and hence this formula does yield integer values of n ). We can now test possible pairs of d , u . The table below gives all cases yielding n ≤ 1 0 0 : ( d , u ) ( 1 , 1 ) ( 1 , 7 ) ( 2 , 2 ) ( 2 , 8 ) ( 3 , 5 ) ( 5 , 1 ) ( 6 , 2 ) n 1 3 7 8 9 8 6 3 3 5 7 2 ( d , u ) ( 1 , 3 ) ( 1 , 9 ) ( 2 , 4 ) ( 3 , 1 ) ( 4 , 2 ) ( 5 , 3 ) ( 7 , 1 ) n 7 6 1 2 6 9 2 8 6 5 9 1 ( d , u ) ( 1 , 5 ) ( 1 , 1 1 ) ( 2 , 6 ) ( 3 , 3 ) ( 4 , 4 ) ( 5 , 5 ) n 1 9 9 1 5 6 2 7 6 4 1 0 0 This list does not contain the numbers 2 , 1 6 , 5 4 , which can only be written as the sum of equal cubes 2 = 1 3 + 1 3 , 1 6 = 2 3 + 2 3 , 5 4 = 3 3 + 3 3 . Since 9 1 is the only number which appears twice in this table, we are done.
9 1 = 6 3 + ( − 5 ) 3 = 3 3 + 4 3
Please clarify that it is a positive number.
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Yeah, even I assumed all the numbers to be positive
Can you prove that 9 1 is the smallest (positive) number with this property?
You may need to ask for the smallest such positive number, otherwise the number could be negative, e.g., ( − 1 2 ) 3 + ( − 1 ) 3 = ( − 9 ) 3 + ( − 1 0 ) 3 .
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We are interested in which positive integers can be written as the sum of two integer cubes. If n = a 3 + b 3 then a + b = d is a divisor of n , and then a + b = d a b = 3 1 ( d 2 − d n ) and hence a , b are the roots of the quadratic equation X 2 − d X + 3 1 ( d 2 − d n ) = 0 It is easy to check this quadratic for integer roots for any (without loss of generality, positive) divisor of n , and hence find all the integer pairs a , b such that a 3 + b 3 = n .
It turns out that the only numbers between 1 and 1 0 0 that can be written as the sum of two integer cubes are 1 , 2 , 7 , 8 , 9 , 1 6 , 1 9 , 2 6 , 2 7 , 2 8 , 3 5 , 3 7 , 5 4 , 5 6 , 6 1 , 6 3 , 6 4 , 6 5 , 7 2 , 9 1 , 9 8 Of these, only 9 1 can be written as a sum of cubes in two ways: 9 1 = 6 3 + ( − 5 ) 3 = 3 3 + 4 3 .