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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

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

nonn,dead,changed

More terms from Michel Marcus, Feb 01 2014

proposed

approved

editing

proposed

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

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

More terms from Michel Marcus, Feb 01 2014

approved

editing

proposed

approved

editing

proposed

It remains to prove that every prime of the form 8k + 1 has a representation as x^2 + 4xy - 4y^2. _Vladimir Shevelev_ and _Jan Orwat_ have both presented arguments to this effect in SeqFan. - 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