29 Sep
2014
29 Sep
'14
1:38 p.m.
On 29/09/2014 20:26, W. Edwin Clark wrote:
Say that n ~ m if for each of the first 1000 primes p we have legendre(n,p) = legendre(m,p)
Each line below is an equivalence class with more than two elements where n and m are each <= 500.
If m,n are multiples of the same set of primes and their ratio is the square of a rational number, then legendre(m,p) = legendre(n,p) for *all* p. It seems like it's worth filtering out these cases. Brief inspection suggests that in fact all your equivalence classes are of this form, but I haven't done the actual calculations to check in more than a handful of easy-by-eye cases. -- g