## Solution: Day 3, problem 3

Let

*h*(

*x*) = [ 1 +

*x*

^{m}+ (1 -

*x*)

^{m}]/2 be a polynomial in

**F**

_{p}[

*x*]. Each number

*a*in

**F**

_{p}such that both

*a*and 1 -

*a*are quadratic non - residues modulo

*p*is a root of both

*h*

_{p}(

*x*) and

*h*

_{p}'(

*x*), and there are exactly

*N*= [

*m*/2] such numbers. (Counting them is easily seen to be equivalent tocounting elements a in

**F**

_{p}- {0, 1} for which both a and a - 1 are quadratic residues, and such

*a*are parametrized by

*a*= (

*t*+

*t*

^{-1})

^{2}/4 for

*t*in

**F**

_{p}with

*t*

^{2}≠ -1 mod

*p*.) But

*h*

_{p}has degree 2

*N*and constant term 1, so we must have that

*h*

_{p}(

*x*) is the product over

*a*of (1 -

*x*/

*a*)

^{2}.

Another proof is to note that *h*(*x*) = *F*(*x*)^{2} + *O*(*x*^{m}) in **F**_{p}[[*x*]], where *F*(*x*) = √(1/2 + √(1 - *x*)/2) = 1 - 1/8*x* - 5/128*x*^{2} - ... in **Z**[^{1}/_{2}][[*x*]].

From the differential equation *x*(1 - *x*)*F*'' + (1/2 - *x*)*F*' + ^{1}/_{16} *F* = 0 we find that the coefficient of *x*^{n} in *F* equals -2^{1-4n} *binom*(4*n* - 3, 2*n* - 1)/*n*, which is 0 modulo *p* for N < n < m. Hence *F*(*x*) = *f*(*x*) + *O*(*x*^{m}) for some polynomial *f*(*x*) in **F**_{p}[*x*] of degree ≤ *N*. From *h*(*x*) = *f*(*x*)^{2} mod *x*^{m} and *deg*(*h*) = 2*N* ≤ *m* it immediately follows that *h* = *f*^{2} if *m* is odd, while for *m* even we must use the above formula for the coefficients of *f*(*x*) to verify that the leading coefficients of *h*(*x*) and *f*(*x*)^{2} are both 1.

A third proof, provided during the Colloquium, is even simpler: the product of *h*(*x*) and *g*(*x*) = [ 1 + *x*^{m} - (1 - *x*)^{m}]/2 in **F**_{p}[*x*] is *x*(1 - *x*^{m})^{2}/4, which is *x* times a square. Since the polynomials *f*(*x*) and *g*(*x*)/*x* are coprime, they must both be squares (up to a common scalar multiple, but in fact without it since *f*(0) = 1).