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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?

9 comments:

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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Devadatta S. Rajadhyaksha said...

Even without imaginary numbers, you can have many more answers. Just include negative numbers as well, as the squares will all be positive.

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