the 332-232 problem:
really, really, trivial. just a repeated application of (a2-b2) = (a-b)(a+b)
from that procedure, however, you get only 3 prime numbers under 100: 5, 13 and 97. now there's the less trivial part of finding a 4th prime. unless you count 1 as a prime, which you really shouldn't do.
if i were in a high school math competition, due to time pressure, i would take advantage of the question saying 4 prime numbers under 100. the easiest thing would be to calculate the next factor, which is 38+28. this is reasonably fast to calculate even without a calculator, and you get 6561+256=6817. this number happens to be easily factorizable, by inspection--clearly, 6817=17*401. (both 68 and 17 are multiples of 17). and i would move on to the next question, having gotten what the question asks for: 5, 13, 17 and 97.
there might be more sub-100 factors left in the 216+316 part, but in a competition you would just move on. i verified that this number has factors 3041 and 14177, i.e. no sub-100 primes. there might be an elegant way of showing that 216+316 does not have a sub-100 prime.
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