Question

Let be the integers a,b,c with

Answer 1

A^2 - 4b = c^2 <=> a^2 - c^2 = 4b <=> ( a + c )( a - c ) = 4b, cu a, b, c nr. intregi ;
Avem doua posibilitati :
a) a = 2x si b = 2y;
Atunci ( a + c )( a - c ) = 4b <=> x^2 - y^2 = b;
Relatia  a^2 -2b devine 2x^2 +2y^2 = ($\sqrt{2} x$)^2 + ($\sqrt{2} y$)^2, adica suma a doua patrate perfecte ;

b) a = 2x + 1 si b = 2y + 1 ;
In mod analog. obtii ca x^2 - y^2 + x - y = b;
si, dupa ce prelucrezi, ai ca a^2 - 2b = [$\sqrt{2} ( x + 1 / 2 )$]^2 + []$\sqrt{2} ( y + 1 / 2 )$^2, adica, suma a doua patrate perfecte .

Bafta!

Answer 2

${ a }^{ 2 }-4b={ c }^{ 2 }\\ \\ { a }^{ 2 }-4b+2b={ c }^{ 2 }+2b\\ \\ { a }^{ 2 }-2b={ c }^{ 2 }+{ \left( \sqrt { 2b } \right) }^{ 2 }$

$b={ 2 }^{ n }\\ \\ n>0$

(n) is the set of odd natural numbers greater than 0.