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How do I see if $a^3+b^3=c^3+d^3$ has any solutions where $ 1 \le a,b,c,d \in \mathbb{Z} \le 1000$ and $a \ne b \ne c \ne d$ ?

I know I can write a program to brute force this and find out, but is there a way I can determine this through algebra?

I thought I could use the Difference of Powers formula:

$(a-c)(a^2+ac+c^2)=a^3-c^3=d^3-b^3=(d-b)(d^2+db+b^2)$.

At this point I am stuck.

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  • $\begingroup$ it has lots, probably en.wikipedia.org/wiki/Fermat_cubic $\endgroup$ – K B Dave May 30 '19 at 3:56
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    $\begingroup$ The first of all the numbers which can be expressed as $a^3 + b^3 = c^3 + d^3$ is $1729$, which was pointed out by Mr. Srinivas Ramanujan, as far as I remember. $$1729 = 1^3 + 12^3 = 9^3 + 10^3$$ To know more, visit this :- en.wikipedia.org/wiki/1729_(number) $\endgroup$ – Soumalya Pramanik May 30 '19 at 4:01
  • $\begingroup$ oeis.org/A001235 $\endgroup$ – vadim123 May 30 '19 at 4:09
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    $\begingroup$ Some of the solutions:$ (1 , 12 , 9,10) (1 , 12 , 10,9) (1 , 103 , 64,94) (1 , 103 , 94,64) (1 , 150 , 73,144) (1 , 150 , 144,73) (1 , 249 , 135,235) (1 , 249 , 235,135) (1 , 495 , 334,438) (1 , 495 , 438,334)$ $\endgroup$ – NoChance May 30 '19 at 4:10
  • $\begingroup$ Note that the largest it (the common sum) could be is $1000^3+1000^3$, or 2 billion. Hence, there are at most 2184 of them (see this list from the oeis link I posted). $\endgroup$ – vadim123 May 30 '19 at 4:11

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