**Tony's bubble universe » R**, and kindly contributed to R-bloggers)

Just had supper. My stomach is full of cabbage, carrot and noodle. I’d like to solve the ninth problem to stretch my mind. This one is about Pythagorean theorem.

A Pythagorean triplet is a set of three natural numbers,

a<b<c, for which,a^{2}+b^{2}=c^{2}. For example, 3^{2}+ 4^{2}= 9 + 16 = 25 = 5^{2}. There exists exactly one Pythagorean triplet for whicha+b+c= 1000. Find the productabc.

I admit this is a hard one for me, based on my almost totally forgotten knowledge in algebra. I was thinking about it all day long, even when I was eating or walking. I could try out each combination of a, b and c less than 1,000 to search for the triplet. However, there is only one right answer, thus it would be like finding a needle in a haystack by using brute force algorithm. So I had to search the internet for some help.

Wikipedia on Pythagorean triple provides me the very clue for this problem. Given a^{2} + b^{2} = c^{2}, we have a = m^{2} – n^{2}, b = 2*m*n, c = m^{2} + n^{2}, in which m > n and both are positive integers. Considering another condition a + b + c = 1000, we get the equation 2*m*(m+n) = 1000. Thus, I know 500 can be evenly divided by m and (m+n), and 250 < m^2 < 500. After the analysis, I write down the solution in R.

^{?}View Code RSPLUS

1 2 3 4 5 6 7 8 9 10 | m.high <- floor(sqrt(500)) m.low <- ceiling(sqrt(250)) m <- m.low:m.high m <- m[which(500 %% m == 0)] n <- 500 / m - m a <- m ^ 2 - n ^ 2 b <- 2 * m * n c <- m ^ 2 + n ^ 2 result <- a * b * c cat("The result is:", result, "\n") |

After years in experimental research, I realize I have forgotten so much fundanmental knowledge in other disciplines. Mathematics is great for mental exercises. I should schedule some time for it.

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