Let m and n be two given integers. Alice thinks of a pair of real numbers x, y and then she tells Bob the values of xm+ym and xn+yn, in this order. Bob's goal is to determine the value of xy using that information. Find all values of m and n for which it is possible for Bob to fulfill his wish, whatever numbers that Alice had chosen.
I think that (2k+1, 4k+2) and (4k+2,2k+1) work for all integer values of k >= 0.
Let m = 2k+1, and n = 4k+2.
Let a = x^m+ y^m (Alice's first answer) and b = x^n + y^n (alice's 2nd answer)
then a^2 - b = 2(xy)^m
So xy can be calculated as the (m)th root of (a^2 - b)/2.
m must be an odd in order that the mth root is unambiguous.
Similarly, if m = 2n, then xy = the (n)th root of (b^2 - a)/2
If the problem had specified that x and y cannot be zero, then (2k+1, 4k+2) and (4k+2,2k+1) also work for k < 0. But the problem requires that it must work for all x and y that Alice can choose, and negative values of m and n do not work if x or y are 0, so the negative values almost (but not quite) satisfy the problem conditions.
Maybe I am splitting hairs, but if the problem had said that it must work for whatever numbers Alice tells Bob, than I think negative m and n would be work, because if she can tell him two numbers even when m and n are negative, then this means that x and y are non-zero.
Edited on December 18, 2020, 8:43 pm
Values that work (answer)
