Az Eszterházy Károly Tanárképző Főiskola Tudományos Közleményei. 2004. Sectio Mathematicae. (Acta Academiae Paedagogicae Agriensis : Nova series ; Tom. 31)

LUCA, F. and SZALAY, L., Linear diophantine equations with three consecutive binomial coefficients

56 F. Luca, L. Szalay Note that we may divide by D = B

2 — 4AC, because D ^ 0. With the above formulas for a and ß, we get that + Baß + Cß

2) + Aa + Cß= + C

)

, and so equation (7) becomes , 2 0 „ 2 -AC(A -B + C) An

2 + Buv + C»

2 = This last equation implies that (2 + and since 2Au + Bv = (2Ax + By) - (!2Aa + Bß) , 2ACÍB - 2.4) + AB(B - 2

C) = (2Az + By) -i d

2 }J

aa

c

[ -1 = 2

Ax + By - A, while (B

2 - 4AC)y - A(B - 2C) v = y - ß = B

2 - 4AC it follows that if we write X := {B

2 - 4AC)y - A(B - 2C), Y := 2Ax + By- A, E :=4A

2C{A- B + C), we get that X, Y 6 X and (8) X

2 - DY

2 = E. We thus see that if £) < 0, then the diophantine equation (3) has at most finitely integer solutions 1 < k < A; + 2 < n — 1. We now assume that D > 0. If E = 0, then since AC / 0, it follows that B = A + C. In this case, D = B

2-4AC = (A — C)

2, and so pairs of integers A', Y satisfying equation (8) satisfy either X = (C - A)Y or X = {A- C)Y. In terms of the variables x and y, the above lines become x + y = 1 or Ax + Cy — 0.