## Equation : A^2 + B^2 + 1 = C^2 + D^2 —- (Part 4)

$A^2 \; + \; B^2 \; + \; 1 \; = \; C^2 \; + \; D^2$

and

$(n + 1)^2 \; + \; (2 \, n)^2 \; + \; 1 \; = \; (2 \, n + 1)^2 \; + \; (n - 1)^2$

We can write

$C^2 \; + \; D^2 \; - \; ( \,A^2 \; + \; B^2 \,) \; = \; 1$

$A^2 \; = \; (n + 1)^2$             $B^2 \; = \; (2 \, n)^2$

$C^2 \; = \; (2 \, n + 1)^2$           $D^2 \; = \; (n - 1)^2$

$X \; = \; A^2 \; + \; B^2$             $Y \; = \; C^2 \; + \; D^2$

$X \; = \; a \, d$                          $Y \; = \; b \, c$

$b \, c \; - \; a \, d \; = \; 1$

Fibonacci’s identity

$(a^2 + b^2) \,(c^2 + d^2) = (a \,c - b \,d)^2 + (a \,d + b \,c)^2 = (a \,c + b \,d)^2 + (a \,d - b \,c)^2$

$(a \, c + b \, d)^2 + (a \, d - b \, c)^2 = (a \, c - b \, d)^2 + (a \, d + b \, c)^2$

We can use this identity by letting   $b \, c \; - \; a \, d \; = \; 1$

For example,

$13^2 \; + \; 24^2 \; + \; 1 \; = \; 11^2 \; + \; 25^2 \; = \; 746$

$(2\times 149 \; + \; 5\times 373)^2 \; + \; (2\times 373 \; - \; 5\times 149)^2$

$= \; (2\times 149 \; - \; 5\times 373)^2 \; + \; (2\times 373 \; + \; 5\times 149)^2$

$= \; (2^2 \; + \; 5^2) \,(149^2 \; + \; 373^2)$

$= \; 4678570$