Question

Demonstrati ca nu exista numere naturale de forma abc astfel incat:
$a,b^{2}$-$b,c^{2}$=a,bc

Answer 1

Relatia se mai poate scrie $ab^{2}$-$bc^{2}$=abc
$ab^{2}$100a+bc·(bc+1)
U($ab^{2}$)={0; 1; 4; 5; 6; 9} , U(100a+bc(bc+1))={0; 2; 6} si b$\neq$0⇒b{4; 6}
b=4⇒$a4^{2}$=100a+$4c^{2}$+4c
a=2⇒c∈{2; 7}
a=5⇒U($4c^{2}$+4c)$\neq$5
a=6⇒$64^{2}$>900+$49^{2}$+49
b=6⇒$a6^{2}$=100a+$6c^{2}$+6c
a=6⇒c∈{2; 7} si $66^{2}$$\neq$500+6c·(6c+1)
a=7⇒$76^{2}$>900+$69^{2}$+69⇒nu se verifica relatia
~Sper ca te-am ajutat!~