Erdős 699: a hypothetical cubic counterexample gives an irreducible cubic

P699 research · Erdős problem 699 · The uniform proof attempt

---

This note proves a uniform restriction on a hypothetical counterexample with index three. After removing its exact power-of-two content, the associated cubic is irreducible over the rationals. The proof uses no finite search. It does not yet prove nonexistence or global finiteness of the counterexamples.

The subsequent elliptic-curve note proves that a specified twist of $y^2=P(z)$ has a point of infinite order, and explains why the change to the monic cubic gives an elliptic quartic.

1. The integer cubic supplied by badness

Suppose that no odd prime divides both $\binom n3$ and $\binom nj$, with $4\le j\le n/2$. The earlier cubic reductions give

$$n=2^h u,\qquad h\ge2,\qquad u\text{ odd},\qquad j<n/2.$$

Put $c=n-j$ and

$$W=\frac{u(n-1)(n-2)}6=\frac{\binom n3}{2^h}.$$

Then $W$ is an odd positive integer. Badness implies that $W$ divides every coefficient of

$$R(z)=\sum_{r=0}^3\binom j{3-r}\binom cr z^r.\tag{1}$$

All four coefficients are positive, and Vandermonde's identity gives

$$R(1)=\binom n3=2^h W.\tag{2}$$

We will only use this coefficient divisibility, so the result also applies to any admissible pair satisfying the necessary endpoint divisibilities from the uniform proof attempt.

2. The column has smaller two-adic valuation than the row

Lemma. One has $t:=v_2(j)<h$, and $v_2(c)=t$.

Proof. Suppose instead that $2^h\mid j$. Then $2^h\mid c$ as well. For an integer $x$ divisible by four,

$$v_2\!\binom x3=v_2(x),\qquad v_2\!\binom x2=v_2(x)-1.$$

Consequently the two endpoint coefficients in (1) are divisible by $2^h$. Each middle coefficient has valuation $v_2(j)+v_2(c)-1\ge2h-1\ge h$. Thus every coefficient of $R$ is divisible by $2^hW$. Since all four are positive, this contradicts (2). Therefore $t<h$. The equality $v_2(c)=t$ now follows from $c=n-j$. $\square$

Lemma. The largest power of two dividing every coefficient of $R$ is exactly $2^t$.

Proof. There is nothing to prove for the lower bound when $t=0$. For $t\ge1$, the middle coefficients have valuation $2t-1\ge t$. For $x=j,c$, the endpoint valuation is

$$v_2\!\binom x3=t+v_2(x-2)-1\ge t.$$

Thus $2^t$ divides all coefficients. On the other hand,

$$R'(1)=c\binom{n-1}{2} =\frac{c(n-1)(n-2)}2$$

has valuation exactly $t$. If every coefficient had valuation greater than $t$, its derivative at one would too. This proves exactness, including when $t=0$. $\square$

Define

$$P(z)=\frac{R(z)}{2^tW}\in\mathbb Z[z].\tag{3}$$

It follows that

$$P(1)=2^{h-t},\qquad P'(1)=\frac{3c}{2^t u}\text{ is an odd integer}.\tag{4}$$

The integrality of the derivative follows from (3); its oddness follows either from the exact valuation of $R'(1)$ or directly from the displayed quotient. Thus the reduction of $P$ modulo two has a simple root at one. Moreover, $P$ is primitive: any common divisor of its coefficients divides $P(1)$, hence is a power of two, and the exact-content lemma excludes two.

3. All three roots are negative and real

For completeness, this needs no general theorem about Jacobi polynomials. Expansion of the cubic discriminant gives

$$\operatorname{Disc}R =\frac{j^2c^2(j-1)(c-1)(n-1)(n-2)^2}{12}>0.\tag{5}$$

This is also the cubic specialization of the discriminant identity in the cutoff and cubic note. A real cubic with positive discriminant has three distinct real roots. The positive coefficients of $R$ exclude nonnegative roots. Therefore all three roots of $R$, and hence of $P$, are strictly negative.

They in fact all lie in $(-1,0)$. Write $d=c-j$. The earlier alternating coefficient argument gives $d^2>3n-2$, so

$$\begin{aligned} R(-1)&=-d(d^2-3n+2)/6<0,\\ R'(-1)&=\frac c2\bigl((d-1)^2-(n-1)\bigr)>0,\\ R''(-1)&=c(c-1)(2-d)<0. \end{aligned}$$

For the middle inequality, $d^2-2d-n+2>2(n-d)>0$. Also $R'''(-1)>0$. Consequently every coefficient of $-R(-1-z)$ is positive, excluding any root of $R$ at or below $-1$.

4. A general irreducibility observation

Lemma. Let $F\in\mathbb Z[z]$ be primitive, with positive leading coefficient and all roots strictly negative and real. If $F(1)$ is a power of a prime $p$ and $p\nmid F'(1)$, then $F$ is irreducible over $\mathbb Q$.

