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

On a problem of W. Sierpinski DARIUSZ FREJMAN and ALEKSANDER GRYTCZUK Abstract. In this paper we dealt with the Diophantine équation (X

2 — 1) (;

y

2 - 1) = (z

2 - 1) . We prove: if X > l,y > 1 and Z = y + 1 then the only solution is (x, y, z) = (10. 13, 14). Let t

n = ^+ll be the n-th triangular number. K. Zarankiewicz (see [2], p. 53) has asked whether there exists a Pytagorean triangle whose sides are triangular numbers, i. e. (1) t

2

a + t

2=t

2

c. The ans wer to this question is affirmative, because for a — 132, b — 143, c = 164 the triangular numbers Í132, Í143, íi64 satisfy (1). On the other hand we have 8t

n = (2n + 1) — 1 and we see that (1) is équivalent to (2) ((2a+ 1)

2 -l)

2+ ((26+ 1)

2 - i)' = ((2c + l)

2 - l)'. Thus the équation (3) (X' - L)

2 + (»' - L)

1 = - 1)' has a solution in odd natural numbers x, y

, namely x — 2a + 1 = 265, y - 26 + 1 = 287, z = 2c -h 1 = 329. The équation (3) has also another solution in which not all numbers x, y, z axe odd, namely x = 10. y = 13, z = y + l = U (Cf. [2], p.

54). W. Sierpinski (see [2], p.

54) writes that we do not know whether the équation (3) has infinitely many solutions in natural numbers greater than one. In this connection we prove the following theorem: Theorem. The Diophantine équation (X

2-1)

2+(

Î

/

2-1)

2 = (Z

2-1)

2