Problem:
For how many positive integers n less than or equal to 1000 is
(sint+icost)n=sinnt+icosnt
true for all real t?
Solution:
Note that
(sint+icost)nβ=[cos(2Οββt)+isin(2Οββt)]n=cosn(2Οββt)+isinn(2Οββt)=cos(2nΟββnt)+isin(2nΟββnt)β
and that sinnt+icosnt=cos(2Οββnt)+isin(2Οββnt). Thus the given condition is equivalent to
cos(2nΟββnt)=cos(2Οββnt) and sin(2nΟββnt)=sin(2Οββnt)
In general, cosΞ±=cosΞ² and sinΞ±=sinΞ² if and only if Ξ±βΞ²=2Οk. Thus
2nΟββntβ2Οβ+nt=2Οk
which yields n=4k+1. Because 1β€nβ€1000, conclude that 0β€kβ€249, so there are 250β values of n that satisfy the given conditions.
OR
Observe that
β(sint+icost)n=[i(costβisint)]n=in(cosntβisinnt), and that sinnt+icosnt=i(cosntβisinnt)β
Thus the given equation is equivalent to in(cosntβisinnt)=i(cosntβisinnt). This is true for all real t when in=i. Thus n must be 1 more than a multiple of 4, so there are 250β values of n that satisfy the given conditions.
The problems on this page are the property of the MAA's American Mathematics Competitions