Coefficient growth in square chains

TL;DR

This paper investigates the growth of coefficients in iterated quadratic polynomials that factor over integers, showing they grow exponentially and become too large for explicit construction as the number of iterations increases.

Contribution

It establishes new bounds on prime divisors and exponential growth of coefficients in square chains, highlighting limitations in constructing such polynomials explicitly.

Findings

Prime divisors of coefficients are bounded by powers of 2.

Coefficients grow exponentially, exceeding manageable sizes for large chains.

Growth estimates suggest explicit construction of large chains is infeasible.

Abstract

Suppose $((\cdots((x^{2}-c_{1})^{2}-c_{2})^{2}\cdots)^{2}-c_{k-1})^{2}-c_{k}$ splits into linear factors over $\mathbb{Z}$ and $c_{k}\neq0$. We show that for each $j$ and each prime $p$, if $p\leq2^{j-1}$ then $p$ divides $c_{j}$. Consequently, $$\ln c_{j}>\frac{1}{4}\cdot2^{j}\,\,\mathrm{for}\,j\geq5$$ If we also have $p\equiv3\,(\mathrm{mod\,4)}$ then $p^{2^{j-\left\lceil \lg p\right\rceil }}$ divides $c_{j}$. Consequently, if $k\geq3$, there exists some absolute constant $\lambda>0$ so that, $$\ln c_{j}>\lambda k2^{j}\mathrm{\,\,for\,all\,}j$$ These estimates argue against the possibility of explicitly constructing polynomials of the given form for large $k$, as the coefficients quickly become too large to manipulate.

Equations

$$\ln c_j>\frac{1}{4}\cdot 2^j\mathrm{for}j\ge 5$$

$$\ln c_j>\lambda k{2}^j\mathrm{for}\mathrm{all}j$$

$$P(x)=((\cdots ((x^2-c_1{)}^2-c_2{)}^2\cdots {)}^2-c_{k-1}{)}^2-c_k$$

$$P(x)=((\cdots (x^2-c_1{)}^2-\cdots {)}^2-c_{k-1}-\sqrt{c_k})((\cdots (x^2-c_1{)}^2-\cdots {)}^2-c_{k-1}+\sqrt{c_k})$$

$$|T_{j}|=\left\{\begin{array}[]{rl}2(|T_{j+1}|-1)+1&\mathrm{if\,}0\in T_{j+1}\\ 2|T_{j+1}|&\mathrm{if\,}0\not\in T_{j+1}\end{array}\right\}\geq 2(|T_{j+1}|-1)+1$$

$$|T_0|\ge 2(|T_1|-1)+1\ge 2^2(|T_2|-1)+1\ge \cdots \ge 2^{k-1}(|T_{k-1}|-1)+1$$

$$P(x)$$

$$=(x^{2^{k-1}}-2^{-1}{)}^2-2^{-2}$$

$$=(x^{2^{k-1}}-1)(x^{2^{k-1}}-0)$$

$$c_j=c_{j+1}=\cdots =c_k=0_F$$

$$c_{j-1}\pm t\equiv 0,1,\mathrm{or}4(\mathrm{mod}8)$$

$$(c_{j-1}+t)-(c_{j-1}-t)\equiv 2t\equiv 0,1,3,4,5,\mathrm{or}7(\mathrm{mod}8),$$

$$t\equiv 0,2,4,\mathrm{or}6(\mathrm{mod}8).$$

$$m\#=\prod_{\begin{array}[]{c}p\mathrm{\,prime}\\ p\leq m\end{array}}p$$

$$(\cdots (x^2-c_1{)}^2-\cdots {)}^2=c_i\pm \sqrt{\cdots \pm \sqrt{c_k}}$$

$$\begin{array}[]{cc}x(1-\frac{1}{2\ln x})\leq\ln(x\#)&\mathrm{for\,}x\geq 563\end{array}$$

$$\begin{array}[]{lr}\frac{1}{2}x<\ln(x\#)&\mathrm{for\,}11\leq x\leq 563\end{array}$$

$$\begin{array}[]{lr}\frac{1}{2}2^{j-1}<\ln(2^{j-1}\#)\leq\ln c_{j}&\mathrm{if\,}11\leq 2^{j-1}\end{array}$$

