Local information and representation first moments

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Status

The lemmas and theorem in this note are proved. They describe limitations of local first-moment arguments, not solutions of the density or infinitude questions.

Every residue contains two powerful summands

For every $r\ge1$, every modulus $M\ge1$, and every residue $a\pmod M$, there are infinitely many sums of exactly two positive $r$-powerful integers in that residue.

For each prime power $p^k\parallel M$, write $a=b_p+c_p$, with each summand either a unit or zero modulo $p^k$. For $p\ge3$, choose $b_p\bmod p$ outside $\{0,a\}$, so both summands are units. At $p=2$, use two odd summands when $a$ is even, and (0+a) when $a$ is odd. CRT gives positive representatives $b,c$. Choose $E\ge r$ with $E\equiv1\pmod{\varphi(M)}$. Then

$$b^E+c^E\equiv a\pmod M.$$

Both summands are $r$-powerful. Replacing $b$ by $b+tM$ yields infinitely many distinct values. The case $M=1$ is immediate.

For $r\ge3$, the two-summand set nevertheless has density zero, by the bound $O_r(X^{2/r})$ for its candidate tuples. Thus surjectivity modulo every integer is far weaker than positive density.

Large good primes cannot produce arbitrary savings when $r\ge5$

For $F(\mathbf x)=\sum_{i=1}^r d_i x_i^r$, define

$$\rho_{p^k}(a;F)=p^{-k(r-1)}\#\{\mathbf x\bmod p^k:F(\mathbf x)\equiv a\pmod{p^k}\}.$$

If $p\nmid r\prod_i d_i$, then, uniformly in $a,k$,

$$\rho_{p^k}(a;F)\ge 1-(r-1)^r p^{1-r/2}-p^{1-r}. \tag{1}$$

For $h\ne0\pmod p$, expansion into multiplicative characters gives

$$\left|\sum_{x\bmod p}e_p(hd_i x^r)\right|\le(r-1)\sqrt p.$$

Additive orthogonality therefore bounds the deviation of the number of solutions modulo $p$ from $p^{r-1}$ by $(r-1)^r p^{r/2}$. The only possibly singular solution is the zero vector. Remove it, and lift each nonsingular solution to exactly $p^{(k-1)(r-1)}$ solutions modulo $p^k$. This proves (1).

When $r\ge5$, the errors in (1) are summable over primes. For sufficiently large $P$, products of these lower bounds over any set of primes greater than $P$ are bounded below by a positive constant tending to (1) as $P\to\infty$. Allowing high prime powers or highly divisible targets does not alter this conclusion.

Nonzero good-prime targets: a sharper bound

The distinction in Section 4 of the independent record is also valid: for unit coefficients and $a\not\equiv0\pmod p$,

$$|\rho_p(a;F)-1|\le(r-1)^r p^{-(r-1)/2}. \tag{1a}$$

Indeed, put $g=\gcd(r,p-1)$. Each nonzero-frequency power sum expands into $g-1$ Gauss sums of absolute value $\sqrt p$. Expanding the product over the $r$ coordinates leaves at most $(g-1)^r$ terms. The remaining frequency sum is a multiplicative Gauss sum twisted by the nonzero target: its absolute value is $\sqrt p$ for a nonprincipal character and $1$ for the principal character. Division by the orthogonality factor $p$ and by the normalizing factor $p^{r-1}$ proves (1a). At target zero that frequency sum can instead have magnitude $p-1$, giving the weaker bound used in (1).

If also $p\nmid r$, every solution at a nonzero target is nonsingular, so $\rho_{p^k}(a;F)=\rho_p(a;F)$ whenever $p\nmid a$. Thus for a fixed form, using only sufficiently large good primes with nonzero targets cannot make the product arbitrarily small already for $r\ge4$. The successful quartic proof uses targets $p\pmod{p^2}$, which are zero modulo $p$, and is not subject to this sharper summability obstruction.

