Pythagorean triples (a, b, c); where c – b = 8 (a, b, c); where c – b = 8

 
The integers

a \; = \; 4 \, n
b \; = \; (n^2 - 4)
c \; = \; (n^2 + 4)

with   a \; < \; b

form a Pythagorean triple

 

a \; + \; c \; = \; 4 \, n \; + \; (n^2 + 4) \; = \; (n + 2)^2

b \; + \; c \; = \; (n^2 - 4) \; + \; (n^2 + 4) \; = \; 2 \, n^2
a^2 \; = \; (4 \, n)^2 \; = \; 8 \, (2 \, n^2)

that is,     a^2 \; = \; 8 \,(b + c)
 

a \, b \, c \; = \; 4 \, n \,(n^2 - 4) \,(n^2 + 4) \; = \; 4 \, n^5 \; - \; 64 \, n \; = \; 2^2 \, n^5 \; - \; 2^6 \, n
 

Area   A :
A \; = \; 2 \, n \,(n^2 - 4) \; = \; 2 \, n^3 \; - \; 2^3 \, n

Perimeter   P :
P \; = \; 4 \, n \; + \; 2 \, n^2 \; = \; 2 \, n \,(2 + n)

semiperimeter   s :
s \; = \; n \,(n + 2) \; = \; (n + 1)^2 \; - \; 1

 
c-b=8 A1

 
c-b=8 A2

 
 
 
 
 
 
 
 

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About benvitalis

math grad - Interest: Number theory
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