Cel Mai Mare Divizor Comun ; Numere prime_intre_ele

          In Ion CUCUREZEANU - Probleme de Aritmetica si Teoria Numerelor vedem aceasta problema (pag 26):

          In Wacław Franciszek Sierpiński -250 Problems in Elementary Number Theory aflam (pag 3):

     II. RELATIVELY PRIME NUMBERS  

41. Prove that for every integer k the numbers 2k+l and 9k+4 are

 relatively prime, and for numbers 2k-l and' 9k+4 find their

 greatest common  divisor as a function of k.

    Problemele au enunturi similare (or fi avand o "sursa" comuna ? sau e "folclor" ?)

Reformulam:

I.127 (i) Demonstrati ca $gcd(3\cdot k+1,14\cdot k+5)=1$;

        (ii) Aflati $gcd(3\cdot k-1,14\cdot k+5)$.

II.41 (i) Prove that $gcd(2\cdot k+1,9\cdot k+4)=1$;

       (ii) Find $gcd(2\cdot k-1,9\cdot k+4)$.

Vedem ca aceste probleme sunt inrudite cu postarea de aici.

    Solutia problemei I.127 arata asa (pag 54):

 

 iar solutia problemei II.41 este (pag 93; n.cip-erata k s; 9 vrea sa zica $k\neq9$):

II. RELATIVELY PRIME NUMBERS  

41. Numbers 2k+l and 9k+4 are relatively prime since 

 9(2k+l)- -2(9k+4) = I. Since 9k+4 = 4(2k-I)+(k+8), 

 while 2k-1 = 2(k+8)- 17, we have  

(9k+4, 2k-l) = (2k-l, k+8) = (k+8, 17). If k = 9  (mod 17),

 then (k+8, 17) = 17; in the contrary case, we have 17Ik+8, 

 hence (k+8, 17) = 1. Thus, (9k+4, 2k-l) = 17 if k = 9 (mod 17) 

 and  (9k+4, 2k-l) = 1 if k ;s 9 (mod 17) . 

?! Din nou...asemanari...