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

 

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

 
In   A^2 + B^2 + 1 = C^2 + D^2 —- (Part 1)

and

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

(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
 
SQRS+1 Part3

(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

 
 
 

 
 
 
 
 
 
 
 
 
 
 
 
 
 

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About benvitalis

math grad - Interest: Number theory
This entry was posted in Number Puzzles and tagged . Bookmark the permalink.

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