Az Eszterházy Károly Tanárképző Főiskola Tudományos Közleményei. 1994. Sectio Mathematicae. (Acta Academiae Paedagogicae Agriensis : Nova series ; Tom. 22)

FREJMAN, D. and GRYTCZUK, A., On a problem of W. Sierpinski

98 Ùariusz Frejman and Aleksander Grytczuk has exactly one solution in natural numbers x > 1, y > 1, z, such that z = y -f 1, namely < x,y,z > = < 10,13,14 > . In the proof of the Theorem we use the following: Lemma. The Diophantine équation 3w

4 - 2v

2 = 1 has exactly two solutions in natural numbers u, v namely < u, v > = < 1,1 >, < 3,11 > . The proof of this Lemma has been given by R. T. Bumby in [1], PROOF of the Theorem. Let z = y + 1 and suppose that the equaiton (3) has a solution in natural numbers x, y, z. Then we have (x

2-1)

2=(2î/ + 1) (2y

2+2y-l). Let d= (2

y + 1, 2

y

2 + 2y ~ Í) then we have (5) 2y + l = dA, 2y

2 + 2y-l = dB- (A, B) = 1. From (5) we have (6) d

2A

2 -3 = 2d-B. By (6) it follows that d | 3, thus d = 1 or d = 3. Let us consider the case d = 1. Then by (5) and (4) it follows that (7) {x

2 — l)

2 = A • B] (A,B) = 1. From (7) we obtain that there exist integers a,ß such that (a,/?) = 1 and (8) A = a

2,B = ß

2,x

2 -1 = a- ß. By (8) and (6) it follows that (9) a

4 - 3 = 2ß

2

, {a,ß) = 1. From (9) we have that a; is an odd number. If ß is also an odd number then by (9) we have 2ß

2 +3 = 5 (mod 8) and so a

2 = 1 (mod 8) and we get a contradiction.