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Can you solve this variant of Dragan K's equation?
Find distinct positive integers a,b,c such that a^3+b^3 = c^3+11c.
@Dragan K: Nice job finding that solutions. There is at least one more "small" solution.
@Dragan K: Bravo! Now the real question is: can some kind of pattern be found that will allow one to easily generate more (hopefully infinitely many) solutions? Unfortunately, I do not know the answer.
@mathteacher: Thank you for those interesting observations. There is much more to find out about this problem, but unfortunately no time left.
2 Answers
- Dragan KLv 51 decade agoFavorite Answer
I can find only one triple:
755^3 + 6913^3 = 6916^3 + 11*6916
Probably there are more solutions, but they are huge.
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I found it! :)
1135^3 + 1157^3 = 1444^3 + 11*1444
- StringLv 41 decade ago
I know this is too late, as I haven't solved the problem but only have one minor observation that might start off some other bright head:
c³ + 11c = (c-1)·c·(c+1) + 12c
and since c-1, c and c+1 are three consecutive integers at least one of them is divisible by 2 and at least one is divisible by 3. Since 12c is always divisible by 6 as well the entire expression is divisible by 6. Hence if c³+11c = a³+b³ is divisible by 6 then aâ¡-b mod 6.
If c-1 or c+1 is divisible by 12 then the expression is divisible by 12c obviously.
