Waring's Problem with Polynomial Summands

Let f ( x ) be an integer‐valued polynomial with no fixed integer divisor ⩾ 2, that is, for no integer d ⩾ 2 does d ∣ f ( x ) for all integers x . One generalization of the famous Waring problem is to determine whether, for large enough s , the equation f ( x 1 )+ f ( x 2 )+…+ f ( x s )= n (1.1) is solvable in positive integers x 1 , …, x s for sufficiently large integers n . The existence of such s for every f was established by Kamke [ 5 ] in 1921. Subsequent authors (Pillai, Hua [ 2–4 ], Vinogradov, Načaev [ 7 ], and others) have studied the problem of bounding G ( f ), the least s for which (1.1) is solvable for all large n . Questions of local solubility of (1.1), that is, solubility of the congruence f ( x 1 )+ f ( x 2 )+…+ f ( x s )≡ n (mod q ) (1.2) play a more important and complicated role in this problem than in the classical Waring problem. Let Γ 0 ( f ) denote the least number s so that (1.2) is solvable for every pair n , q . It is well known that Γ 0 ( x k ) ⩽ 4 k for every k , but Hua [ 4 ] found that for every k , the polynomialf k ( x ) = ∑ j = 1 k( ‐ 1 )k ‐ j2 j ‐ 1(xj)satisfies Γ 0 ( f k ) ⩾ 2 k − 1 (take s = 2 k −2, q = 2 k and n = (−1) k in (1.2)). Clearly G ( f ) ⩾ Γ 0 ( f ), but one can say more by restricting the values of n under consideration, as has been done by several authors in the case f ( x ) = x 4 (for example [ 1 , 6 ]). The singular seriesS s , f( n ) =∑ q = 1 ∞∑a = 1( a ,   q ) = 1q( 1 q ∑ r = 1 q e (a f ( r ) / q ))s e (‐ a n / q ) ,where e ( z ) = e 2π iz , encapsulates the local solubility information. In particular, S s, f ( n ) ⩾ 0 for every n and S s, f ( n ) > 0 if and only if (1.2) is soluble for every q . Define G( f ) to be the least number s so that for every δ >0 and every n > n 0 (δ) with S s, f ( n ) ⩾ δ, (1.1) is soluble. The reason for taking S s, f ( n ) ⩾ δ instead of S s, f ( n ) > 0 is that we wish to exclude from consideration certain n lying in sparse sequences for which (1.1) is insoluble but S s, f ( n ) > 0. For example, taking f ( x ) = x 4 , s = 15 and n j = 79·16 j ( j = 0, 1, …), it can be shown that (1.1) is not soluble for n = n j , that S s, f ( n j ) > 0 for all j , and that S s, f ( n j ) → 0 as j → ∞. It is known that G ( x 4 ) = 16 (see [ 1 ]) and that G( x 4 ) ⩾ 11 almost holds (see [ 6 ]).