6
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Based on this.

Can you find 4 distinct integers a, b, c, and d such that a+b, b+c, c+d, a+b+c, b+c+d and a+b+c+d are all perfect cubes?

If not, prove it.

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  • 1
    $\begingroup$ May one of the integers be zero? $\endgroup$ – 355durch113 Jun 12 '16 at 18:32
  • $\begingroup$ Yes. So pajonks answer is valid. $\endgroup$ – ev3commander Jun 12 '16 at 18:58
  • $\begingroup$ That's good, you made a question out of it :) $\endgroup$ – ABcDexter Jun 13 '16 at 12:47
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This answer describes all of the solutions to the desired equation. In particular, it proposes three fundamentally different classes of solutions, with representatives $(a,b,c,d)$ given as $(0,−8,9,−1)$ and $(1,7,-7,-1)$ and $(-728,728,1,999)$, each among an larger class of similarly constructed examples.

Let $a+b=x^3$ and $c+d=y^3$ and $a+b+c+d=z^3$. Then $$x^3+y^3=z^3.$$ This is impossible for non-zero $x,y,z$ by Fermat's Last Theorem (or special cases thereof which Fermat actually did prove). So, we need one of the summands to be zero.

Case 1: $a+b+c+d=0$

In this case, we get that both $a$ and $d$ and $a+d$ are perfect cubes since their negatives are $b+c+d$ and $a+b+c$ and $b+c$, which are each perfect cubes. So, now we need that one of $a$ and $d$ and $a+d$ is zero.

Case 1a: $a=0$

Now, we get that $b$ is a perfect cube, since $a+b$ is. However, then one of $b$ and $c+d$ and $b+c+d$ must be zero. However, it must be that $b+c+d$ is zero, since $b=-c-d$ cannot be zero since $a$ is. At this point, we can solve the system $$a=0$$ $$b+c=\alpha^3$$ $$c+d=\beta^3$$ $$b+c+d=0$$ which has solutions for all $\alpha,\beta$. For instance $(0,-8,9,-1)$ is of this form.

Case 1b: $d=0$

This case, by symmetry, will yield the same result as the previous case, except with $(a,b,c,d)$ swapped to $(d,c,b,a)$.

Case 1c: $a+d=0$

Note that this implies $b+c=0$ as well, so $a=-d$ and $b=-c$ follow. Moreover, since $a+b+c=a$ is a cube, $a$ must be a non-zero cube. Then, we get the following family of solutions $$a=\alpha^3$$ $$a+b=\beta^3$$ $$c=-b$$ $$d=-a$$ which can be solved for all $\alpha,\beta$ input into the system. For instance, the tuple $(1,7,-7,-1)$ is of this form

Case 2: $a+b=0$

Since $a+b+c$ is a perfect cube, we get that $c$ is a perfect cube. It suffices to have $c$ and $b+c$ and $c+d$ and $b+c+d$ to all be cubes. In particular, if you have a solution to $$\alpha^3+\beta^3=\kappa^3+\gamma^3$$ then you can generate a solution as $$c=\alpha^3$$ $$b+c+d=\beta^3$$ $$b+c=\kappa^3$$ $$c+d=\gamma^3$$ $$b=-a$$ Which looks overdetermined, but one can see that the first four equations are linearly dependent, so is not overdetermined. More work is necessary to finish classifying this class of solutions. One can use an identity that Ramanujan was fond of $1729=1^3+12^3=9^3+10^3$ to get the solution $(-728,728,1,999)$. It should be noted that $\alpha,\beta,\kappa,\gamma$ must be distinct for this to give distinct $a,b,c,d$. I'm not sure how to characterize the general set of such solutions, but it is notable that one can plug in taxicab numbers to get a lot of solutions of this form.

Case 3: $c+d=0$

This is, by symmetry, the same as Case 2.

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  • $\begingroup$ +1. With this someone could easily make a program to generate thousands of solutions. $\endgroup$ – ev3commander Jun 12 '16 at 19:57
  • $\begingroup$ That's a very good answer, explanation is crystal clear and especially the mention of Hardy-Ramanujan numbers :) $\endgroup$ – ABcDexter Jun 13 '16 at 12:50
  • $\begingroup$ Is there a solution with non-negative integers only? $\endgroup$ – Yakk Jun 13 '16 at 13:39
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    $\begingroup$ @Yakk No - notice that the only way to have one of $a+b$ or $c+d$ or $a+b+c+d$ be zero for non-negative integers only would be to have either $a=b=0$ or $c=d=0$ (or both). Both of these possibilities violate the condition that the numbers be distinct. $\endgroup$ – Milo Brandt Jun 13 '16 at 16:14
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With no restrictions to integers it's

$a=0$
$b=1$
$c=7$
$d=-8$

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  • $\begingroup$ -1 is a perfect cube of what? i? Isn't that a stretch? $\endgroup$ – user1717828 Jun 12 '16 at 20:51
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    $\begingroup$ @user1717828 Of -1. $\endgroup$ – Scimonster Jun 12 '16 at 21:21
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    $\begingroup$ @Scimonster, wow, read and wrote "cube" while thinking "square". Leaving comment up for the next Sunday-brained guy that comes along. $\endgroup$ – user1717828 Jun 12 '16 at 21:28
1
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Partial answer: By Fermat's Last Conjecture, there are no positive integers x, y, z, such that x^3 + y^3 = z^3

Let a+b = x, and c+d = y, and it follows that a+b+c+d cannot be a perfect cube.

