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?
Thursday, January 31, 2008
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5 comments:
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.
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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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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