(a^2 + b^2)(c^2 + d^2)(e^2 + f^2) as the sum of two squares

 
 
We know that

(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

so, applying this recursively,

(a^2 \; + \; b^2) \, (c^2 \; + \; d^2) \, (e^2 \; + \; f^2)
= \; (a c e + b d e + a d f - b c f)^2 \; + \; (a c f + b d f - a d e + b c e)^2
= \; (a c e + b d e - a d f + b c f)^2 \; + \; (a c f + b d f + a d e - b c e)^2
= \; (a c e - b d e + a d f + b c f)^2 \; + \; (-a c f + b d f + a d e + b c e)^2
= \; (-a c e + b d e + a d f + b c f)^2 \; + \; (a c f - b d f + a d e + b c e)^2

 

if

a \; = \; n
b \; = \; c \; = \; (n+1)
d \; = \; e \; = \; (n+2)
f \; = \; (n+3)

then

(n^2 \; + \; (n+1)^2) \, ((n+1)^2 \; + \; (n+2)^2) \, ((n+2)^2 \; + \; (n+3)^2)
= \; (2 \, n^3 + 8 \, n^2 + 9 \, n + 1)^2 \; + \; (2 \, n^3 + 10 \, n^2 + 15 \, n + 8)^2
= \; (2 \, n^3 + 8 \, n^2 + 11 \, n + 7)^2 \; + \; (2 \, n^3 + 10 \, n^2 + 13 \, n + 4)^2
= \; (2 \, n^3 + 8 \, n^2 + 7 \, n - 1)^2 \; + \; (2 \, n^3 + 10 \, n^2 + 17 \, n + 8)^2
= \; (2 \, n^3 + 12 \, n^2 + 19 \, n + 7)^2 \; + \; (2 \, n^3 + 6 \, n^2 + n - 4)^2

 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 

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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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