Want to share your content on R-bloggers? click here if you have a blog, or here if you don't. A square Le Monde mathematical puzzle:

Given a triplet (a,b,c) of integers, with a? Can you find the triplet (a,b,c) that produces the sum a+b+c closest to 1000?

This is a rather interesting challenge and a brute force resolution does not produce interesting results. For instance, using the function is.whole from the package Rpmfr, the R functions

```ess <- function(a,b,k){
#assumes a<b<k
ess=is.whole(sqrt(a+b))&
is.whole(sqrt(b+k))&
is.whole(sqrt(a+k))&
is.whole(sqrt(a+b+k))
mezo=is.whole(sqrt(c((a+k+1):(b+k-1),(b+k+1):(a+b+k-1))))
return(ess&(sum(mezo==0)))
}
```

and

```quest1<-function(a){
b=a+1
while (b<1000*a){
if (is.whole(sqrt(a+b))){
k=b+1
while (k<100*b){
if (is.whole(sqrt(a+k))&is.whole(b+k))
if (ess(a,b,k)) break()
k=k+1}}
b=b+1}
return(c(a,b,k))
}
```

do not return any solution when a=1,2,3,4,5

Looking at the property that a+b,a+c,b+c, and a+b+c are perfect squares α²,β²,γ², and δ². This implies that

a=(δ+β)(δ-β), b=(δ+γ)(δ-γ), and c=(δ+α)(δ-α)

with 1<α<β<γ<δ. If we assume β²,γ², and δ² consecutive squares, this means β=γ-1 and δ=γ+1, hence

a=4γ, b=2γ+1, and c=(γ+1+α)(γ+1-α)

which leads to only two terms to examine. Hence writing another R function

```abc=function(al,ga){
a=4*ga
b=2*ga+1
k=(ga+al+1)*(ga-al+1)
return(c(a,b,k))}
```

and running a check for the smallest values of α and γ leads to the few solutions available:

```> for (ga in 3:1e4)
for(al in 1:(ga-2))
if (ess(abc(al,ga))) print(abc(al,ga))
 80 41 320
 112 57 672
 192 97 2112
 240 121 3360
 352 177 7392
 416 209 10400
 560 281 19040
 640 321 24960
 816 409 40800
 912 457 51072
```

Filed under: Kids, R Tagged: Alice and Bob, is.whole, Le Monde, mathematical puzzle, R, recursive function, Rmpfr  