welshfella one year ago 1729 is the smallest integer that can be written as the sum of 2 cubes of positive integers in 2 different ways. What is the smallest if negative integers are allowed?

1. welshfella

The result is also positive.

2. welshfella

I don't know if there is an analytical way of doing this . I did it by trial and error.

3. welshfella

I think its a fair assumption that 2 of the numbers will be close together and the other quite far apart as is the case with 1729

4. welshfella

so it was not all trial an error!

5. welshfella

maybe...

6. ganeshie8

how do you know that the smallest number is positive ?

7. welshfella

I don't I assumed that.

8. ganeshie8

there can be some negative integers, which can be expressible as sum of two cubes in two different ways right

9. welshfella

yes

10. ganeshie8

11. ganeshie8

$-|u| = a^3+b^3=c^3+d^3$

12. welshfella

yes Well maybe i should change the question to lowest positive number.

13. ganeshie8

Ahh you should change the question, because there is no "least" element in the set of integers.

14. welshfella

I also assumed (for no reason other than its the case with 1729) that one of the pairs of numbers differed by only 1)

15. welshfella

yes

16. ganeshie8

1729 is the smallest integer that can be written as the sum of 2 cubes of positive integers in 2 different ways. What is the smallest $$\color{red}{\text{positive}}$$ integer if negative integers are allowed? does that look good ?

17. welshfella

yes

18. ganeshie8

by trial and error i got 91 is that what you have too ?

19. welshfella

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

20. welshfella

I guess there must be a lot of higher numbers which have the same property as 1729. I'll have to ask my grandson if he can find some. He is presently learning the programming language Python in college . I expect that would find some.

21. ganeshie8

Hey, I found this generating formula \begin{align}(3a^2+5ab−5b^2)^3+(4a^2−4ab+6b^2)^3&= (6a^2−4ab+4b^2)^3\\&+(-5a^2+5ab+3b^2)^3 \end{align}

22. ganeshie8

plugin $$a=1, b=0$$ and we get \begin{align}(3*1^2+0-9)^3+(4*1^2−0+6=0)^3&= (6*1^2−0+0)^3\\&+(-5*1^2+0+0)^3 \end{align} $3^3+4^3 = 6^3+(-5)^3$

23. ganeshie8
24. welshfella

brilliant