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# (a) Prove that if n + 1 is prime, then fn is irriducible over Q

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(1)

Problem 11798

(American Mathematical Monthly, Vol.121, October 2014) Proposed by F. Holland (Ireland).

For positive integers n, let fn be the polynomial given by

fn(x) =

n

X

k=0

n k

 x⌊k/2⌋.

(a) Prove that if n + 1 is prime, then fn is irriducible over Q.

(b) Prove for all n,

fn(1 + x) =

⌊n/2⌋

X

k=0

n − k k



2n−2kxk.

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

(a) We have that

fn(x) =

⌊n/2⌋

X

k=0

 n 2k

 +

 n

2k + 1



xk =

⌊n/2⌋

X

k=0

 n + 1 2k + 1

 xk.

If p = 2 then fp−1(x) = 2 which is irriducible. If p > 2 is a prime then

fp−1(x) =

(p−1)/2

X

k=0

 p

2k + 1



xk = x(p−1)/2+

(p−1)/2

X

k=1

 p

2k + 1

 xk+ p.

Since the leading coefficient is 1, p divides the coefficients 2k+1p 

for k = 1, . . . , (p − 1)/2 and fp−1(0) = p, it follows by Eisenstein’s criterion that fp−1 is irriducible over Q.

(b) The polynomial

Pn(x) :=

⌊n/2⌋

X

k=0

n − k k



2n−2kxk satisfies the linear recurrence

P0(x) = 1, P1(x) = 2, Pn(x) = 2Pn−1(x) + xPn−2(x) for n ≥ 2.

Hence, for x > −1

Pn(x) = un+1− vn+1 u − v where u = 1 +√

1 + x, v = 1 −√

1 + x are the solutions of the equation z2= 2z + x. Moreover, un+1− vn+1

u − v = 1

2√ 1 + x

n+1

X

k=0

n + 1 k



(1 + x)k/2

n+1

X

k=0

(−1)kn + 1 k



(1 + x)k/2

!

=

⌊n/2⌋

X

k=0

 n + 1 2k + 1



(1 + x)k = fn(1 + x)

Hence the required identity holds for x > −1. Since we are comparing two polynomials, it follows that the identity holds for all x ∈ C.

Riferimenti

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