|
| |
| |
|
|
|
17, 41, 73, 89, 97, 113, 137, 193, 233, 241, 257, 281, 313, 337, 353, 401, 409, 433, 449, 457, 521, 569, 577, 593, 601, 617, 641, 673, 761, 769, 809, 857, 881, 929, 937, 953, 977, 1009, 1033, 1049, 1097, 1129, 1153, 1193, 1201, 1217, 1249, 1289, 1297, 1321
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
1,1
|
|
|
COMMENTS
|
Originally "Primes of the form x^2 + 4xy - 4y^2 (as well as of the form x^2 + 6xy + y^2)."
R. J. Mathar was the first to wonder whether these are also primes of the form 8k + 1. I did the easy part, proving that all primes of the form x^2 + 4xy - 4y^2 are congruent to 1 mod 8. Since x^2 + 4xy - 4y^2 = 2 or -2 is impossible, x must be odd. And since x is odd, x^2 = 1 mod 8.
If y is even, then both 4xy and 4y^2 are multiples of 8. If y is odd, then 4xy = 4 mod 8, but so is 4y^2, cancelling out the effect and leaving x^2 = 1 mod 8.
It remains to prove that every prime of the form 8k + 1 has a representation as x^2 + 4xy - 4y^2. - Alonso del Arte, Jan 28 2017
A necessary and sufficient condition of representation of p = 8n + 1 in your quadratic form is {8y^2 + 8n + 1 is perfect square}, since only in this case solving square equation for x, we have x = -2y + sqrt(8y^2 + 8n + 1) is [an] integer. For this a sufficient condition is { n has a form n = k^2 - k + i(4k + i - 1)/2, i >= 0, k >= 1}. In this case x = 2i + 2k - 1. y = k." - Vladimir Shevelev, Jan 26 2017
|
|
|
LINKS
|
|
|
|
CROSSREFS
|
|
|
|
KEYWORD
|
dead
|
|
|
AUTHOR
|
Laura Caballero Fernandez, Lourdes Calvo Moguer, Maria Josefa Cano Marquez, Oscar Jesus Falcon Ganfornina and Sergio Garrido Morales (oscfalgan(AT)yahoo.es), Jun 12 2008
|
|
|
EXTENSIONS
|
|
|
|
STATUS
|
approved
|
| |
|
|
