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Source. Pipeline-math, Erdős problem 477, commit 99d916ff32a90e77c98eb004537ccda409262346 (29 June 2026), Lemma 1.4, printed/PDF pp. 2-4 of the manuscript.

Statement

Let $N\in\mathbb Q^\times$ and let

$$X_N:\quad X^{13}+Y^{13}+Z^{13}=NW^{13}\subset\mathbb P^3_{\mathbb Q}.$$

If $N\notin\mathbb Q^{13}$, there is no nonconstant rational map $\mathbb P^1_{\mathbb Q}\dashrightarrow X_N$. When $N=d^{13}$ for some $d\in\mathbb Q^\times$, every nonconstant morphism $\mathbb P^1_{\mathbb Q}\to X_N$ has one of the following three lines as its geometric image, and each of the three arises in this way:

$$X=-Y,\ Z=dW;\qquad X=-Z,\ Y=dW;\qquad Y=-Z,\ X=dW.$$

Here a rational parametrized curve means the image of a nonconstant one-parameter rational map defined over $\mathbb Q$. It need not parametrize every rational point of its image. Rational maps to this projective surface extend to morphisms by the coordinate argument below.

External premise

We use the genus-zero three- and four-term unit bounds recalled by Corvaja and Zannier (2011), printed p. 438, PDF p. 3, at the recalled S-unit bounds. Write $\mathrm h$ for that page's projective height. Over an algebraically closed field $k$ of characteristic zero, let $\Sigma$ be a finite subset of $\mathbb P^1_k$ with $|\Sigma|\ge2$. If nonzero $\Sigma$-units $f_1,\ldots,f_r$ sum to zero, their projective tuple is nonconstant, and no proper nonempty subsum vanishes, the needed bounds are

$$\mathrm h(f_1:f_2:f_3)\le |\Sigma|-2, \qquad \mathrm h(f_1:f_2:f_3:f_4)\le3(|\Sigma|-2).$$

The first is the recalled Mason-Stothers inequality. The second is the recalled Brownawell-Masser inequality. The four-term recalled statement also allows constant normalized tuples; the stronger nonconstancy hypothesis above holds in our application. To translate the source's normalized formulas, divide by $f_1$, and in the four-term case put $z=-f_4/f_1=1+f_2/f_1+f_3/f_1$. The no-subsum conditions then agree. Multiplication of every coordinate by a rational function, and of any coordinate by a nonzero constant, preserves projective height.

Only these two cases of the manuscript's Theorem 1.2 are used. Their external proofs are not part of the reconstruction.

Proof

Coordinates and height

Suppose a nonconstant rational map to $X_N$ is given. On the generic affine parameter $s$, its coordinates are rational functions over $\mathbb Q$. Clear denominators, homogenize to a common degree in $[S:T]$, and divide out their common homogeneous factors. This yields

$$\phi=[F:G:H:R],\qquad F,G,H,R\in\mathbb Q[S,T],$$

with all nonzero forms of the same degree $e$ and no common geometric zero. Indeed a common zero in $\mathbb P^1_{\overline{\mathbb Q}}$ would give a common linear factor there; taking its conjugates gives a common factor over $\mathbb Q$, contrary to cancellation. Some of the four forms may be identically zero. The tuple defines a morphism on all of $\mathbb P^1_{\mathbb Q}$. Its image still lies in $X_N$, since the homogeneous identity valid at the generic point is a polynomial identity:

$$F^{13}+G^{13}+H^{13}-NR^{13}=0. \tag{1}$$

This also represents any initially given morphism, since it agrees on the dense open set where the original rational coordinates were chosen. If $e=0$, all coordinates would be constant, so nonconstancy gives $e\ge1$.

Work now over $k=\overline{\mathbb Q}$. For a tuple of nonzero rational functions on $\mathbb P^1_k$, define

$$\mathrm h(q_1:\cdots:q_r) =-\sum_{P\in\mathbb P^1_k}\min_i\operatorname{ord}_P(q_i).$$

Orders of poles are negative. Multiplying every coordinate by $q$ changes this expression by $-\sum_P\operatorname{ord}_P(q)=0$: a rational function has equally many zeros and poles with multiplicity. Nonzero constant factors have order zero at every point.

Let $P_1,\ldots,P_r$ be the nonzero forms among $F,G,H,R$. These forms have no common zero, since the omitted forms are identically zero. For their dehomogenizations $p_i(s)=P_i(s,1)$, at every finite point the minimum order is zero. Also at least one $p_i$ has degree $e$, since otherwise all $P_i$ would vanish at $[1:0]$. Therefore the minimum order at infinity among $p_i^{13}$ is $-13e$, and

$$\mathrm h(p_1^{13}:\cdots:p_r^{13})=13e. \tag{2}$$

This notation gives the height of the corresponding homogeneous coordinate tuple; dividing by an active coordinate or multiplying by its nonzero coefficient in (1) does not change (2).

