If a and b are positive integers such that N=(a+ib)3−107i is a positive integer, then the value of N is
Since N is positive integer, its imaginary part equals zero.
Since 107 is prime, hence it has exactly two divisors which are 1 and 107
Hence, b=1 and 3a2−b2=107
If 3a2−b2=1 and b=107, then a is not an integer.
⇒a=6 (Neglecting a=−6)