122 II Divisibility
Putα= 1 +ai+bj.ThenpdividesN(α)=αα ̄= 1 +a^2 +b^2 inZand hence
also inH.However,pdoes not divide eitherαorα ̄inH,sinceαp−^1 andα ̄p−^1 are
not inH.
Letγ=(p,α)r.Thenp=βγfor someβ∈H.Ifβwere a unit,pwould be a
right divisor ofγand hence also ofα, which is a contradiction. ThereforeN(β) >1.
Evidentlyγα ̄is a common right divisor ofpα ̄andαα ̄,andtheB ́ezout representation
forγimplies thatγα ̄ =(pα,α ̄ α) ̄r.Sincepα ̄ = ̄αpandpdividesαα ̄, it follows
thatpis a right divisor ofγα ̄.Sincepdoes not divideα,γ ̄ is not a unit and hence
N(γ) >1. Since
N(β)N(γ)=N(p)=p^2 ,we must haveN(β)=N(γ)=p.
Thus ifγ=c 0 +c 1 i+c 2 j+c 3 k,thenc 02 +c^21 +c 22 +c^23 =p.Ifc 0 ,...,c 3 are
all integers, we are finished. Otherwisec 0 ,...,c 3 are all halves of odd integers. Hence
we can writecv= 2 dv+ev,wheredv∈Zandev=± 1 /2. If we put
δ=d 0 +d 1 i+d 2 j+d 3 k,ε=e 0 +e 1 i+e 2 j+e 3 k,thenγ = 2 δ+εandN(ε)=1. Henceθ :=γε ̄= 2 δε ̄+1 has all its coordinates
integers andN(θ)=N(γ)=p.
In hisMeditationes Algebraicae, which also contains the first statement in print of
Wilson’s theorem, Waring (1770) stated that every positive integer is a sum of at most
4 positive integral squares, of at most 9 positive integral cubes and of at most 19 posi-
tive integral fourth powers. The statement concerning squares was proved by Lagrange
in the same year, as we have seen. The statement concerning cubes was first proved by
Wieferich (1909), with a gap filled by Kempner (1912), and the statement concerning
fourth powers was first proved by Balasubramanian, Deshouillers and Dress (1986).
In a later edition of his book, Waring (1782) raised the same question for higher
powers.Waring’s problemwas first solved by Hilbert (1909), who showed that, for
eachk∈N, there existsγk∈Nsuch that every positive integer is a sum of at most
γkk-th powers. The least possible value ofγkis traditionally denoted byg(k).For
example,g( 2 )=4, since 7= 22 + 3 · 12 is not a sum of less than 4 squares.
A lower bound forg(k)was already derived by Euler (c. 1772). Letm= ( 3 / 2 )k
denote the greatest integer≤( 3 / 2 )kand take
n= 2 km− 1.Since 1≤n< 3 k, the onlyk-th powers of whichncan be the sum are 0k, 1 kand 2k.
Since the number of powers 2kmust be less thanm, and sincen=(m− 1 ) 2 k+
( 2 k− 1 ) 1 k, the least number ofk-th powers with sumnism+ 2 k−2. Hence
g(k)≥w(k),where
w(k)= ( 3 / 2 )k + 2 k− 2.In particular,
w( 2 )= 4 ,w( 3 )= 9 ,w( 4 )= 19 ,w( 5 )= 37 ,w( 6 )= 73.