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))
[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 51072

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