Problem:
For integers a and b consider the complex number
ab+100ab+2016βββ(ab+100β£a+bβ£ββ)i
Find the number of ordered pairs of integers (a,b) such that this complex number is a real number.
Solution:
The complex number is real if either ab+100ab+2016ββ is real and ab+100β£a+bβ£ββ=0 or ab+100ab+2016ββ is imaginary and equals ab+100β£a+bβ£ββi. In the former case, a=βb, and ab+100ab+2016ββ must be real. This will occur if and only if ab+2016β₯0 and abξ =β100. So a2β€2016, and there are 2β β2016ββ+1β2=87 possible values for a (including 0, and excluding Β±10). In the latter case, ab+2016β=ββ£a+bβ£β. If a+b>0, then ab+2016=β(a+b), which can be rewritten as (a+1)(b+1)=β2015=β5β 13β 31. There are 8 integer solutions to this equation with a+b>0. Similarly if a+b<0, then ab+2016=a+b, which can be rewritten as (aβ1)(bβ1)=β2015. Again, there are 8 integer solutions to this equation with a+b<0. This yields a total of 87+8+8=103β possible ordered pairs.