The Friday puzzle in Le Monde this week is about “friendly perfect squares”, namely perfect squares x²>10 and y²>10 with the same number of digits and such that, when drifting all digits of x² by the same value a (modulo 10), one recovers y². For instance, 121 is “friend” with 676. Here is my R code:

xtrct=function(x){

  x=as.integer(x)
  digs=NULL
  for (i in 0:trunc(log(x,10))){
    digs[i+1]=trunc((x-sum(digs[1:i]*10^(trunc(log(x,10)):(trunc(log(x,10))-
            i+1))))/10^(trunc(log(x,10))-i))}

  return(digs)
  }

pdfct=(4:999)^2

for (t in 2:6){
  pfctsq=pdfct[(pdfct>=10^t)&(pdfct0)
       print(c(pfctsq[i],pfctsq[
       ((i+1):dim(rstrct)[2])[(dive==1)]]))
    }
  }

which returns

[1] 121 676
[1] 1156 4489
[1] 2025 3136
[1] 13225 24336
[1] 111556 444889

namely the pairs (121,676), (1156,4489), (2025,3136), (13225,24336), and (111556,444889) as the solutions. The strange line of R code

    if (is.matrix(dive))
       dive=lapply(seq_len(ncol(dive)), function(i) dive[,i])

is due to the fact that, when the above result is a matrix, turning it into a list means each entry of the matrix is an entry of the list. After trying to solve the problem on my own for a long while (!), I found the above trick on stackoverflow. (As usual, the puzzle is used as an exercise in [basic] R programming. There always exists a neat mathematical solution!)