Erdős 699: symmetric moments reduce to the 81/80 singular points

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This is a retained earlier research note. The current reading guide gives the latest status and distinguishes completed proofs from remaining cases.

P699 research · Erdős problem 699

The following exact reduction explains a proposed simultaneous treatment of $n,n-1,n-2$. It gives a valid construction and stronger divisibility than a first inspection suggests. No new effective index bound follows here.

For nonnegative integers $a,b,d$, with $a,b$ even, put

$$W(x)=x^a(x-2)^a(x-1)^b(10-x)^d(x+8)^d,$$ $$J_i=\int_i^{10}W(x)\,dx\quad(i=0,1,2),\qquad I=\int_0^1W(x)\,dx.$$

A common integer normalization may be applied throughout. Since $W(2-x)=W(x)$,

$$J_0-J_1=J_1-J_2=I.$$

Consequently, if $X_i=n-i$, then

$$X_0J_1-X_1J_0=X_1J_2-X_2J_1=J_0-nI.\tag{1}$$

If the integer-normalized $X_i,J_i$ both have the appropriate three pairwise coprime prime-power block divisors, (1) is divisible by their product. This uses one moment row. Its size bound involves $J_0+nI$, rather than products of two moment sizes.

Nonvanishing also has an exact proof. Replace $W$ by $(x-1)^2W$ to obtain $I',J_i'$. Write $H=\int_2^{10}W$. The multiplier is strictly below 1 on $(0,1)$ and strictly above 1 on $(2,10)$, so

$$H'/I'>H/I.$$

Since $J_0/I=2+H/I$, the two numbers $J_0-nI$ and $J_0'-nI'$ cannot both vanish. The extra factor changes the degrees by a fixed amount and leaves the original block divisors available.

The quadratic reduction

Set $u=x-1$. Then

$$W(x)=(u^2-1)^a u^b(81-u^2)^d.$$

For the odd antiderivative

$$F(v)=\int_0^v u^b(u^2-1)^a(81-u^2)^d\,du,$$

we have exactly

$$J_0=F(9)+F(1),\qquad J_1=F(9),\qquad J_2=F(9)-F(1), \qquad I=F(1).$$

A second substitution, $y=u^2$, gives

$$\boxed{F(v)=\frac12\int_0^{v^2} y^{(b-1)/2}(y-1)^a(81-y)^d\,dy.}\tag{2}$$

Thus the five-root integrand reduces to three singular points $0,1,81$. These are the same singular points as the $81-80=1$ Padé seed. Because $b$ is even, the first exponent in (2) is a half-integer. The reduction alone does not justify importing the integer-parameter coefficient-gcd formula: its denominator normalization must be established separately. Nor does identifying the same singular points prove that every resulting cofactor inequality is already implied by an existing source.

The full block exponents

Let $L=2a+b+2d$. Integration of an integer polynomial can lose at most $v_p(\operatorname{lcm}(1,\ldots,L+1))=O(\log L)$ powers of a fixed prime. The following substitutions therefore identify the exponential parts of the block divisors, before any common normalization.

For $J_0$, put $x=10t$: the integrand and differential contain $10^{a+d+1}$. Hence its 5-adic valuation is at least

$$a+d+1-\lfloor\log_5(L+1)\rfloor.$$

For $J_1=F(9)$, put $u=9t$. The factors $u^b$, $(81-u^2)^d$, and $du$ contain $9^{b+2d+1}$. Its 3-adic valuation is therefore at least

$$2b+4d+2-\lfloor\log_3(L+1)\rfloor.$$

Finally, in $J_2$ put $x=2+8t$. The factors $x^a$, $(x-2)^a$, $(10-x)^d$, $(x+8)^d$, and $dx$ contribute, respectively, $2^a,2^{3a},2^{3d},2^d,2^3$. Thus its 2-adic valuation is at least

$$4(a+d)+3-\lfloor\log_2(L+1)\rfloor.$$

In directions $(a,b,d)=(Am,Bm,Dm)$, the block rates are consequently

$$\boxed{5^{(A+D)m},\quad 3^{(2B+4D)m},\quad 2^{4(A+D)m},}$$

up to factors of polynomial size. The middle and 2-adic blocks are stronger than $9^{(B+D)m}$ and $8^{(A+D)m}$ obtained by looking only at the integration endpoints. The endpoint ratio is exactly the $2^4\cdot5$ factorization of $80$; the middle block has the $3^4$ factorization of $81$.

The direct five-factor multinomial normalization is expensive. Small exact computations confirm that its evaluated common divisor removes a substantial part of that expense, so its generic coefficient-gcd lower bound should not be mistaken for a sharp bound. The exploratory computations did not yield a strict inequality excluding the remaining balanced index-six profile.