Uniform lower bound for all progression first moments, $r\ge5$

Let

$$R_r(X;a,M)=\#\{(n_1,\ldots,n_r)\in\mathcal P_r^r: \textstyle\sum_i n_i\le X,\ \sum_i n_i\equiv a\pmod M\}.$$

Theorem. For every $r\ge5$, there is $c_r>0$, independent of $a,M$, such that

$$\liminf_{X\to\infty}\frac{M}{X}R_r(X;a,M)\ge c_r. \tag{2}$$

The limit here is taken for each fixed $a,M$.

A finite family handles all small-prime targets

Choose $B\ge r$ so large that, for every prime $p>B$, both

$$\gamma_p=1-(r-1)^r p^{1-r/2}-p^{1-r},\qquad \gamma'_p=1-(r-1)^{r-1}p^{(3-r)/2}-p^{2-r}$$

exceed (1/2). Put

$$h_p=2v_p(r)+1,\qquad Q=\prod_{p\le B}p^{h_p}.$$

Consider the finite family

$$F_u(\mathbf x)=u^{r+1}x_1^r+x_2^r+\cdots+x_r^r, \qquad 1\le u\le Q,\quad (u,Q)=1.$$

All positive summands of every $F_u$ are $r$-powerful.

Given an integer $a$, choose $t\pmod Q$ that is $0\pmod{p^{h_p}}$ if $p\nmid a$, and $1\pmod{p^{h_p}}$ otherwise. Then $a-t^r$ is a unit modulo $Q$. Take its representative $u\in[1,Q]$, and let $v\equiv u^{-1}\pmod Q$. We have

$$F_u(v,t,0,\ldots,0)\equiv a\pmod Q.$$

At every $p\le B$, the derivative with respect to $x_1$ has valuation $v_p(r)$. For $k\ge h_p$, fix the other coordinates modulo $p^{h_p}$, and vary them modulo $p^k$. Each of the $p^{(k-h_p)(r-1)}$ choices admits a lift of $x_1$, by the Hensel criterion $h_p>2v_p(r)$. For $k<h_p$, reduction gives at least one solution. In either case,

$$\rho_{p^k}(a;F_u)\ge p^{-h_p(r-1)}. \tag{3}$$

Large primes and the uniform product

For $p>B$ not dividing $u$, (1) gives the lower bound $\gamma_p$. If $p\mid u$, use only the remaining $r-1$ unit-coefficient variables. Their nonsingular solution count, normalized by $p^{r-2}$, is at least $\gamma'_p$, by the same Gauss-sum proof with $r-1$ variables. The free first coordinate and nonsingular lifting give

$$\rho_{p^k}(a;F_u)\ge\gamma'_p.$$

Every prime dividing $u$ is at most $Q$. As $r\ge5$, $\prod_{p>B}\gamma_p>0$. Since all the factors below lie in ((0,1]), equations (1) and (3) imply

$$\prod_{p^k\parallel M}\rho_{p^k}(a;F_u)\ge\eta_r>0,$$

where one may take

$$\eta_r= \prod_{p\le B}p^{-h_p(r-1)} \prod_{p>B}\gamma_p \prod_{B<p\le Q}\gamma'_p.$$

Lattice counting

For fixed $M,u$, counting positive lattice points in the region $F_u(\mathbf x)\le X$, residue class by residue class, gives

$$\#\{\mathbf x\in\mathbb N^r:F_u(\mathbf x)\le X, F_u(\mathbf x)\equiv a\pmod M\} =\frac{\Gamma(1+1/r)^r X}{M u^{(r+1)/r}} \prod_{p^k\parallel M}\rho_{p^k}(a;F_u) +O_{M,u,r}(X^{1-1/r}).$$

The map to the ordered summand tuple $(u^{r+1}x_1^r,x_2^r,\ldots,x_r^r)$ is injective. Taking $u\le Q$ proves (2), with

$$c_r=\Gamma(1+1/r)^r Q^{-(r+1)/r}\eta_r.$$

Scope. This theorem rules out extending the cubic proof by making ordered representation first moments arbitrarily small on progressions. By itself it does not exclude a useful fixed saving, and it gives no density bound for distinct values. The subsequent quintic finite-certificate argument, extended at the prime five, proves the stronger lower bound $385732347/156250000>2.468$ for the unordered first moment when $r=5$, ruling out a deficit below one in that case.

