Verified Answer :

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

\(b={ 2 }^{ n }\\ \\ n>0\)

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

crisforp

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!


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