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  • 1
    $\begingroup$ Unless $z=0$... $\endgroup$ – Milo Brandt Jun 12 '16 at 18:32
  • $\begingroup$ Yeah- whoops! Restricting to positive integers resolves that, but the positive restriction isn't in the question. $\endgroup$ – Tim C Jun 12 '16 at 18:33
1
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A quick python script spit out every answer where a, b, c, and d were integers between -30 and 30 inclusive. I didn't filter out any answers that were just reordered versions of other answers though:

! (-28, 1, 26, -26)
(-28, 27, -26, 26)
(-28, 28, -27, 0)
(-28, 28, -27, 26)
(-28, 28, -1, -26)
(-28, 28, -1, 0)
(-27, 19, -19, 27)
(-27, 19, 8, 0)
(-27, 26, -26, 27)
(-27, 26, 1, 0)
(-27, 28, -28, 27)
(-27, 28, -1, 0)
(-26, -1, 28, -28)
(-26, 26, -27, 0)
(-26, 26, -27, 28)
(-26, 26, 1, -28)
(-26, 26, 1, 0)
(-26, 27, -28, 28)
(-19, 19, -27, 0)
(-19, 19, 8, 0)
(-16, 16, -8, 0)
(-9, 1, 7, -7)
(-9, 8, -7, 7)
(-9, 9, -8, 0)
(-9, 9, -8, 7)
(-9, 9, -1, -7)
(-9, 9, -1, 0)
(-8, -19, 19, 8)
(-8, -19, 27, 0)
(-8, 7, -7, 8)
(-8, 7, 1, 0)
(-8, 9, -9, 8)
(-8, 9, -1, 0)
(-8, 16, -16, 8)
(-7, -1, 9, -9)
(-7, 7, -8, 0)
(-7, 7, -8, 9)
(-7, 7, 1, -9)
(-7, 7, 1, 0)
(-7, 8, -9, 9)
(-2, 2, -1, 0)
(-1, -26, 26, 1)
(-1, -26, 27, 0)
(-1, -7, 7, 1)
(-1, -7, 8, 0)
(-1, 2, -2, 1)
(-1, 9, -9, 1)
(-1, 9, -8, 0)
(-1, 28, -28, 1)
(-1, 28, -27, 0)
(0, -27, 19, -19)
(0, -27, 19, 8)
(0, -27, 26, -26)
(0, -27, 26, 1)
(0, -27, 28, -28)
(0, -27, 28, -1)
(0, -8, -19, 19)
(0, -8, -19, 27)
(0, -8, 7, -7)
(0, -8, 7, 1)
(0, -8, 9, -9)
(0, -8, 9, -1)
(0, -8, 16, -16)
(0, -1, -26, 26)
(0, -1, -26, 27)
(0, -1, -7, 7)
(0, -1, -7, 8)
(0, -1, 2, -2)
(0, -1, 9, -9)
(0, -1, 9, -8)
(0, -1, 28, -28)
(0, -1, 28, -27)
(0, 1, -28, 27)
(0, 1, -28, 28)
(0, 1, -9, 8)
(0, 1, -9, 9)
(0, 1, -2, 2)
(0, 1, 7, -8)
(0, 1, 7, -7)
(0, 1, 26, -27)
(0, 1, 26, -26)
(0, 8, -16, 16)
(0, 8, -9, 1)
(0, 8, -9, 9)
(0, 8, -7, -1)
(0, 8, -7, 7)
(0, 8, 19, -27)
(0, 8, 19, -19)
(0, 27, -28, 1)
(0, 27, -28, 28)
(0, 27, -26, -1)
(0, 27, -26, 26)
(0, 27, -19, -8)
(0, 27, -19, 19)
(1, -28, 27, 0)
(1, -28, 28, -1)
(1, -9, 8, 0)
(1, -9, 9, -1)
(1, -2, 2, -1)
(1, 7, -8, 0)
(1, 7, -7, -1)
(1, 26, -27, 0)
(1, 26, -26, -1)
(2, -2, 1, 0)
(7, -8, 9, -9)
(7, -7, -1, 0)
(7, -7, -1, 9)
(7, -7, 8, -9)
(7, -7, 8, 0)
(7, 1, -9, 9)
(8, -16, 16, -8)
(8, -9, 1, 0)
(8, -9, 9, -8)
(8, -7, -1, 0)
(8, -7, 7, -8)
(8, 19, -27, 0)
(8, 19, -19, -8)
(9, -9, 1, 0)
(9, -9, 1, 7)
(9, -9, 8, -7)
(9, -9, 8, 0)
(9, -8, 7, -7)
(9, -1, -7, 7)
(16, -16, 8, 0)
(19, -19, -8, 0)
(19, -19, 27, 0)

Here's the script: http://pastebin.com/7nQjrxKM

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  • $\begingroup$ Brute forcing this feels a bit like cheating, but maybe giving this many answers will help someone else identify patterns. $\endgroup$ – Stephen Lujan Jun 13 '16 at 15:38

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