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## Thursday, January 31, 2008

### Three Squares in 2008

Here's another request from a reader:

"How can I write 2008 as the sum of three squares?"

There are four ways you can write 2008 as the sum of three squares, can you find all of them?

CT said...

To find all possible solutions, we can determine the largest squares from the ceilings after each term.

For example:
45^2 > 2008, we use 44^2 as the first term. 2008 - 44^2 = 72

9^2 > 72, we can use 8^2 as 2nd term, but the difference is not a square

look down until 72 - 6^2 = 36 = 6^2. Hence we have a way: 2008 = 44^2 + 6^2 + 6^2

Formula for searching should be:
2008 - x^2 - y^2 = z^2, where x^2 < 2008 and y^2 < (2008-x^2), x, y, z are positive natural numbers.

We can write a program to calculate this. There are 2 loops:
1. for(i=1;i^2<2008;i++)
2. for(j=1;j^2<(2008-i^2);j++)
And check whether (2008-i^2-j^2) is a square or not.

If there is no function to check that, we can use 3 loops to check that (2008-i^2-j^2-k^2)==0

PS: General program should have a user input of "2008" variable.

Math Genius 900 said...

I also used a computer program to go thru all possible answers and found all 4 solutions.
6,6,44
6,26,36
10,12,42
18,28,30
Also, as a side note, there are actually infinity different answers if you include imaginary numbers.
for example,
28,35,i

because
28^2 + 35^2 + i^2
784 + 1225 + -1
2009 - 1 = 2008

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