$${\nu }_p(r^2+s^2)$$

$$={\nu }_p(c_{j-1})$$

$$\ge 2^{j-1-\lceil \lg p\rceil }$$

$$2{\nu }_p(r+is)$$

$$={\nu }_p((r+is)(r-is))$$

$$={\nu }_p(r^2+s^2)$$

$$\ge 2^{j-1-\lceil \lg p\rceil }$$

$${\nu }_p(r+is)\ge \frac{1}{2}{2}^{j-1-\lceil \lg p\rceil }$$

$${\nu }_p(r),{\nu }_p(s)\ge \frac{1}{2}{2}^{j-1-\lceil \lg p\rceil }$$

$${\nu }_p(r^2),{\nu }_p(s^2)\ge 2^{j-1-\lceil \lg p\rceil }$$

$${\nu }_p(2t)$$

$$\ge 2^{j-1-\lceil \lg p\rceil }$$

$${\nu }_p(t)\ge 2^{j-1-\lceil \lg p\rceil }$$

$${\nu }_p(t^2)\ge 2^{j-\lceil \lg p\rceil }$$

$${\nu }_p(c_j+u),{\nu }_p(c_j-u)\ge 2^{j-\lceil \lg p\rceil }$$

$${\nu }_p(2c_j)$$

Full text

[TABLE]

If we also have $p\equiv 3\,(\mathrm{mod\,4)}$ then $p^{2^{j-\left\lceil\lg p\right\rceil}}$ divides $c_{j}$. Consequently, if $k\geq 3$, there exists some absolute constant $\lambda>0$ so that,

[TABLE]

These estimates argue against the possibility of explicitly constructing polynomials of the given form for large $k$, as the coefficients quickly become too large to manipulate.

1. Motivation: factoring integers with square

chains

Call a polynomial of the form

[TABLE]

a square chain of length $k$. Some square chains of lengths $k=3,4$ are presented in Crandall and Pomerance (2005, research problem 6.18) which have the property that they have $2^{k}$ distinct integer roots. Crandall and Pomerance then ask about the existence of longer square chains, suggesting that sufficiently long chains might be useful for factoring large integers. Indeed, a simple scheme shows promise:

Suppose $n=pq$ is an odd semiprime, and that $p$ and $q$ are approximately the same size. If $P(x)\in\mathbb{Z}[x]$ has about $\frac{p}{2}\approx\frac{q}{2}$ distinct roots, one can reasonably hope that $P(m)\equiv 0\,(\mathrm{mod\ }p)$ for about half of $m\in\{0,1,\ldots,n-1\}$, and similarly hope $P(m)\equiv 0\,(\mathrm{mod\ }q)$ for about half of $m\in\{0,1,\ldots,n-1\}$. If we assume heuristically that these are independent events, then about a quarter of the choices of $m$ yield $\gcd(P(m),n)=p$, and about a quarter yield $\gcd(P(m),n)=q$. For a more rigorous analysis, see Lipton (1994).

To make this a tractable factoring algorithm, we need a polynomial $P$ which can be efficiently evaluated and has sufficiently many distinct roots. Square chains are nearly ideal from the standpoint of efficient evaluation. As a polynomial of degree $2^{k}$, a square chain of length $k$ may have up to $2^{k}$ roots and may be evaluated using only $k$ multiplications. No polynomial of degree $2^{k}$ can be evaluated with fewer multiplications (Borchert et al. (2013)). An obstacle is finding square chains with many distinct roots.

Let us set aside the question of existence. Suppose that there exists a square chain of length $k$ which has exactly $2^{k}$ roots, counting multiplicity, even if those roots are not all distinct. What properties might such a square chain have?

We say a polynomial $P$ crumbles over the unique factorization domain $D$ if $P$ may be written as the product of (not necessarily distinct) linear polynomials in $D[x]$. Unless otherwise indicated by context, we will assume $D=\mathbb{Z}.$ Clearly $x^{2^{k}}=(\cdots(x^{2}-0)^{2}\cdots-0)^{2}-0$ crumbles over any UFD. Unfortunately, it has only 1 distinct root. More generally, if we have any crumbling square chain $P$, then $P(x)^{2}-0$ is a crumbling square chain longer than $P$. But extending a chain in this manner doesn’t create any new roots. Conversely, any square chain whose final coefficient is 0 has exactly the same root set as a square chain of shorter length, and we can instead consider the shorter square chain. To this end, call a square chain whose final coefficient is nonzero a fundamental square chain.

