Numbers that are the sum of three nonzero squares

 

Legendre’s three-square theorem states that a natural number can be represented as the sum of three squares of integers

N \; = \; x^2 \; + \; y^2 \; + \; z^2

if and only if   N   is not of the form    N \; = \; 4^{a} \; (8 \, b \; + \; 7)

for integers   a   and   b  

 
 

m   and   n   are integers that are the sum of three nonzero squares :

m \; = \; a^2_1 \; + \; a^2_2 \; + \; a^2_3
n \; = \; b^2_1 \; + \; b^2_2 \; + \; b^2_3

do not imply that    m \cdot n \; = \; c^2_1 \; + \; c^2_2 \; + \; c^2_3

 

Here are the first few counterexamples:

 
  3 \; = \; 1^2 + 1^2 + 1^2
  6 \; = \; 1^2 + 1^2 + 2^2

  9 \; = \; 1^2 + 2^2 + 2^2

11 \; = \; 1^2 + 1^2 + 3^2
12 \; = \; 2^2 + 2^2 + 2^2
17 \; = \; 2^2 + 2^2 + 3^2
18 \; = \; 1^2 + 1^2 + 4^2
19 \; = \; 1^2 + 3^2 + 3^2

27 \; = \; 3 \times 9 \; = \; 1^2 + 1^2 + 5^2 \; = \; 3^2 + 3^2 + 3^2
33 \; = \; 3 \times 11 \; = \; 2^2 + 2^2 + 5^2 \; = \; 1^2 + 4^2 + 4^2

66 \; = \; 6 \times 11
66 \; = \; 8^2 + 1^2 + 1^2
66 \; = \; 7^2 + 4^2 + 1^2
66 \; = \; 5^2 + 5^2 + 4^2

187 \; = \; 11 \times 17
187 \; = \; 13^2 + 3^2 + 3^2
187 \; = \; 9^2 + 9^2 + 5^2

209 \; = \; 11 \times 19
209 \; = \; 14^2 + 3^2 + 2^2
209 \; = \; 13^2 + 6^2 + 2^2
209 \; = \; 12^2 + 8^2 + 1^2
209 \; = \; 12^2 + 7^2 + 4^2
209 \; = \; 10^2 + 10^2 + 3^2
209 \; = \; 9^2 + 8^2 + 8^2

 
 
 
 

 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 

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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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