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The Friday puzzle in Le Monde this week is about “friendly perfect squares”, namely perfect squares x^{2}>10 and y^{2}>10 with the same number of digits and such that, when drifting all digits of x^{2} by the same value a (modulo 10), one recovers y^{2}. 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)]]))
}
}

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!)