From link:
Find integers a, b and c such that:
987654321a + 123456789b + c = (a + b + c)³
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$\begingroup$ The link you provided has solutions on it, together with code for a brute force search. Are you asking for a general algorithm to find integer points on cubic surfaces? $\endgroup$– S. Carnahan ♦Commented Oct 12, 2009 at 0:57
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$\begingroup$ How are the limits in that program derived? $\endgroup$– Cade RouxCommented Oct 12, 2009 at 2:36
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$\begingroup$ As far as I can tell, 31427 was chosen because it is the smallest number whose square is larger than 987654321. This covers all cases where a,b, and c are nonnegative. I would not be surprised if there were additional solutions where a and b have opposite signs. $\endgroup$– S. Carnahan ♦Commented Oct 12, 2009 at 6:09
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2 Answers
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Note that x^3 - x = (x-1)x(x+1). Now let x = (a+b+c)
and rewrite the equation as
(x-1)x(x+1) = 987654320*a + 123456788*b.
Let D be gcd(987654320,123456788) = 16. There are
integers A, B so that 987654320*A + 123456788*B = D,
e.g (1, -8). Pick your favorite x so that x^3 - x is a
multiple of D, say kD, let a = kA and b = kB, and then
set c to (x - (a+b)). If you need a and b to be
positive, choose x large enough so that kD is big
enough so that you can subtract multiples of
987654320*123456788/D^2 from, say ka and add them to kb.
If you need c to be positive as well, then pick x not
too large, as RHS < 10^9 * (a+b) < 10^9 x, so x larger
than 10^5 will not get you positive values of c.
Gerhard "Ask Me About System Design" Paseman, 2010.01.12
answered Jan 12, 2010 at 17:18
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Well, the numbers cannot be more then something like 10^5, so a simple program that cycles through all possible a and b and sees if there's a corresponding c might give you an answer pretty fast -- I dunno, from a minute to a day depending on your computer and programming language.
This also depends on whether you also implement some trivial checks, like try it modulo 2, 3 and 5 first.
answered Oct 11, 2009 at 23:50
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$\begingroup$ Can you prove analytically that the answers have to be divisible by 10 (not including the trivial answers)? $\endgroup$ Commented Oct 12, 2009 at 2:37
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$\begingroup$ No, I can't. The answer I posted goes in a completely different direction, but of course it's quite stupid. If you know and are able to prove that these numbers indeed must be divisible by 10, then you're solving it in a smarter way, congrats! $\endgroup$ Commented Oct 12, 2009 at 17:49
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$\begingroup$ I believe it should be possible, and I think it has something to do with 123456789 + 987654321 = 1111111110. I've looked at transforming using c -> x - a - b and looking at the solutions of the standard cubic for x in terms of a and b. $\endgroup$ Commented Oct 13, 2009 at 18:23