Problem:
Find c if a,b and c are positive integers which satisfy c=(a+bi)3−107i, where i2=−1.
Solution:
First note that one may write c in the form
c=a(a2−3b2)+i[b(3a2−b2)−107](1)
From this, in view of the fact that c is real, one may conclude that
b(3a2−b2)=107(2)
Since a and b are positive integers, and since 107 is prime, two possible cases arise from (2):
Either b=107or b=1 and and 3a2−b2=1,3a2−b2=107.
In the first case, 3a2=1072+1 would follow. But this is impossible, since 1072+1 is not a multiple of 3.
In the second case, one finds that a=6 and hence, in view of (1), c=a(a2−3b2)=6(62−3⋅12)=198.
The problems on this page are the property of the MAA's American Mathematics Competitions