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        riddles >> medium >> An Ingenious Cubic Puzzle
        
(Message started by: K Sengupta on Dec 17^th, 2006, 6:51am)

Title: An Ingenious Cubic Puzzle
Post by K Sengupta on Dec 17^th, 2006, 6:51am

Determine analytically the minimum value of a positive integer constant S such that the equation A*(B^3) - B^3 + A + B = S has precisely :
(i) Three distinct solutions in positive integers.
(ii) Four distinct solutions in positive integers.

Title: Re: An Ingenious Cubic Puzzle
Post by THUDandBLUNDER on Dec 17^th, 2006, 11:22am

Did anyone do your last such problem?

Title: Re: An Ingenious Cubic Puzzle
Post by K Sengupta on Dec 17^th, 2006, 11:43pm

I would like advise that I was unable to solve Part II ( five solution case) of "An Ingenious Quadratic Puzzle" (Reference:
http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_easy;action=display;num=1161617444) but an accurate solution to the foregoing problem was achieved on the basis of comments from the members.

Title: Re: An Ingenious Cubic Puzzle
Post by THUDandBLUNDER on Dec 18^th, 2006, 8:51am

Oh, I meant An All Possible Value Puzzle.
http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_medium;action=display;num=1162566342

I see that no one has. Even Eigenray diidn't finish it. :o He must have got distracted by a harder one. :D

There the variables were exponents. Therefore more intractable, IMHO.

Title: Re: An Ingenious Cubic Puzzle
Post by Barukh on Dec 18^th, 2006, 11:20am

For the 3-solutions part, I get [hide]200[/hide].

Title: Re: An Ingenious Cubic Puzzle
Post by Barukh on Dec 19^th, 2006, 4:46am

...and for 4-solutions: [hide]9300[/hide].

:-/

Title: Re: An Ingenious Cubic Puzzle
Post by Eigenray on Dec 19^th, 2006, 9:35am

Barukh, I think you are off by one: [hide]200[/hide] and [hide]9300[/hide] have 4 and 5 solutions, respectively. I get record values at
S = [hide]2, 4, 12, 200, 4008, 274404, 1663680[/hide],
going up to 7 solutions, but I don't know how to do it "analytically" (one could certainly check up through [hide]200[/hide] by hand, but I don't know if that counts).

Basically, [hide]for each divisor (b+1) of S, check if a-1 = (S-(b+1))/(b³+1) is an integer. One only needs to check those b satisfying b³+1 <= S-(b+1), and then add the solution (a, b) = (1, S-1).[/hide]

Title: Re: An Ingenious Cubic Puzzle
Post by Barukh on Dec 19^th, 2006, 10:39am

on 12/19/06 at 09:35:31, Eigenray wrote:

Barukh, I think you are off by one: [hide]200[/hide] and [hide]9300[/hide] have 4 and 5 solutions

Yes, of course! I missed one simple solution for [hide]A=1[/hide]!

Quote:

S = [hide]2, 4, 12, 200, 4008, 274404, 1663680[/hide]

Interesting: my 4-solution (actually, 5-solution) number is not on your list! Must see how I miscalculated it.

Title: Re: An Ingenious Cubic Puzzle
Post by Eigenray on Dec 19^th, 2006, 12:16pm

on 12/19/06 at 10:39:29, Barukh wrote:

Interesting: my 4-solution (actually, 5-solution) number is not on your list! Must see how I miscalculated it.

I guess you missed 4008, 5016, and 8040, as these all have 5 solutions. How did you get your answers?

Title: Re: An Ingenious Cubic Puzzle
Post by Eigenray on Dec 19^th, 2006, 2:17pm

Ah, I see. If there are n+1 solutions for S, then there are [hideb]1<b₁<b₂<...<b_n < S-1 such that
S = b_i+1 mod b_i³+1.

Taking B = (b_i) = (1,2,3,4) gives your answer 9300. This is certainly a good way to construct values of S with any given number of solutions, but verifying minimality is tricky. For example, S(1,2,3,11) = 4008 is not too surprising, since (2,3,4,12) have lots of factors in common, but S(1,2,3,10) = 5016 and S(1,2,3,9) = 8040 are also lower.

However, the problem only asked for 4 solutions, and this is much easier. We may easily solve S = 2 mod 2, 3 mod 9, and 4 mod 28, to get S(1,2,3) = 228. On the other hand, if S < 228, then S >= (b+1)+(b³+1) implies b < 7, so it's not too much work to find S(1,3,4) = 200 as the smallest S with 4 solutions. (In fact, if S < 200, then b<6, so we need only check (1,2,4), (1,2,5), (1,3,5), and (1,4,5).[/hideb]

Title: Re: An Ingenious Cubic Puzzle
Post by Barukh on Dec 19^th, 2006, 11:24pm

on 12/19/06 at 14:17:45, Eigenray wrote:

Ah, I see.

That's exactly how I got my answers. I just missed the obvious solution [hide]b_n = S-1[/hide].
:D

Nice problem!