Le Monde puzzle [#954]

A square Le Monde mathematical puzzle:

Given a triplet (a,b,c) of integers, with a<b<c, it satisfies the S property when a+b, a+c, b+c, a+b+c are perfect squares such that a+c, b+c, and a+b+c are consecutive squares. For a given a, is it always possible to find a pair (b,c) such (a,b,c) satisfies S? 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 Rmpfr, 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+γ)(δ-γ), b=(δ+β)(δ-β), 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=2γ+1, b=4γ, 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=2*ga+1
 b=4*ga
 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] 41 80 41 320
[1] 57 112 672
[1] 97 192 2112
[1] 121 240 3360
[1] 177 352 7392
[1] 209 416 10400
[1] 281 560 19040
[1] 321 640 24960
[1] 409 816 40800
[1] 457 912 51072

5 Responses to “Le Monde puzzle [#954]”

  1. It strikes me that this is a situation where my (shameless plug) ktsolver package might help. While ktsolver is designed to back-solve a general set of equations, I would expect in the case of all-integer equations like these it should converge to the exact solution. If I remember :-) , I’ll try to try putting your equations into ktsolver myself soon.

  2. misspelled, package “Rpmfr” name, should be “Rmpfr”

  3. Your results do not satisfy the requirement a<b<c (due to a minor error)

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