Originally posted by FabianFnas Consider this equation:
n^4 + 4*m^4 = p
where x and y is an positive integer and p is a prime.
Does this equation have
(a) no solutions,
(b) one solution,
(c) many solutions, or
(d) infinitely many solutions?
If possible, show the a, b and p in the solution(s).
perhaps a silly question... but 3 pairs of variables? n-m x-y or a-b? i assume you mean to have the question be consistent in choosing one of these pairings (i.e. n^3 + 4*m^4 ... where n and m are pos int.... find n and m)?
in terms of finding the solution, i have to think for a bit 🙂
Originally posted by sonhouse Why isn't P=257 if m and n is 1? 1^4+ 4*1^4=257
Oh, I think the equation should have been written
That makes it =5. Otherwise its 257
I took it to be 1^4 + (4*1)^4 =P
I'd keep the things geepamoogle mentioned in mind and then try factorising b^4 over the ring of integers in Q[sqrt(p)], ie the ring D = Z[(1+sqrt(p))/2]. This gives, where d = (1+sqrt(p))/2:
4b^4 = (2d -a^2 -1)(2d + a^2 - 1)
Now let (a,b) be an integer solution for some given prime p and consider the ideals H = (2d - a^2 -1) and I = (2d + a^2 -1) in the ring where we live. Then there's some business about checking the ideal class group of Q[sqrt(p)]. You ultimately get some other equation whose integral solutions give rise to the integral solutions of the equation you want. I forget the details right now, but I can refresh my memory later perhaps.
EDIT On second thought, it isn't at all easy to see what the ideal class group will look like in general, so this probably isn't the approach to take.