While it may be a big ask, suppose we are able to find some fundamental crumbling square chain $P$. Then it is guaranteed to have plenty of distinct roots.

Proposition 1.1**.**

Let $D$ be a unique factorization domain with $1_{D}+1_{D}\neq 0_{D}$. Suppose $P(x)=(\cdots(x^{2}-c_{1})^{2}-\cdots)^{2}-c_{k}\in D[x]$ crumbles over $D$. If $c_{k}\neq 0_{D}$, then $P$ has at least $2^{k-1}+1$ distinct roots.

Proof.

Let $T_{k}=\{c_{k}\}$, and call this the $k$-th tail square set of $P$. Note that $c_{k}$ must be a perfect square; indeed, for any root $r$ of $P$, we have $c_{k}=(\cdots(r^{2}-c_{1})^{2}\cdots-c_{k-1})^{2}$. So we may use the identity $a^{2}-b^{2}=(a+b)(a-b)$, to split $P(x)$ into two square chains:

[TABLE]

Since $P$ crumbles and $D[x]$ is a unique factorization domain, each of these factors must crumble. Let $T_{k-1}=\{c_{k-1}\pm t:t^{2}\in T_{k}\}$ be the tail squares of the two factors. Exactly as was the case with $T_{k}$, elements of $T_{k-1}$ must be perfect squares.

We then repeat this splitting with each factor, splitting $P$ into 4 factors with tail square set $T_{k-2}=\{c_{k-2}\pm t:t^{2}\in T_{k-1}\}$, then 8 factors with tail square set $T_{k-3}=\{c_{k-3}\pm t:t^{2}\in T_{k-2}\}$, and so forth onto $2^{k-1}$ factors of the form $(x^{2}-a)$ with $a\in T_{1}=\{c_{1}\pm t:t^{2}\in T_{2}\}$. If we make the convention that $c_{0}=0$, we may split the previous set of factors one more time to get $2^{k}$ factors of the form $(x-r)$ with $r$ in the set $T_{0}=\{0\pm t:t^{2}\in T_{1}\}$. That is, $T_{0}$ is the set of roots of $P$.

Suppose $j\in\{0,1,\ldots,k-1\}$. What can we say about the size of $T_{j}$? Distinct elements of $T_{j+1}$ yield distinct elements of $T_{j}$: if $\{c_{j}\pm t\}\cap\{c_{j}\pm s\}\neq\emptyset$ then $t=\pm s$, and so $t^{2}=s^{2}$. Each nonzero element of $T_{j+1}$ yields two distinct elements of $T_{j}$: if $c_{j}+t=c_{j}-t$, then $2t=0$, and so $t=0$ since $D$ is an integral domain and $2\neq 0$. As such,

[TABLE]

Consequently,

[TABLE]

To complete the argument, we note that $|T_{k-1}|=2|T_{k}|=2$ since $0\not\in T_{k}$ by hypothesis. ∎

As an aside, the bound of proposition 1.1 is met exactly when $D=\mathbb{Z}_{q}$ for a prime $q$ satisfying $q\equiv 1\,(\mathrm{mod\ }2^{k-1})$, and when the square chain $P$ of length $k$ is given by

[TABLE]

In this case, [math] is a root of $P$ with multiplicity $2^{k-1}$, and each $2^{k-1}$-th root of unity $(\mathrm{mod\ }q)$ is a root of $P$ with multiplicity 1.

2. Coefficient growth

Perhaps more interesting than proposition 1.1 is a specialization of its contrapositive: for an odd characteristic finite field, fundamental crumbling square chains must be of strictly bounded length.

Corollary 2.1**.**

Suppose $P(x)=(\cdots(x^{2}-c_{1})^{2}-\cdots)^{2}-c_{k}\in F[x]$ crumbles over the finite field $F$ with $\mathrm{char}(F)\neq 2$. If $|F|\leq 2^{j-1}$ then

[TABLE]

Proof.

