**A** digit puzzle as Le weekly Monde current mathematical puzzle that sounds close to some earlier versions:

Perfect squares are pairs (a²,b²) with the same number of digits such that a²b² is itself a square. What is the pair providing a²b² less than 10⁶? Is there a solution with both integers enjoying ten digits?

The run of a brute force R code like

cek<-function(a,b){ u<-trunc if ((n<-u(log(a^2,ba=10)))==u(log(b^2,ba=10))& (u(sqrt(a^2*10^(n+1)+b^2))^2==(a^2*10^(n+1)+b^2))) print(c(a,b))}

provides solutions to the first question.

[1] 2 3 [1] 4 9 [1] 12 20 [1] 15 25 [1] 18 30 [1] 49 99 [1] 126 155 [1] 154 300 [1] 159 281 [1] 177 277 [1] 228 100 [1] 252 310 [1] 285 125

with the (demonstrable) conclusion that the only pairs with an *even* number of digits are of the form (49…9²,9…9²), as for instance (49999²,99999²) with ten digits each.