The p=3 case is indeed an elliptic curve, by a special hack: The equation becomes [n(n+1)/2]^2 = m^3, so m must be a square, say m = k^2; then n(n+1)/2 = k^3, which is an elliptic curve. Pushing a tad further gives 4n^2 + 4n + 1 = (2n+1)^2 = (2k)^3 + 1. The only solutions are (+-1)^2 = 0^3+1 and (+-3)^2 = 2^3+1, which lead only to trivial solutions of the original problem. Rich PS: When doing number theory, try to reserve "p" for primes. It saves trouble for those of us who skip over details. --R ----- Quoting Warren D Smith <warren.wds@gmail.com>:
Asimov: It's known that the equation 1^2 + . . . + n^2 = m^2 has only one non-trivial solution: (n,m) = (24,70).
4) What solutions (n,m) exist if the exponent 2 is replaced everywhere by an integer p > 2 ???
--the way uniqueness was shown for exponent p=2 was (a) computer finds all small solutions, (b) this problem is an "elliptic curve" which known technology will resolve.
The case p=3 should be settleable by the same method since it too yields an elliptic curve? Or is this not elliptic? Anyhow, in general Faltings' theorem, solving the Mordell conjecture, should be applicable to show that all of these problems (for each p) have only a finite number of solutions each (at most)?
I have not actually gone thru the details so caveat emptor -- I'm merely trying to point you in right directions. If you can find a modulus N under which some such equation is not soluble (easy to find such N using computer if exist) then of course no integer solutions can exist.
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