Sum of four Squares

In this video you will learn that if and two integer can be written as sum of two squares their product can also be written as the sum of squares of four integers using Euler's identity. The proof of Euler's identity is simple algebraic manipulation that even a 8th grader can do!

[jeffw]

I would simply go straight to 7839 and find the largest square that is less than that. 88^2 = 7744 and (7839 - 7744) = 95. However, 95 is a member of the (8n+7) series, which can only be represented by four non-zero squares. Since I'm already using 88^2 as one of my squares, I only have three left and it will not work. So I have to drop down and try 87^2 (7569) as my first square. The difference is 270 (which is not part of either the 8n+7 series or the 32n+28 series, meaning that 270 can certainly be represented with three squares or less), , and the largest square I can use as my second term is 16^2 = 256. (270 - 256) is a difference of 14, and I can't represent that with only the two squares I have left to work with. So I drop down the second square to 15^2 = 225. (270 - 225) is a difference of 45, and I see immediately that this can be represented by (6^2 + 3^2) = 45. My first solution is therefore (87^2 + 15^2 + 6^2 + 3^2) = 7839. (87^2 + 14^2 + 7^2 + 5^2) is another solution, and (87^2 + 13^2 + 10^2 + 1^2) is another. There will be MANY other solutions, though they become tougher to find as I reduce the value of the first square to 86^2, 85^2, etc. The larger the difference after selecting a first square, the harder it is to find a solution using the three remaining squares, as not every number can be represented by three squares, namely the (8n+7) series and the (32n+28) series. It helps to have all the squares up to 99^2 = 9801 memorized, as I do.