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.