p^2 = a^2 + b^2 + c^2, p is an odd prime

 
 

In how many ways can we solve
 

p^2 \; = \; a^2 \; + \; b^2 \; + \; c^2               for      0 < a \leq b \leq c

if   p   is an odd prime?

 
 
 
 

 
 
 
 
 

 
 
 

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About benvitalis

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

2 Responses to p^2 = a^2 + b^2 + c^2, p is an odd prime

  1. K.D. BAJPAI says:

    p^2 = a^2 + b^2 + c^2 for 0 < a <= b <= c
    Some solutions for the problem are: [given here only distinct a,b,c]

    7^2 = 2^2 + 3^2 + 6^2
    11^2 = 2^2 + 6^2 + 9^2
    13^2 = 3^2 + 4^2 + 12^2
    17^2 = 8^2 + 9^2 + 12^2
    19^2 = 1^2 + 6^2 + 18^2
    19^2 = 6^2 + 10^2 + 15^2

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