Proof. Otherwise Gauss's lemma supplies a factorization $F=AB$ into nonconstant integer polynomials with positive leading coefficients. Every root of either factor is negative and real, so every coefficient of either factor is strictly positive. Hence $A(1),B(1)\ge2$. Their product is the power $F(1)$ of $p$, so both are divisible by $p$. But then

$$F'(1)=A'(1)B(1)+A(1)B'(1)\equiv0\pmod p,$$

a contradiction. $\square$

Applying this lemma with $p=2$ and using (4) proves:

Theorem. Every hypothetical bad triple with index three gives a primitive irreducible cubic $P$ as in (3), with three distinct negative real roots in $(-1,0)$, $P(1)=2^{h-t}$, and $P'(1)$ odd.

The same conclusion holds under the necessary endpoint divisibilities alone; the higher Lucas digit conditions have not been used.

5. An associated monic integer cubic

The same hypotheses also give a monic irreducible cubic

$$H(y)=y^3-ky+2q,\qquad k=\frac{3jc}{2^{2t}(n-1)},\quad q=\frac{jcd}{2^{3t}(n-1)(n-2)},\tag{6}$$

where $k,q$ are positive odd integers. Here is an integrality argument that also covers the possible canceled factor of three. The unnormalized integer moments from the uniform note are

$$m=\frac{3j}{u},\qquad \ell=\frac{3j(j-1)}{u(n-1)},\qquad a=\frac{3j(j-1)(j-2)}{u(n-1)(n-2)},\qquad 3\mid a.$$

Thus

$$K:=\frac{3jc}{n-1}=u(m-\ell)\in\mathbb Z,\qquad B:=\frac{2jcd}{(n-1)(n-2)} =\frac{2u}{3}(2a+m-3\ell)\in\mathbb Z.$$

The latter is integral because $um=3j$ and $3\mid a$. The four positive integer coefficients of $P$ sum to $2^{h-t}$, so $t\le h-2$. It follows that $v_2(d)=t+1$ and hence

$$v_2(K)=2t,\qquad v_2(B)=3t+1.$$

Dividing $K$ by $2^{2t}$ and $B$ by $2^{3t}$ proves the assertions about $k,q$. The exact change of variable is

$$P(z)=\frac{2^{2t}(1-z)^3}{u n^2} H\!\left(\frac{j+cz}{2^t(1-z)}\right).\tag{7}$$

This invertible rational change proves that $H$ is irreducible and has three real roots. Its discriminant satisfies

$$k^3-27q^2 =2^{2(h-t)-2}u^6\operatorname{Disc}P>0.$$

This gives a monic formulation of the simultaneous real and two-adic restrictions. It is still a necessary condition: arbitrary odd $k,q$ satisfying a discriminant congruence need not reconstruct admissible integers $n,j,u$.

6. What this settles, and what it leaves open

This excludes every hypothetical pair for which the associated cubic has a rational root, or any other rational factorization. It also identifies a totally real cubic field associated with every remaining pair. The simple root at one modulo two is compatible with irreducibility over the rationals, so it is not itself a contradiction.

For reference, the exact discriminant after normalization is

$$\operatorname{Disc}P =\frac{108\,j^2c^2(j-1)(c-1)} {2^{4t}u^4(n-1)^3(n-2)^2}.\tag{8}$$

Its positivity and integrality can sharpen necessary size inequalities. They do not supply an upper bound on $h$: when $t$ and $u$ are fixed and $j/n$ stays inside $(0,1/2)$, the displayed expression grows proportionally to $n$. Thus neither the elementary discriminant lower bound nor irreducibility alone completes a finiteness argument.

There is also a direct way to see the limitation of rational approximation. Homogenize with the orientation

$$F(x,y)=x^3P(y/x),\qquad r=jx+cy,\quad s=x-y.$$

Expansion gives the exact identity

$$F(x,y)=\frac{1}{2^t u n^2}\left( r^3-\frac{3jc}{n-1}rs^2 +\frac{2jc(c-j)}{(n-1)(n-2)}s^3\right).\tag{9}$$

The integer pairs $(r,s)$ form a lattice of determinant $n$, since $r=ny+js$. Dirichlet approximation, with $Q=\lfloor n^{1/4}\rfloor$, gives integers $s,y$ with $1\le |s|\le Q$ and $|ny+js|\le n/(Q+1)$. Set $x=y+s$. Since $n>2j\ge8$, the two coefficients inside (9) satisfy

$$\frac{3jc}{n-1}\le n,\qquad \frac{2jc(c-j)}{(n-1)(n-2)}\le n.$$

Thus (9) yields a nonzero integer value with

$$1\le |F(x,y)|\le \frac{2n^{1/4}+n^{-1/4}}{2^t u}.\tag{10}$$

Nonvanishing follows from irreducibility and $(x,y)\ne(0,0)$. The numerator scale in (10) still grows when $u,t$ are fixed. Moreover, the distinguished value $F(1,1)=2^{h-t}$ does not imply that the new value $F(x,y)$ is a power of two. This approximation therefore does not give a descent preserving the original hypotheses. Identities (5), (7), and (9), and the derivative identity used above, were checked symbolically.

The unresolved objective remains to rule out these irreducible cubics, or to prove that only finitely many can arise from bad triples. Even the infinite subfamily $u=1$ is not settled by the arguments in these notes.