Two Math Professors

Even though this isn't specifically mentioned, and I perfectly understand that this probably wasn't intended to be a hint, but because the arrangement fits, I think a logical solution is possible with this extra clue. That exact hint is that the split was done in the middle and no further rearrangement necessary.

(10a + b)² = 100x + 10y + z

(10c + d)² = {one of the other 5 permutations of 100x + 10y + z}

[10(a + c) + (b + d)]² = 1000a + 100b + 10m + n , with m ≥ c

n, z, {x @ y} = { 1, 4, 5, 6, 9 }
0 is excluded because of it's double appearance, m would also be 0 and if y = z = 0, then you cannot have another 3 digit square with a different order.

The then-ages must be older than 10 and younger than 32 to be within the number of digits mentioned.

Since the oldest possible the younger professor could be THEN by number of digit limitations is 29 (31 is the older guy's oldest possible younger age and we already discarded 0 for z, so 30 is rejected due to the impossibility of two 3-digit numbers), then our 4-digit numbers must be less than 3000. By this, their then-ages must sum at < √3000 = 55, with the younger prof oldest then-age = (55 - 1) / 2 - 1 = 27 - 1 = 26. And by repeatedly using this iteration, we get the his (younger prof) oldest possible younger age to be 24 and their maximum then-sum = 49.

Conversely, if we set one of the then-ages to be 11, their then-sum must be larger than √1100 = 33+, so the other then-age is at least 23.

For their now-sum to be a 3-digit numbers, with 49 as max then-sum, we would need the first meeting to have happened AT LEAST (99 - 49) / 2 + 1 = 25 + 1 = 26 years ago. With that, one of the now-ages must at least be older than 25 + 26 = 51 ≤ 10m + n.

Since the maximal then-sum is around 50, we can actually use 50 special squares. (50 - k)²
= 2500 - 2(50)(k) + k²
= 2500 - 100k + k²
= 100(25 - k) + k²

As we can see, the first term is a multiple of 100, so the tens and ones places are only affected by k², so considering that k² mod 100 ≥ 51, then k = { 8, 9, 13, 14, ... } . Trying these in increasing order gets us a solution of 41² = 1681 = (16 + 25)² from k = 9 with 16² = 256 as a permutation of 25² = 625. (The previous one of 42² = 1764 = (17 + 25)², but 17² = 289 with no square permutation).

Now-sum = 16 - 25 + 2(81) = 153

Answer = 1³ + 5³ + 3³ = 153

For the first condition: there are 3 sets of 2-digit numbers (whose squares are 3-digit numbers) where their squares have the same digits but in different orders: $(12,21),(13,14,31),(16,25)$ .

Doing trial and error on the 5 possible pairs $(12,21),(13,14),(14,31),(13,31),(16,25)$ fulfilling the first condition to test the second condition, we see that only the pair $(16,25)$ works, with the square of their sum being $1681$ .

We thus have $81$ being Prof. Uno's current age, $25$ his age then and $16$ being Prof. Dos's age then. Prof' Dos's current age is then $16+(81-25)=72$ .

The sum of their ages is then $81+72=153$ , the sum of the cubes of their digits is then $1^3+5^3+3^3=\boxed{153}$ .

Interesting note: As noticed, $153$ is the sum of the cubes of its digits. It is in fact one of the four possible 3-digit numbers that have this property.

If we look at the squares of two digit numbers, we can see that the square of 16 is 256, the square of 25 is 625 ans the square of 16+25=41 is 1681 ( wich contains one of the two ages, as the condition states, "16" and "81"). Since the difference beetween the two is 25-16=9, ans that Professor Dos said that the square of the sum of their ages was giving proffesor one's age at that time, wich happens to be 16, then professor one is 81-9=72 year old, and professor two 81 year old. Then, if we sum these two ages, we get 153, as a=1; b=5 and c=3. then, we have to cube each of these, we have a^3=1 ; b^3=125 and c^3=27. 125+27+1 = 153.

Their 2 ages are 16 and 25 last time 41^2=1681, so it was 56 years ago Total age = 153 1^3+5^3+3^3=153