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 A141174 Duplicate of A007519. 8

%I

%S 17,41,73,89,97,113,137,193,233,241,257,281,313,337,353,401,409,433,

%T 449,457,521,569,577,593,601,617,641,673,761,769,809,857,881,929,937,

%U 953,977,1009,1033,1049,1097,1129,1153,1193,1201,1217,1249,1289,1297,1321

%N Duplicate of A007519.

%C Originally "Primes of the form x^2 + 4xy - 4y^2 (as well as of the form x^2 + 6xy + y^2)."

%C _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.

%C 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.

%C 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

%C 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

%H Vincenzo Librandi, <a href="/A141174/b141174.txt">Table of n, a(n) for n = 1..1000</a>