Algebra 4092 (n−^1 )(n−^2 )/^2∏
1 ≤i<j≤n(cosjα−cosiα) = 0.196.Because the five numbers lie in the interval[− 2 , 2 ], we can find corresponding
anglest 1 ,t 2 ,t 3 ,t 4 ,t 5 ∈[ 0 ,π]such thatx =2 cost 1 ,y =2 cost 2 ,z=2 cost 3 ,v =
2 cost 4 , andw=2 cost 5. We would like to translate the third and fifth powers into
trigonometric functions of multiples of the angles. For that we use the polynomials
Tn(a). For example,T 5 (a)=a^5 − 5 a^3 + 5 a. This translates into the trigonometric
identity 2 cos 5θ=(2 cosθ)^5 − 5 (2 cosθ)^3 + 5 (2 cosθ).
Add to the third equation of the system the first multiplied by 5 and the second
multiplied by−5, then use the above-mentioned trigonometric identity to obtain
2 cos 5t 1 +2 cos 5t 2 +2 cos 5t 3 +2 cos 5t 4 +2 cos 5t 5 =− 10.This can happen only if cos 5t 1 =cos 5t 2 =cos 5t 3 =cos 5t 5 =cos 5t 5 =−1. Hence
t 1 ,t 2 ,t 3 ,t 4 ,t 5 ∈{
π
5,
3 π
5,
5 π
5}
.
Using the fact that the roots ofx^5 =1, respectively,x^10 =1, add up to zero, we
deduce that
∑^4k= 0cos
2 kπ
5=0 and∑^9
k= 0cos
kπ
5= 0.
It follows that
cosπ
5+cos3 π
5+cos5 π
5+cos7 π
5+cos9 π
5= 0.
Since cosπ 5 =cos^95 πand cos^35 π=cos^75 π, we find that cosπ 5 +cos^35 π =^12. Also, it
is not hard to see that the equationT 5 (a)=−2 has no rational solutions, which implies
that cosπ 5 is irrational.
The first equation of the system yields∑ 5
i= 1 ti =0, and the above considerations
show that this can happen only when two of thetiare equal toπ 5 , two are equal to^35 π, and
one is equal toπ. Let us show that in this situation the second equation is also satisfied.
Using∑ T 3 (a)=a^3 − 3 a, we see that the first two equations are jointly equivalent to
5
k= 1 costi=0 and
∑ 5
k= 1 cos 3ti=0. Thus we are left to check that this last equality
is satisfied. We have
2 cos3 π
5
+2 cos9 π
5
+cos 3π=2 cos3 π
5
+2 cosπ
5
+cosπ= 0 ,as desired. We conclude that up to permutations, the solution to the system is