Suppose $c_{i}$ is the last non-zero coefficient in $P$, so that $Q(x)=(\cdots(x^{2}-c_{1})^{2}-\cdots)^{2}-c_{i}$ is a fundamental crumbling square chain over $F$. By proposition 1.1, $Q$ must have at least $2^{i-1}+1$ distinct roots in $F$. Of course, $2^{i-1}+1\leq|F|$, as $Q$ can have, at most, all of $F$ as roots. So $2^{i-1}+1\leq|F|\leq 2^{j-1}.$ Consequently, $i<j$, and so all coefficients of index $j$ or larger must be zero. ∎

Our primary interest in corollary 2.1 will be in the case $F=\mathbb{Z}_{p}$ for $p$ any prime. But corollary 2.1 does not cover the case of $\mathbb{Z}_{2}$. Instead, we derive a similar, if weaker, result to cover $\mathbb{Z}_{2}$.

Lemma 2.2**.**

Suppose $P(x)=(\cdots(x^{2}-c_{1})^{2}-\cdots)^{2}-c_{k}\in\mathbb{Z}[x]$ crumbles over $\mathbb{Z}$. Then $c_{2}\equiv c_{3}\equiv\cdots\equiv c_{k}\equiv 0\,(\mathrm{mod\ }2)$

Proof.

Consider the tail square sets $T_{j}$ from proposition 1.1. Let us adopt the convention that $T_{k+1}=\{0\}$. This is consistent with our previous definition, as it makes $T_{k}=\{c_{k}\pm t:t^{2}\in T_{k+1}\}=\{c_{k}\}$.

For an arbitrary $j>1$, choose an arbitrary $t^{2}\in T_{j}$. Then $c_{j-1}\pm t\in T_{j-1}$ by definition, and so $c_{j-1}\pm t$ are both squares. Thus,

[TABLE]

The limited set of congruence classes that these elements fall into lets us somewhat limit what congruence classes their difference falls into. That is,

[TABLE]

This limits the congruence classes $t$ may fall into as well:

[TABLE]

So $t$ must be even. Since $t$ was chosen arbitrarily, each element $t^{2}\in T_{j}$ must be the square of an even number.

For a given $j$ with $1<j<k+1$, choose any $r^{2}\in T_{j+1}$. By the previous argument, $r$ is even. We have $(c_{j}+r)\in T_{j}$ and so $(c_{j}+r)$ is even. Consequently, $c_{j}$ must be even. ∎

As a consequence of corollary 2.1, coefficients in long crumbling square chains over $\mathbb{Z}$ must have many prime factors. This implies a lower bound on the size of those coefficients. To quantify this, define the primorial of $m$ as:

[TABLE]

Proposition 2.3**.**

Suppose $P(x)=(\cdots(x^{2}-c_{1})^{2}-\cdots)^{2}-c_{k}\in\mathbb{Z}[x]$ crumbles over $\mathbb{Z}$. Then for each $j$, we have $2^{j-1}\#$ divides $c_{j}$. If $c_{k}\neq 0$ and $j\geq 5$, then $\ln c_{j}>2^{j-2}$.

Proof.

$c_{j}$ is even for $j\geq 2$ by lemma 2.2. For each odd prime $p\leq 2^{j-1}$, corollary 2.1 tells us that $p$ divides $c_{j}$. Thus $2^{j-1}\#$ divides $c_{j}$.

Suppose $c_{k}>0.$ We claim that $c_{j}>0$ for each $j$. Clearly $c_{j}\geq 0.$ To see that $c_{j}\neq 0$, let $i$ be an arbitrary index $1\leq i<k$. We may write the equation $P(x)=0$ as

[TABLE]

Since the right-hand side of this equation must be non-negative for every choice of signs, we have $c_{i}\geq\sqrt{c_{i+1}+\cdots}\geq\sqrt{c_{i+1}}$. As the choice of $i$ was arbitrary, we may apply this inequality recursively, yielding $c_{j}\geq\sqrt[2^{k-j}]{c_{k}}$. But $c_{k}\neq 0$; it follows that $c_{j}\neq 0$ as well.

Together with the fact that $2^{j-1}\#$ divides $c_{j}$, the positivity of $c_{j}$ implies that $2^{j-1}\#\leq c_{j}$.

