For n ≥ 1, let f be the symmetric polynomial in variables x1

Problem 11736

(American Mathematical Monthly, Vol.120, November 2013) Proposed by Mircea Merca (Romania).

For n ≥ 1, let f be the symmetric polynomial in variables x1, . . . , xn, given by

f (x1, . . . , xn) =

n−1

X

k=0

(−1)^k+1ek(x1+ x²₁, x2+ x²₂, . . . , xn+ x²_n),

where e_k is the kth elementary polynomial in n variables. Also, let ω be a primitive nth root of unity. Prove that

f (1, ω, ω², . . . , ωⁿ⁻¹) = Ln− L0, where Lk is the k-th Lucas number.

Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.

It is known that

P (z) =

n

Y

k=1

(z − z_k) =

n

X

k=0

(−1)^ke_k(z₁, . . . , z_n)z^n−k.

Morever Ln= (τ+)ⁿ+ (τ₋)ⁿ with τ_±= (1 ±√ 5)/2.

Hence, by letting zk = ω^k+ ω^2k, we obtain

f (1, ω, ω², . . . , ωⁿ⁻¹) = P (0) − P (1) = ((−1)ⁿ− 1) − (1 − Ln+ (−1)ⁿ) = Ln− 2 = Ln− L0

because

P (0) =

n−1

Y

k=0

(−(ω^k+ ω^2k)) = ω^n(n−1)/2

n−1

Y

k=0

(−1 − ω^k) = ω^n(n−1)/2((−1)ⁿ− 1) = (−1)ⁿ− 1,

and

P (1) =

n−1

Y

k=0

(1 − (ω^k+ ω^2k)) = (−1)ⁿ

n−1

Y

k=0

(−τ+− ω^k)(−τ−− ω^k)

= (−1)ⁿ

n−1

Y

k=0

(−τ+− ω^k)

n−1

Y

k=0

(−τ₋− ω^k) = (−1)ⁿ((−τ+)ⁿ− 1)((−τ₋)ⁿ− 1)

= (−1)ⁿ((−1)ⁿ− (−1)ⁿ((τ+)ⁿ+ (τ₋)ⁿ) + 1) = 1 − Ln+ (−1)ⁿ.