When the sum of 3 squares is a square – Part 1

Posted on March 20, 2015 by benvitalis

 

Find all integers   N \; \leq \; 200   such that   N^2 = A^2 + B^2 + C^2

A,   B,   C   are positive integers

 

1^2 \; + \; 2^2 \; + \; 2^2 \; = \; 3^2
2^2 \; + \; 3^2 \; + \; 6^2 \; = \; 7^2
2^2 \; + \; 4^2 \; + \; 4^2 \; = \; 6^2

1^2 \; + \; 4^2 \; + \; 8^2 \; = \; 4^2 \; + \; 4^2 \; + \; 7^2 \; = \; 9^2

4^2 \; + \; 8^2 \; + \; 8^2 \; = \; 12^2

4^2 \; + \; 6^2 \; + \; 12^2 \; = \; 14^2
8^2 \; + \; 9^2 \; + \; 12^2 \; = \; 17^2

4^2 \; + \; 12^2 \; + \; 18^2 \; = \; 22^2
3^2 \; + \; 16^2 \; + \; 24^2 \; = \; 29^2

 
 
 

 
 
 
 
 
 
 
 
 
 
 

About benvitalis

math grad - Interest: Number theory
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1 Response to When the sum of 3 squares is a square – Part 1

  1. Paul says:
    March 20, 2015 at 10:02 am

    There are quite a few of them, 2255 in total, so here are a few, the complete solution set is at this link.

    https://www.dropbox.com/s/giukx22cxvqg90k/sum%20of%203%20squares.txt?dl=0

    1^2 + 2^2 + 2^2 = 3^2
    2^2 + 4^2 + 4^2 = 6^2
    2^2 + 3^2 + 6^2 = 7^2
    1^2 + 4^2 + 8^2 = 9^2
    3^2 + 6^2 + 6^2 = 9^2
    4^2 + 4^2 + 7^2 = 9^2
    2^2 + 6^2 + 9^2 = 11^2
    6^2 + 6^2 + 7^2 = 11^2
    4^2 + 8^2 + 8^2 = 12^2
    3^2 + 4^2 + 12^2 = 13^2
    ..
    ..
    51^2 + 90^2 + 170^2 = 199^2
    51^2 + 134^2 + 138^2 = 199^2
    62^2 + 126^2 + 141^2 = 199^2
    78^2 + 99^2 + 154^2 = 199^2
    82^2 + 114^2 + 141^2 = 199^2
    90^2 + 125^2 + 126^2 = 199^2
    93^2 + 114^2 + 134^2 = 199^2
    99^2 + 118^2 + 126^2 = 199^2
    72^2 + 96^2 + 160^2 = 200^2
    96^2 + 120^2 + 128^2 = 200^2

    Paul

    Reply

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