Problem:
Let w=23β+iβ and z=2β1+i3ββ, where i=β1β. Find the number of ordered pairs (r,s) of positive integers not exceeding 100 that satisfy the equation iβ wr=zs.
Solution:
Answer (834):
Note that w=cos6Οβ+iβ sin6Οβ is a primitive twelfth root of unity, so w3=i and wr=wr+12m for all integers m and r. Furthermore, z=cos32Οβ+iβ sin32Οβ is a primitive cube root of unity, so w4=z and
iβ wr=wr+12m+3=zs=w4s.
Hence the given equation requires that there exists an integer m such that r+12m+3=4s, so r+3β‘4s(mod12). In particular, r+3β‘0,4, or 8(mod12).
If r+3β‘0(mod12), then r=9+12m and s=3n, where 0β€mβ€7 and 1β€nβ€33, accounting for 8β 33=264 ordered pairs.
If r+3β‘4(mod12), then r=1+12m and s=1+3n, where 0β€mβ€8 and 0β€nβ€33, accounting for 9β 34=306 ordered pairs.
If r+3β‘8(mod12), then r=5+12m and s=2+3n, where 0β€mβ€7 and 0β€nβ€32, accounting for 8β 33=264 ordered pairs.
The requested number of ordered pairs ( r,s ) is therefore 264+306+264=834.