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

HERENDI, T. and PETHŐ, A., Trinomials, which are divisible by quadratic polynomials

Trinomials, which are divisible by quadratic polynomials 67 and from (5) and (6), using Lemma 1 we obtain a | A and bn-k-1 mod2 ,

/

n_

f

c_

i(

a^

2

) A = a (—1) bk-1 mod 2 . f

k_

1 b

2) Lemma 5. Let k,n G N and A £ Z\{0}. Then there exist only finitely many, efFectively computable a, 6, B G Z such that x

2 ~bx-a \ x

n — Bx

k — A. PROOF. Let a,b £ Z be such that b

k~

l mod 2 • /

k_

x (a,b

2) / 0 and x

2 — bx — a j x

n — Bx

k — A. Then by Lemma 4 (b) a | A and (7) 0 = A • b

k~

l mod 2 • (a, b

2) - a^-l)^"-*1 mod 2 • (a, 6

2) . Because of a | a may assume only finitely many différent values. Let a be fixed. Then the right hand side of (7) is a polynomial in 6, which has only finitely many roots, and the integer roots of it are efFectively computable. So there exist only finitely many possibilités for a, b (and they are efFectively computable). As f^-i (a, b

2) ± 0, by Lemma 4 (b) B is explicitely determinable from a and b so the numbers of the possible B is also finite and the values of B are effectively computable. Let a, 6 now be such that /jt_ 1 (a,b

2) = 0. By Lemma 2 (b) By Lemma 4 (a) a | A and (9) A = a • (b

n~

2 mod 2 • /

n_

2 (a, b

2) - B • b

k~

2 mod 2 • f

k_

2 (a, b

2)) , where b

k~

2 mod 2 • /

f

c_

2 (a, b

2) ± 0. ( Otherwise b

1 mod 2 • f

{ (a, b

2) = 0 would hold for every i and it is possible only when a, 6 = 0.)Asa | A the cardinahty of the possible a-s is finite and by (8) the cardinality of the possible b-s is also finite and efFectively computable. Let fix now a and b. Then (9) is a linear équation in B which has only one solution and the solution is explicitely given. So we obtain that B has only finitely many possible values in both cases and they are efFectively computable. By replacing y with y

2 in the définition of f

n(x

yy) it is easy to prove the following: Lemma 6. y

n mod 2 • f

n (x, y

2) = y

n • F

n {f)

.