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. For example, legendre(2,p) = legendre(8,p)= legendre(32,p) = legendre(128,p) for the first 1000 primes p. 2,8,32,128, 3,27,243, 4,16,64,256, 5,125, 6,24,54,96,216,384,486, 7,343, 9,81, 10,40,160,250, 12,48,108,192,432, 14,56,224, 15,135,375, 18,72,162,288, 20,80,320,500, 21,189, 22,88,352, 26,104,416, 28,112,448, 30,120,270,480, 33,297, 34,136, 36,144,324, 38,152, 39,351, 42,168,378, 44,176, 45,405, 46,184, 50,200, 51,459, 52,208, 58,232, 60,240, 62,248, 66,264, 68,272, 70,280, 74,296, 76,304, 78,312, 82,328, 84,336, 86,344, 90,360, 92,368, 94,376, 98,392, 100,400, 102,408, 106,424, 110,440, 114,456, 116,464, 118,472, 122,488, 124,496, On Mon, Sep 29, 2014 at 12:55 PM, Henry Baker <hbaker1@pipeline.com> wrote:
Oops, that last "N" should have been "a":
Under what conditions can I reconstruct a ?
At 09:49 AM 9/29/2014, Henry Baker wrote:
The following question occurred to me on a long bike ride (?!?) yesterday:
Suppose I know the quadratic character of a wrt N for lots of different N's.
Under what conditions can I reconstruct N ?
I.e., is there a "Chinese Remainder Theorem" for quadratic residues?
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