Problem:
The number of distinct pairs of integers (x,y) such that 0<x<y and 1984​=x​+y​ is
Answer Choices:
A. 0
B. 1
C. 3
D. 4
E. 7
Solution:
Note that the prime factorization of 1984 is 26⋅31, that x<1984 and that y=(1984​−x​)2=1984+x−21984x​. It follows that y is an integer if and only if 1984x is a perfect square, that is, if and only if x is of the form 31t2. Since x is less than 1984, we have 1≤t≤7, yielding the pairs (31,1519),(124,1116) and (279,775) for (x,y), corresponding to t=1,2,3. Since y≤x for t>3, these are the only solutions.