Problem 11909
(American Mathematical Monthly, Vol.123, May 2016) Proposed by H. Ohtsuka (Japan).
Prove that for every positive integer m there exists a polynomial Pmin two variables, with integer coefficients, such that for all integers n and r with0 ≤ r ≤ n,
r
X
k=−r
n r+ k
n r − k
k2m= Pm(n, r) Qm
j=1(2n − 2j + 1)
2n 2r
.
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
We have that for −r ≤ k ≤ r,
2n n
n r+ k
n r − k
=2n 2r
2r r+ k
2(n − r) (n − r) + k
. Hence
2n n
Pm(n, r) Qm
j=1(2n − 2j + 1)=2n n
2n 2r
−1 r
X
k=−r
n r+ k
n r − k
k2m
=
r
X
k=−r
2r r+ k
2(n − r) (n − r) + k
k2m.
Moreover
(r2−k2)
2r r+ k
= 2r(2r − 1) 2(r − 1) r −1 + k
implies that
r
X
k=−r
(r2−k2)
2r r+ k
2(n − r) (n − r) + k
k2m= 2r(2r−1)
r−1
X
k=−(r−1)
2(r − 1) r −1 + k
2((n − 1) − (r − 1)) (n − 1) − (r − 1) + k
k2m
and the following recurrence holds for m ≥ 0,
2n n
r2Pm(n, r) Qm
j=1(2n − 2j + 1)−2n n
Pm+1(n, r) Qm+1
j=1 (2n − 2j + 1)=2(n − 1) n −1
2r(2r − 1)Pm(n − 1, r − 1) Qm
j=1(2(n − 1) − 2j + 1) that is
Pm+1(n, r) = r2(2n − 2m − 1)Pm(n, r) − r(2r − 1)nPm(n − 1, r − 1).
where we note that r2(2n − 2m − 1) and r(2r − 1)n are polynomials with integer coefficients. Since
P0(n, r) =2n 2r
−1 r
X
k=−r
n r+ k
n r − k
=2n 2r
−12n 2r
= 1,
it follows, by the above recurrence,
P1(n, r) = r2(2n − 1) − r(2r − 1)n = r(n − r), P2(n, r) = r(n − r)(3nr − n − 3r2)
and, inductively, we have that Pmis always a polynomial with integer coefficients.
