Az Eszterházy Károly Tanárképző Főiskola Tudományos Közleményei. 1998. Sectio Mathematicae. (Acta Academiae Paedagogicae Agriensis : Nova series ; Tom. 25)

TSANGARIS, PANAYIOTIS G., A sieve for all primes of the form R2+(X-)-l)2 ....

46 Fanayiotis G. Tsangaris (i) Let X -f YV

2 be a (non-negative) integral solution of (Fk), with Y > k + 1. Let x = (X - l)/2, y = (y + jfe - l)/2 and z = (Y - k - l)/2. Then (x, y, z) is a triad of positive integral solutions of (T). (ii) Let (x, y, z) be a triad of positive integral solutions of (T) with y > z. Let k = y - z, X ~ 2x + 1 and Y = 2y - (k - 1). Then X + y^ is a (non-negative) integral solution of (Fk) with Y > k -f 1. Proof. By using Theorem 2.1, Lemma 4.1 and the fact that the Diophantine equation (T) is equivalent to the equation (E). Proposition 4.3. Let k be a natural number. Let X + Y\J

2 be a non-negative integral solution of(Fk). Then the following hold true: (i) Let 0 < y < k - 1. Then X + Yy/

2 is a fundamental solution of a class of integral solutions of (Fk)(ii) Y f k. (Hi) Let Y = k + 1. Then X = 2k + 1. Moreover, X + Yy/

2 = (2k + 1) + (k + l)\/2 is obtained írom the fundamentcd solution (X

* = 2k — l,y* = k — 1) as follows: X + yy/2 - (2k - 1 + (k - 1)a/2)(3 + 2y/2) for k = 1 and X + yv/

2 = {2k-l-{k - l)\/2)(3 + 2>/2) for fc > 1. Proof. By direct computations. Proposition 4.4. Consider the Diophantine equation (Fk), where k > 1. Let X

" + Y*y/2 be the fundamental solution of a class A of (Fk) with X" > 0. Let 3 + 2\/2 be the fundamental solution of the equation x

2 - 2y

2 = I. Z

n = X

n + Y

nV

2 - (X* + Y*V2)(3 + 2\/2)

n for all n = 0,1,..., and z'

n = x'

n + Y'

nV2 = (X* - y V2)(3 -f 2\/2)

n for all n = 1,2,.... Then the following hold true: (i) Let A be genuine. Then the only (non-negative) integral solutions X -f Y\/

2 of (Fk) which belong to A or to A and satisfy the inequality Y > k + 1 are the following: