**R – Xi'an's Og**, and kindly contributed to R-bloggers]. (You can report issue about the content on this page here)

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**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?

**T**his 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 aand

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)) [1] 80 41 320 [1] 112 57 672 [1] 192 97 2112 [1] 240 121 3360 [1] 352 177 7392 [1] 416 209 10400 [1] 560 281 19040 [1] 640 321 24960 [1] 816 409 40800 [1] 912 457 51072Filed under: Kids, R Tagged: Alice and Bob, is.whole, Le Monde, mathematical puzzle, R, recursive function, Rmpfr

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