Let $\Sigma$ be the union of the zeros of $P_1\cdots P_r$ on the complete projective line, including infinity when applicable. Each form has $e$ roots counted with multiplicity, so $|\Sigma|\le re$. After dividing the active terms of (1) by one of them, every resulting function is a $\Sigma$-unit. The normalized tuple is nonconstant: if every ratio of thirteenth powers were constant, every underlying coordinate ratio would have no zeros or poles and would be constant, making $\phi$ constant. A nonconstant rational function has a zero and a pole at distinct points of $\mathbb P^1_k$. Thus $|\Sigma|\ge2$, as required by the external bounds.

Cases with no vanishing subsum

If all four terms of (1) are nonzero and no proper nonempty subsum vanishes, the four-term bound and (2) imply

$$13e\le3(|\Sigma|-2)\le3(4e-2)=12e-6,$$

which is impossible for $e\ge1$.

Next let exactly one coordinate form be identically zero, so that (1) has three nonzero terms, and suppose no proper nonempty subsum of them is zero. The three-term bound then gives

$$13e\le |\Sigma|-2\le3e-2,$$

which fails for every $e\ge1$. Otherwise some proper nonempty subsum of the three nonzero terms is zero; it has two terms, because each term alone is nonzero, and (1) would then make the third term zero, which it is not. So (1) never has exactly three nonzero terms, whichever coordinate form is the zero one.

There cannot be zero active terms in projective coordinates, and one active term cannot satisfy (1). With exactly two active terms, (1) makes the thirteenth power of their coordinate ratio a nonzero constant. For any rational function $q$ with $q^{13}$ constant, every point order satisfies $13\operatorname{ord}_P(q)=0$. Thus $q$ has no zeros or poles and is constant. With all other coordinates zero, $\phi$ would be a constant map. This excludes the cases with at most two active terms.

The remaining pairings

So a nonconstant map has all four terms of (1) nonzero, together with some proper nonempty subsum equal to zero. That subsum has exactly two terms: one term alone is nonzero, and if three terms summed to zero, (1) would make the fourth term zero. By (1) the other two terms then also sum to zero, so the four terms form two pairs with zero sums.

Consider first

$$F^{13}+G^{13}=0,\qquad H^{13}=NR^{13}. \tag{3}$$

The rational function $F/G$ has constant thirteenth power $-1$, so it is constant. A constant in $\mathbb Q(s)$ lies in $\mathbb Q$: comparing a nonzero coefficient in a relation $p(s)=dq(s)$ with $p,q\in\mathbb Q[s]$ shows $d\in\mathbb Q$. The only rational solution of $d^{13}=-1$ is $d=-1$, since the odd power map is injective on $\mathbb R$. Thus $F=-G$.

Similarly (3) gives

$$13\operatorname{ord}_P(H/R)=\operatorname{ord}_P(N)=0 \quad\text{for every }P\in\mathbb P^1_k.$$

Therefore $H/R=d\in\mathbb Q^\times$, and $N=d^{13}$. The image lies on $X=-Y$, $Z=dW$. The other two pairings are

$$F^{13}+H^{13}=0,\ G^{13}=NR^{13}, \qquad G^{13}+H^{13}=0,\ F^{13}=NR^{13},$$

which give, respectively, $X=-Z$, $Y=dW$ and $Y=-Z$, $X=dW$. These exhaust the possibilities because $-NR^{13}$ must be paired with one of the three other terms. In every case $N\in\mathbb Q^{13}$. This proves the claimed exclusion when $N\notin\mathbb Q^{13}$.

For the full image classification when $N=d^{13}$, the nonconstant map onto its containing line is surjective over $k$. For example, in (3) it is given on that line by $[F:R]$, whose forms have no common zero. For each $[a:b]\in\mathbb P^1_k$, the nonzero homogeneous polynomial $bF-aR$ of positive degree has a root, at which $[F:R]=[a:b]$. It cannot be the zero polynomial for some $[a:b]$, since that would make the map constant. Thus the image is the whole line. Conversely the maps

$$[S:T]\mapsto[S:-S:dT:T],\qquad [S:T]\mapsto[S:dT:-S:T],\qquad [S:T]\mapsto[dT:S:-S:T]$$

have no common coordinate zero and parametrize the three lines inside $X_N$. They are nonconstant and defined over $\mathbb Q$, completing the lemma.

Dependencies and current verification

The external height convention and recalled bounds were checked in text and rendered images at Corvaja-Zannier printed p. 438. Their proofs and the original Brownawell-Masser and Mason-Stothers proofs are not locally reconstructed. The coordinate normalization and height calculation above make explicit steps compressed in the manuscript; they are not an author-issued correction.

This complete reconstruction of Lemma 1.4 received [[diophantine_problems/pipeline_math_2026_tiling_complement/evidence/verify/compilation_review|independent compilation review]]. No material defect was found in its exact frozen statement, essential deductions or consumed unit-bound interface. The source's pp. 2-4 were read in text and rendered images. Attack selection was partly pre-directed; the derivations were independently performed. The six-result review is relative to the Corvaja-Zannier-recalled unit bounds and Heath-Brown's journal Theorem 2, with the recorded nonconstant-family qualification. The external proofs were not independently reviewed; no formal verification is claimed. The current source versions and overall reading scope are recorded in the [[diophantine_problems/pipeline_math_2026_tiling_complement/_index|source digest]].

Bears on. Its non-power case is consumed by Corollary 1.5, then by the proof of Problem 477.