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 ....

A sieve for all primes of the form r

2 + (a-+l)

2 51 Also, by Theorem 4.7 (iv), we deduce that X

n

+i -f V

n+

1 V2, where n > 1, are the only (non-negative) integral solutions of (F\

) such that Y

n+i > k +

1 = 2. Hence, according to Theorem 4.9, the numbers R

2

: ... are all composite. (iii) We have J?

n = R

n for every n = 0,1,... because Y* — 0. By Theorem 4.7 (iii) the numbers + Y

n+i>/2, where n > 0, are the only (non-negative) integral solutions of (F

k) such that Y"

n+

1 > k + 1. This completes the proof by invoking Theorems 1.3 and 4.9. (iv) By Theorem 4.7 (ii) the numbers X

n

+i +Y

n

+1\/2 with n > 0, together with the numbers X

n+

1 -f Y

n+

ls/

2, with n > 1, are the only (nonnegative) integral solutions of (F

k) such that Y

n

+\ > k + 1 and Y

n+

l > k -f 1. Thus the proof is completed by Theorem 1.3 and 4.9. (v) By Theorem 4.7 (i), the numbers X

n

+\ -f Y

n+\\/2 together with the numbers X

n+

1 -f Y

n+

l \/2, where n > 0, are the only (non-negative) integral solutions of (F\) such that Y

n

+1 > k + 1 and Y'

n,

l > k + 1. This finishes the proof of the whole Theorem, again in view of Theorems 1.3 and 4.9. Theorem 4.11. Consider the Diophantine equation (F

k), k — 0,1 Let X: + Y*V

2, (where r — 1, 2,.. ., m) be the only non-negative integral solutions of (F

k) such that: 0 < Y* < k - 1 for k > 1, While

, for k = 0 we have: X* = Y

r* = 1 for ail r = 1, 2,.. ., m. Let R

n, R'

n be the sequences, defined by the recursive formulae: R

n+

l = 34

R

n - - S(2k

2 + 1) for all n = 1,2,..., where R

0 = Y

r*

2 + k

2

, R

x = [2X; + 3Y;)

2 + k

2 (for a typical r). R

n

+i = 34R

n - R

n_

! - 8(2/c

2 + 1) for all n = 1,2,..., where = ^*

2 + R'i = (

2

Xr ~ + ^

2 (for a typical r). Suppose that the number N(x) = x

2 + (x -f I)

2 is composite. Then iV(x) is equal to some of the composite numbers R

n or R

n, for a suitable index, as stated in cases (i)-(v) of Theorem 4.10 (for some value of k). Proof. Since N(x) is composite it follows from Theorem 1.1 that there exist natural numbers y, z such that T(x) = T(y) + T(z).