Pairwise Cube Sums
{ 6 , 1 9 , 3 0 } ⟹ 6 + 1 9 = 2 5 , 6 + 3 0 = 3 6 , 1 9 + 3 0 = 4 9
Shown above is an example of a set of three numbers such that every pair of numbers in the set sums to the square of a distinct rational number. There are infinitely many of these sets of 3 numbers; unfortunately there are no sets with 4 numbers where this property holds. There are, however, infinitely many sets containing 4 numbers such that each pair of numbers sums to the cube of a distinct rational number. If one of these sets can be written as { w , x , y , z } , where w , x , y , z ∈ Q and w < x < y < z , what is the minimum value of ∣ w + x + y + z ∣ ?
The answer is 0.
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1 solution
I'll be honest, neither have I :/ I've done computer search after computer search looking for counter examples but w + x + y + z = 0 in all cases so far, so I felt safe in posting the problem. Hopefully someone will post a proof to this soon!
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I think the problem is false.
Suppose I can find three solutions ( a , b ) to the equation a 3 + b 3 = 2 N for some N . Call them ( a 1 , b 1 ) , ( a 2 , b 2 ) , ( a 3 , b 3 ) .
Then let w x y z = 2 1 ( a 1 3 + a 2 3 + a 3 3 ) − N = 2 1 ( a 1 3 − a 2 3 − a 3 3 ) + N = 2 1 ( − a 1 3 + a 2 3 − a 3 3 ) + N = 2 1 ( − a 1 3 − a 2 3 + a 3 3 ) + N Then if I did this right, we should have the six sums each summing to one of the six cubes. (Please check this!)
There are lots of rational N for which the equation x 3 + y 3 = 2 N has infinitely many rational solutions. I believe N = 3 is one of them (the associated elliptic curve has rank 1).
Oh, and w + x + y + z = 2 N .
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Patrick, I'm very impressed with your work. Here's a solution of the form you've described.
w = ( 2 3 + 9 3 + 1 8 3 ) / 2 − 2 0 5 2 = 1 2 3 2 . 5 x = ( 2 3 − 9 3 − 1 8 3 ) / 2 + 2 0 5 2 = − 1 2 2 4 . 5 y = ( − 2 3 + 9 3 − 1 8 3 ) / 2 + 2 0 5 2 = − 5 0 3 . 5 z = ( − 2 3 − 9 3 + 1 8 3 ) / 2 + 2 0 5 2 = 4 5 9 9 . 5
w + x + y + z = 4 1 0 4 = 2 ( 2 0 5 2 )
I'm going to spend tonight trying to salvage this problem, reword it to try to get the intended answer of 0 , or maybe think about changing the question to ask for the minimum value of ∣ w + x + y + z ∣ , something else entirely, or maybe I'll simply have to delete the problem. Thank you so much for your input.
I'm kinda disappointed with this problem. Once you put in the restriction w,x,y,z are rational then the sum can be as small as you want (so essentially the problem boils down to figuring out whether 0 is a possible answer) Ok, so I assumed that you meant w,x,y,z are integers, an interesting problem on its own(I think the answer would be 32832), but still didn't get the right answer :( Finally, seeing 0 as the answer after viewing the report made me facepalm. I was further confused when I see people actually getting this right.... Anyway, I have been enjoying your problems lately so keep up the good work :)
On a different note, note that problems involving finding 4 numbers such that each pairwise sum is a fixed power boils down to finding a number that can be expressed as the sum of two fixed power three ways(Can you see why?) I'm assuming you don't know this since the description should ring the taxicab number 8 7 5 3 9 3 1 9 , this is enough to establish that ∣ w + x + y + z ∣ isn't always 0 .
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Thanks Xuming, to be honest I'm disappointed in myself for the execution of this problem. This whole concept intrigued me, and when I tried to find solutions myself, the only answer I could come up with was 0 . Over and over again I would find solutions and w + x + y + z = 0 . Clearly I must not have been looking in the right place. I would like to salvage this problem as I feel that it has something worth noting. Any suggestions? Why do you feel that the minimum solution should be 3 2 8 3 2 when w , x , y , z ∈ Z ? If you can prove this then I think that would be a great thing to change the question to!
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@Garrett Clarke – Here's how I thought about this concept: We let w + x = a 3 , w + y = b 3 , x + y = c 3 , w + z = d 3 , x + z = e 3 , y + z = f 3 , what I'm doing is solving for w , x , y , z given a 2 , b 3 , . . . (I think this is called parametizing or something like that). Notice this system has solution in integers if and only if a 3 + f 3 = b 3 + e 3 = c 3 + d 3 as well as some parity condition on a , b , c . . . which I will omit; the important here is finding integers that can be espressed as the sum of two cubes three ways. At this point, you have several ways to create a problem. I recommend restricting a , b , c , . . . to be positive, so then the answer is the famous 8 7 5 3 9 3 1 9 . The number is well-known enough among the community(I think) that we don't have to deal with the higher mathematics of finding these special numbers.
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@Xuming Liang – Perfect, your solution makes sense to me, and the relationship between this problem and the taxicab numbers is great. I honestly think think the best thing to do in this situation is to delete this problem (it's honestly a mess) and I'll get working on posting the new one. Thanks a bunch!
Let a > b > c be some positive rational numbers and z y x w = 4 ( a 3 + b 3 − c 3 ) = 4 ( a 3 − b 3 + c 3 ) = 4 ( − a 3 + b 3 + c 3 ) = 4 ( − a 3 − b 3 − c 3 ) Watch that w < x < y < z and z + y z + x z + w y + x y + w x + w = 8 a 3 = 8 b 3 = − 8 c 3 = 8 c 3 = − 8 b 3 = − 8 a 3 Which are cubes of distinct rational numbers. It gives w + x + y + z = 0 . I haven't proved that there are no other values for w + x + y + z yet.