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Cubic AND Quartic Challenge (Posted on 2004-02-06) Difficulty: 4 of 5
What is the smallest positive integer that is the sum of two different pairs of (non-zero, positive) cubes?

What is the smallest positive integer that is the sum of two different pairs of integers raised to the 4th power? and how did you find it?

In other words what is the smallest x such that:
x = a^4 + b^4 = c^4 + d^4
(where x, a, b, c, and d are all different, non-zero, positive integers)?

Are you able to determine the answer without looking it up on the internet?

See The Solution Submitted by SilverKnight    
Rating: 2.5000 (2 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
re(2): A way to do it.===> NO WAY | Comment 13 of 15 |
(In reply to re: A way to do it.===> NO WAY by Ady TZIDON)

Gamer's comment is insightful, and allows for the quickest solving of the cubic problem.

There is not a single error in the statement, and the logic used is, well, obvious without explanation, but, in the hopes of un-obfuscating Ady, here goes:

Listing only the first twelve perfect cubes, one sees both the cube of 12, which is 1728, and the consecutive cubes of 9 and 10, namely, 729 and 1000. The sum of 729 and 1000 is 1729, about which Gamer commented, "If, from the number 1728, you add the one at the beginning (which is a perfect cube, hence the comment is relevant and not random) to the end, you also have a perfect cube (or rather, a sum of two obvious perfect cubes).

In short, the only thing we should be striving to eschew in this case is comments that are simultaneously rude and incorrect.

  Posted by Frank Riddle on 2004-02-24 21:33:04
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