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Subject: "Diophantine equation"     Previous Topic | Next Topic
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Conferences The CTK Exchange College math Topic #512
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Oskar
guest
May-19-05, 10:14 PM (EST)
 
"Diophantine equation"
 
   I friend told me about this problem (which kinda looks like a special case of Fermat's last theorem). Solve this equation in the positive integers:

a^3 + b^3 = 22c^3

My idea was making a substitution a = m + n, b = m - n which gives:

(m-n)^3 + (m+n)^3 = 22c^3
(m^3 - 3nm^2 + 3mn^2 - n^3) + (m^3 + 3nm^2 + 3mn^2 + n^3) = 22c^3
2m^3 + 6mn^2 = 22c^3
m^3 + 3mn^2 = 11c^3
m(m^2 + 3n^2) = 11c^3

I've done a bunch of other stuff aswell (been working on this for some time :P), but none that appeared to show promise, like I think this one does (mostly because you can factor it). Another thing I was wondering was if it is possible to drag cubic-roots out of a congruence, ie:

a^3 + b^3 = 0 (mod 22)
a^3 = -b^3 (mod 22)
a = -b (mod 22)
a + b = 0 (mod 22)

This would be nice as it would imply that a+b = 22k for some k. Any help would be appriciated!

Oskar Sigvardsson

PS. I am really a senior in high school, but I posted this in the college section anyway, it'seems like such a problem. If this is incorrect, I apologise.


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Kris
Member since Feb-24-02
May-27-05, 08:52 AM (EST)
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1. "RE: Diophantine equation"
In response to message #0
 
   perhaps dividing by c^3 might do the trick. You then get p^3+q^3=22.
the solution where c=0 could then be considered afterwards.


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