Sum of Squares

The eighteenth-century mathematician Joseph Louis Lagrange proved that every positive integer is the sum of, at most, four squares. Find at least one way to write 2008 as the sum of three squares.
Source: NCTM Mathematics Teacher 2008

SOLUTION
\sqrt{2008}=44.81
2008-44^2=72
\sqrt{72}=8.49
2008-44^2-8^2=8
2008=44^2+8^2+2^2+2^2 a sum of four squares
We want a sum of three squares, so we try 7^2,6^2, etc.
2008-44^2-7^2=23 not a perfect square
2008-44^2-6^2=36
2008=44^2+6^2+6^2
Continuing in this slow and tedious fashion we will find 3 more solutions
2008=42^2+12^2+10^2
2008=36^2+26^2+6^2
2008=30^2+28^2+18^2

Answer:
44^2+6^2+6^2
42^2+12^2+10^2
36^2+26^2+6^2
30^2+28^2+18^2

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About mvtrinh

Retired high school math teacher.
This entry was posted in Problem solving and tagged , , , , , . Bookmark the permalink.

One Response to Sum of Squares

  1. Also from Lagrange:
    A number n is the sum of two squares if and only if the primefactors >2 of n that are == 3 mod(4) have even multiplicity. So after you have substracted the first square (44² etc.) and get 72=2*2*2*3*3 you know 72 is the sum of two squares. (3 appear and even number of times)
    If you had gotten 62=3*31 instead etc, you know it was not possible, since 31 is =3 (mod4) and have odd multiplicity.

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