Question

# If a and b are positive integers such that N=(a+ib)3−107i is a positive integer, then the value of N is

Solution

## N=(a+ib)3−107i=a3+3a2bi+3ab2i2+b3i3−107i=(a3−3ab2)+i(3a2b−b3−107) Since N is positive integer, its imaginary part equals zero. b(3a2−b2)=107 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.  ∴3a2−b2=107 ⇒3a2=108⇒a2=36 ⇒a=6   (Neglecting a=−6) ∴N=216−3⋅6⋅1=198 Mathematics

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