Question

Let be the integers a,b,c with


`a^2-4b=c^2`

To be shown that the number
`a^2-2b` can be written as a sum of 2 perfect squares.

Answer 1

`{ 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.

Answer 2

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 = (
`√(2) x`)^2 + (
`√(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 = [
`√(2) ( x + 1 / 2 )`]^2 + []
`√(2) ( y + 1 / 2 )`^2, adica, suma a doua patrate perfecte .

Bafta!