Ramanujan number

What is the smallest positive number which can be expressed as sum of 2 2 different cubes in 2 2 different ways?

Note : a = b 3 + c 3 = c 3 + b 3 a=b^3+c^3=c^3+b^3 is a single way.


The answer is 91.

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2 solutions

Mark Hennings
Apr 27, 2019

We are interested in which positive integers can be written as the sum of two integer cubes. If n = a 3 + b 3 n = a^3 + b^3 then a + b = d a+b = d is a divisor of n n , and then a + b = d a b = 1 3 ( d 2 n d ) a+b \; = \; d \hspace{2cm} ab \; = \; \tfrac13\big(d^2 - \tfrac{n}{d}\big) and hence a , b a,b are the roots of the quadratic equation X 2 d X + 1 3 ( d 2 n d ) = 0 X^2 - dX + \tfrac13\big(d^2 - \tfrac{n}{d}\big) \; = \; 0 It is easy to check this quadratic for integer roots for any (without loss of generality, positive) divisor of n n , and hence find all the integer pairs a , b a,b such that a 3 + b 3 = n a^3 + b^3 = n .

It turns out that the only numbers between 1 1 and 100 100 that can be written as the sum of two integer cubes are 1 , 2 , 7 , 8 , 9 , 16 , 19 , 26 , 27 , 28 , 35 , 37 , 54 , 56 , 61 , 63 , 64 , 65 , 72 , 91 , 98 1, 2, 7, 8, 9, 16, 19, 26, 27, 28, 35, 37, 54, 56, 61, 63, 64, 65, 72, 91, 98 Of these, only 91 \boxed{91} can be written as a sum of cubes in two ways: 91 = 6 3 + ( 5 ) 3 = 3 3 + 4 3 91 = 6^3 + (-5)^3 = 3^3 + 4^3 .

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?

Mr. India - 2 years, 1 month ago

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We can improve the above argument a bit. We now that a , b a,b must be the roots of a quadratic, and those roots are 1 2 [ d ± 1 3 ( 4 n d d 2 ) ] \tfrac12\left[d \pm \sqrt{\tfrac13\big(\tfrac{4n}{d} - d^2\big)}\right] Since these roots have to be integers we must have that 1 3 ( 4 n d d 2 ) = u 2 \tfrac13\big(\tfrac{4n}{d} - d^2\big) \; = \; u^2 where u u is a non-negative integer which has the same parity as d d . Indeed, since we want a b a \neq b , we know that u u must be a positive integer. All this tells us that n = 1 4 d ( d 2 + 3 u 2 ) n \; = \; \tfrac14d(d^2 + 3u^2) where d , u d,u are positive integers with the same parity (note that, since d d and u u have the same parity, d 2 + 3 u 2 d^2 + 3u^2 is divisible by 4 4 , and hence this formula does yield integer values of n n ). We can now test possible pairs of d , u d,u . The table below gives all cases yielding n 100 n \le 100 : ( d , u ) n ( d , u ) n ( d , u ) n ( 1 , 1 ) 1 ( 1 , 3 ) 7 ( 1 , 5 ) 19 ( 1 , 7 ) 37 ( 1 , 9 ) 61 ( 1 , 11 ) 91 ( 2 , 2 ) 8 ( 2 , 4 ) 26 ( 2 , 6 ) 56 ( 2 , 8 ) 98 ( 3 , 1 ) 9 ( 3 , 3 ) 27 ( 3 , 5 ) 63 ( 4 , 2 ) 28 ( 4 , 4 ) 64 ( 5 , 1 ) 35 ( 5 , 3 ) 65 ( 5 , 5 ) 100 ( 6 , 2 ) 72 ( 7 , 1 ) 91 \begin{array}{cccccccc} (d,u) & n & \hspace{1cm} & (d,u) & n & \hspace{1cm} & (d,u) & n \\ \hline (1,1) & 1 & & (1,3) & 7 & & (1,5) & 19 \\ (1,7) & 37 && (1,9) & 61 & & (1,11) & 91 \\ (2,2) & 8 & & (2,4) & 26 & & (2,6) & 56 \\ (2,8) & 98 & & (3,1) & 9 & & (3,3) & 27 \\ (3,5) & 63 & & (4,2) & 28 & & (4,4) & 64 \\ (5,1) & 35 & & (5,3) & 65 & & (5,5) & 100 \\ (6,2) & 72 & & (7,1) & 91 \end{array} This list does not contain the numbers 2 , 16 , 54 2,16,54 , which can only be written as the sum of equal cubes 2 = 1 3 + 1 3 2 = 1^3 + 1^3 , 16 = 2 3 + 2 3 16 = 2^3 + 2^3 , 54 = 3 3 + 3 3 54 = 3^3 + 3^3 . Since 91 91 is the only number which appears twice in this table, we are done.

Mark Hennings - 2 years, 1 month ago
Mr. India
Apr 26, 2019

91 = 6 3 + ( 5 ) 3 = 3 3 + 4 3 91=6^3+(-5)^3=3^3+4^3

Please clarify that it is a positive number.

NO U - 2 years, 1 month ago

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Yeah, even I assumed all the numbers to be positive

Mahdi Raza - 1 year, 4 months ago

Can you prove that 91 91 is the smallest (positive) number with this property?

Jordan Cahn - 2 years, 1 month ago

You may need to ask for the smallest such positive number, otherwise the number could be negative, e.g., ( 12 ) 3 + ( 1 ) 3 = ( 9 ) 3 + ( 10 ) 3 (-12)^{3} + (-1)^{3} = (-9)^{3} + (-10)^{3} .

Brian Charlesworth - 2 years, 1 month ago

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Ya, done. Sorry for inconvenience

Mr. India - 2 years, 1 month ago

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