Az Eszterházy Károly Tanárképző Főiskola Tudományos Közleményei. 2004. Sectio Mathematicae. (Acta Academiae Paedagogicae Agriensis : Nova series ; Tom. 31)
Bui MINH PHONG and LI DONGDONG, Elementary problems which are equivalent to the Goldbach's conjecture
Elementary problems which are equivalent to the Goldbach's Conjecture 35 Now assume that {2n: 6 < 2n < p k + 3) C G. Let *2N be an even integer with f) < 27V < S(k). If 2A 7 < p k + 3, then we have 2TV G G by our assumption. Letp f e + 3 < 2N < S( Ar). Hence 2N-p 1 > 2TV-p 2 > ••• > 2 TV — PA : > 3. On the other hand, the conditions 2TV < S(k) and S(k) = min A k yield 2 N$A k. Since ,4*. = {2n > pk'. 2n — pi, 2n — />2, ..., 2n — all are composite numbers}, the last relations imply that 2TV - pi is a prime for some p t £ {pi, p 2, P3, • • PA.}Consequently, 27V G G, and so Lemma 1 is proved. Lemma 2. Let k be a positive integer. Then {2n: S(k) < 2n < S(k +1)} C G if and only if S(k) > p k+ 1 + 3. Proof. Assume that 5(A>) ^ 5(it+ 1) and {2n: 5(Ar) < 2n < S(k+ 1)} C G'. Then we have S(k) = p + q for for some primes p and q. Since the numbers S(k) — p and S(k) — q are primes, we infer from the definition of S(k) tliat. p > p k and q > Consequently, S(k) = p + q > 2p k + 4 > p fc+ i + 3. Now assume that S(k) ^ S(k + 1) and S(k) > p k+ 1 + 3. Let 2TV be an even integer for which S(k) < 27V < S(k + 1) is satisfied. As we have seen in the proof of Lemma 1, in this case we also liave 27V ^ Ak+\ and 27V - pi > 27V — p-2 > ... > 2TV - p k > 2TV - Pk+ l > S{k) - Pk+ l > 3. Consequently, 2TV - pi is a prime for some p t G {pi, p 2, P3, • • •, Pk , PFC+i}, which shows that 2TV G G'. Finally, in the case S(Ar) = 5(Ar +1) we also have that S{k) = S{k + 1) > p k +i + 9 > p k+i + 1 by the definition of S(k -j- 1). The proof of Lemma 2 is finished.