A single locally universal core for the fifth powers

For $r=5$, the same uniform positive lower bound already holds for the single form

$$F_5=x_1^5+\cdots+x_5^5.$$

Here is a direct proof. For a prime $p\ne5$, let $A=\{x^5:x\in\mathbb F_p\}$. Then

$$|A|=1+\frac{p-1}{\gcd(5,p-1)},\qquad |5A|\ge\min\{p,5|A|-4\}=p$$

by Cauchy–Davenport. Every nonzero target has a primitive solution. For zero, use $(1,-1,0,0,0)$. Nonsingular lifting gives $\rho_{p^k}(a;F_5)\ge p^{-4}$ for every $k,a$.

At $p=5$, the fifth-power residues modulo $25$ are $0,\pm1,\pm7$. Every centered residue from $-12$ to $12$ is a sum of at most five of these, with a nonzero term: use $0=1-1$; ones for $1,\ldots,5$; $6=7-1$; $7$ plus zero through four ones for $7,\ldots,11$; and $12=7+7-1-1$, then negate. Pad with zeros to five terms.

Choose corresponding input residues $u_i\pmod5$, with $u_1\ne0$. For $k\ge2$, every choice $x_i\equiv u_i\pmod5$ for $i=2,\ldots,5$ leaves the required first fifth power in the correct residue class modulo $25$. The map

$$\{x\bmod5^k:x\equiv u_1\pmod5\}\longrightarrow \{y\bmod5^k:y\equiv u_1^5\pmod{25}\},\qquad x\mapsto x^5,$$

is exactly five-to-one: after writing $x=u_1+5t$, the integer polynomial $((u_1+5t)^5-u_1^5)/25$ has derivative $u_1^4\pmod5$ and is bijective modulo every power of $5$. Consequently

$$\rho_{5^k}(a;F_5)\ge5^{-3}\quad(k\ge2),$$

while the modulus-$5$ density is $1$.

For $p>256$, use the stronger good-prime bound

$$\rho_{p^k}(a;F_5)\ge1-1024p^{-3/2}-p^{-4}=\gamma_p>\tfrac12.$$

The product of these $\gamma_p$ is positive. Combining the finitely many other primes gives $\eta_5>0$ with $\rho_M(a;F_5)\ge\eta_5$ for every $M,a$. Lattice counting therefore gives a normalized progression first moment at least $\Gamma(6/5)^5\eta_5$ for this one core alone. This is a local-distribution result, not positive density of its distinct values.

Many denominator primes do not force a proper zero subsum locally

One attempted descent on targets $tm^r$ was to force a proper collection of summands to add to zero at a prime dividing $m$, and then seek a shorter powerful relation. Congruences alone do not force that first step, even for a fixed coefficient tuple.

Fix $r\ge3$ and $F=\sum_i d_i x_i^r$. For $p\nmid r\prod_i d_i$, at least

$$p^{r-1}-(r-1)^r p^{r/2}-(2^r-2)r^2p^{r-2} \tag{4}$$

roots of $F=0\pmod p$ have no vanishing nonempty proper subsum. The power-sum estimate above gives the first two terms. For a fixed proper subset $I$, a root with its $I$-subsum zero also has its complementary subsum zero. Fixing all but one variable in each block bounds the number of such roots by $r^2p^{r-2}$. Union over the $2^r-2$ subsets proves (4).

For sufficiently large $p$ this lower bound is positive. Each surviving root has every coordinate nonzero, since singleton subsums are nonzero. It is nonsingular and lifts to every $p^k$, including to any prescribed target divisible by $p$, while retaining the proper-subsum condition modulo $p$. CRT combines such roots at arbitrarily many sufficiently large denominator primes avoiding the coefficients.

For canonical $d_i$, the resulting positive root representatives do give $r$-powerful summands satisfying these congruences. Their exact sum need not be the prescribed target, and their sizes are uncontrolled. Thus this excludes only the proposed local forcing of a shorter relation; an argument using exact equality and $d_i x_i^r\le tm^r$ is not ruled out.