gpt 6 astra ultra attempt on Math-204

Exact Asymptotic Waring Function G(k)

Definitions and mathematical statement

For $k\ge3$, define \[ G(k)=\min\{s\ge1:\ \exists N_0\ \forall N\ge N_0\ \exists x_1,\ldots,x_s\in{\mathbb{Z}}_{\ge0},\ N=x_1^k+\cdots+x_s^k\}. \] The problem is to determine this integer for every $k$. Allowing zero summands is equivalent to using at most $s$ positive powers, but not to requiring exactly $s$ positive powers. The cutoff $N_0$ may depend on $k,s$. This asymptotic function differs from the Waring function that must represent every positive integer, including small exceptional values. An exact answer needs both eventual sufficiency and obstructions to every smaller number of summands.

Formulation and concept references: [S1].

Short English statement

For each exponent, what is the smallest number of positive powers needed to represent every sufficiently large integer, allowing fewer summands when necessary?

Research attempt

Status: UNRESOLVED. We prove a counting lower bound $G(k)\ge k+1$, formulate modular obstructions correctly for the eventual-representation target, and give an infinite descent obstruction establishing $G(4)\ge16$. Determining $G(k)$ for every $k$ still requires matching eventual upper bounds.

Context and quantifiers.

The NIST DLMF Waring section [S1], consulted 1 October 2026, distinguishes the function for all positive integers from the function for all sufficiently large integers, and records $G(4)=16$. Only the lower direction for that isolated exponent is reproved below. Small exceptional integers alone cannot lower-bound $G(k)$: the definition permits an exponent-dependent cutoff. A lower-bound obstruction must produce arbitrarily large exceptions for the proposed number of summands.

Counting unordered representations.

Suppose at most $s$ nonnegative $k$th powers are used to represent integers at most $X$. Padding with zeros makes the number of summands exactly $s$, and every base is at most $M=\lfloor X^{1/k}\rfloor$. Each represented integer comes from an unordered multiset of $s$ bases drawn from $\{0,\ldots,M\}$. There are \[ \binom{M+s}{s} \] such multisets. Different multisets may produce the same sum, and some sums exceed $X$, so this is an upper bound on the number of represented integers in $[0,X]$.

For $s<k$ it is $O(X^{s/k})=o(X)$, immediately leaving arbitrarily many exceptions. For $s=k$, \[ \binom{M+k}{k}=\frac{X}{k!}+O_k(X^{1-1/k}). \] Since $k\ge2$ gives $1/k!<1$, a positive fraction of integers up to $X$ remain unrepresented for large $X$. Hence even $k$ summands cannot represent every sufficiently large integer, proving $G(k)\ge k+1$. Using ordered tuples would lose the factorial and would not yield this endpoint improvement.

Modular necessary conditions.

For a modulus $q\ge2$, define $R_q=\{x^k\bmod q:x\in\mathbb Z\}$. If a residue $a$ is absent from the $s$-fold sumset $sR_q$, every positive integer congruent to $a$ modulo $q$ is unrepresentable with $s$ terms. There are arbitrarily large such integers, so $G(k)>s$.

Zero belongs to $R_q$, so the use of exactly $s$ terms here is equivalent to at most $s$. Passing all fixed moduli is only necessary: it does not prove that an actual nonnegative integer representation exists. A finite table of attainable residues cannot replace a theorem of eventual sufficiency.

An infinite obstruction for fourth powers.

Every fourth power is congruent to zero or one modulo sixteen. If a sum of at most fifteen fourth powers is divisible by sixteen, the number of odd bases is between zero and fifteen and is itself divisible by sixteen. It must be zero. Thus every base is even, and dividing the sum by sixteen again produces a representation by at most fifteen fourth powers.

Now consider $N_a=31\cdot16^a$ for integers $a\ge0$. A hypothetical representation of $N_a$ by at most fifteen fourth powers can be divided by sixteen $a$ times using the preceding observation. It would yield such a representation of 31. But the only nonnegative fourth powers at most 31 are $0,1,16$. To sum to 31, either no 16 is used, requiring 31 ones, or one 16 is used, requiring fifteen ones in addition and hence sixteen terms. Two 16s already exceed 31. Therefore 31, and every $N_a$, requires at least sixteen terms.

Since $N_a\to\infty$, this establishes $G(4)\ge16$ for the asymptotic function. The same conclusion could not have been drawn just from the single integer 31. The descent is what propagates the finite obstruction beyond every possible cutoff.

Why a greedy upper-bound attempt is inadequate.

Subtracting the largest available $k$th power always terminates, because the remainder decreases among nonnegative integers. Termination proves representability with some number of summands, but that number may depend on $N$. The target asks for a bound depending only on $k$, uniform over all sufficiently large $N$. Counting the steps by their maximum possible initial value gives no such constant.

Likewise, a recurrence that decreases the remainder by a fixed proportion would give a logarithmic number of steps, still unbounded as $N\to\infty$. A successful upper-bound proof needs additional structure, such as a theorem producing a bounded number of summands in every sufficiently large admissible case. No such general theorem with exact optimal count is obtained here.

Verification and remaining gap.

The counting argument includes zeros and uses unordered multisets to avoid overclaiming their number. The fourth-power descent relies specifically on the modulus sixteen and the cap of fifteen summands; it is not asserted for arbitrary exponents. All checks are symbolic, with no claimed computation over a large range.

The exact value $G(4)=16$ is an attributed source result, with its upper direction external to this proof. For general $k$, the unresolved task is to match all lower obstructions with eventual sufficiency and prove optimality. The worked arguments supply genuine necessary bounds but do not determine the full function requested.

Sources

• S1 DLMF27.13(iii), version1.2.7 page snapshot. https://dlmf.nist.gov/27.13. Formulation source.