@reasonsat yes it’s 126. It was just a question I made up, but I feel like this could be a suitable level 5 question (or maybe a tad harder since most “plugging in” strategies probably don’t work).
A simple algebraic solution is to use difference of squares as mentioned before:
(j-k)*(j+k) = 252
We want to maximize j+k so we minimize j-k (note that they’re both positive ints since j, k are positive ints).
Trying 1*252 = 252, this doesn’t give us a solution because 1 and 252 are of different parity. A simple way to see that there is no solution is to solve j-k = 1, j+k = 252 --> j = 253/2.
Trying 2*126 = 252, since 2 and 126 have the same parity, we obtain a solution where j+k = 126, so 126 is optimal (j=64, k=62).