Problem 9 involves finding the Pythagorean triple \(a^2+b^2=c^2\) where a+b+c=1000.
Pythagorean triples have lots of useful relationships that we can exploit. The ones we shall use to solve this problem are:
\[a = k(m^2-n^2)\]
\[b = k(2mn)\]
\[ c = k(m^2+n^2) \]
In this case, we take \(k=1\), and generate the triples as follows:
# Problem 9 # Pythagorean Triple problem9 <- function() { sum <- 0 a <- 0; b <- 0; c <- 0; m <- 0 for (m in 1:1000) { for (n in m:1000) { a <- m^2-n^2 b <- 2*m*n c <- n^2 + m^2 if (all(a^2 + b^2 == c^2, sum(a,b,c)==1000)) { return(prod(a,b,c)) } } } }