Rosser and Schoenfeld (1962, equation 3.14) establish:

[TABLE]

An exhaustive calculation (omitted) demonstrates the looser bound

[TABLE]

Thus

[TABLE]

∎

3. An asymptotic refinement

It seems unlikely the lower bounds given in proposition 2.3 are the best possible. $\ln(x\#)\sim x$, and so it seems likely, at the very least, that $\ln c_{j}\geq 2^{j-1}$ for $j$ sufficiently large. At the same time, it is plausible that for some primes $p$, not only must $p$ divide $c_{j}$, but possibly $p^{i}$ divides $c_{j}$ for some appropriate condition on $p,$ $i$ and $j$. Indeed, some reflection shows that primes of the form $4n+3$ must be much more prevalent than we’ve heretofore indicated.

Let \nu_{p}(n)=\max\{e\in\mathbb{Z}:p^{e}\big{|}n\} be the exponent of $p$ in the prime factorization of $n$.

As noted by Dilcher (2000), at least some of the coefficients in a crumbling square chain must be expressable as half the sum of two squares. We show that every coefficient in a crumbling square chain must be so expressable. By the sum of two squares theorem, if $p\equiv 3\,(\mathrm{mod\,}4)$, then $\nu_{p}(a^{2}+b^{2})$ must be even. And so $\nu_{p}(c_{j})$ must be even for every coefficient $c_{j}$ in a crumbling square chain.

What’s more, by similar considerations as go into the sum of two squares theorem, we can propagate powers forward to following coefficients, so that $\nu_{p}(c_{j+1})\geq 2\nu_{p}(c_{j})$.

We collect these ideas into the following lemma.

Lemma 3.1**.**

Suppose $P(x)=(\cdots(x^{2}-c_{1})^{2}-\cdots)^{2}-c_{k}\in\mathbb{Z}[x]$ crumbles over $\mathbb{Z}$. If $p$ is prime with $p\equiv 3\,(\mathrm{mod\ }4)$, and $p\leq 2^{j-1}$ then $\nu_{p}(c_{j})\geq 2^{j-\left\lceil\lg p\right\rceil}$ .

Proof.

Suppose $\left\lceil\lg p\right\rceil=h-1$. Then $p$ divides $c_{h}$ by corollary 2.1.

Choose an arbitrary $t^{2}\in T_{h+1}$. By definition, $c_{h}\pm t\in T_{h}$. All members of $T_{h}$ are squares, so there exist $r,s$ so that $c_{h}+t=r^{2}$ and $c_{h}-t=s^{2}$. Then $2c_{h}=r^{2}+s^{2}$, that is $2c_{h}$ is the sum of two squares.

By the sum of two squares theorem, since $p\equiv 3\,(\mathrm{mod\,}4)$ and $p$ divides $r^{2}+s^{2}$, then $p^{2}$ divides both $r^{2}$ and $s^{2}$. It follows that $\nu_{p}(c_{h})\geq 2^{1}=2^{h-\left\lceil\lg p\right\rceil}$.

To handle the more general case, we proceed inductively. Suppose $\nu_{p}(c_{j-1})\geq 2^{j-1-\left\lceil\lg p\right\rceil}$. We will show that $\nu_{p}(c_{j})\geq 2^{j-\left\lceil\lg p\right\rceil}$.

As before, choose an arbitrary $t^{2}\in T_{j}$, making $c_{j-1}\pm t\in T_{j-1}$. Write $c_{j-1}+t=r^{2}$ and $c_{j-1}-t=s^{2}$. Then $2c_{j-1}=r^{2}+s^{2}$. Since $p\neq 2$,

[TABLE]

Also, we may write $r^{2}+s^{2}=(r+is)(r-is)$ as the product of two Gaussian integers. We recall two well known results: the Gaussian integers form a unique factorization domain, and $p$ is a prime Gaussian integer since $p\equiv 3\,(\mathrm{mod\ }4)$. As such, it makes sense to extend our definition of $\nu_{p}$ to Gaussian integers. Now, $\nu_{p}(r+is)=\nu_{p}(r-is)$ since $p$ is its own complex conjugate. Thus

[TABLE]

So

[TABLE]

Any power of $p$ that divides $r+is$ must divide both $r$ and $s$, so

[TABLE]

implying that

[TABLE]

Since $r^{2}$ and $s^{2}$ share a common power of $p$, their difference $r^{2}-s^{2}=2t$ must share the same common power. That is,

[TABLE]

Since $p\neq 2$, this implies that

[TABLE]

and so

[TABLE]

As $t^{2}$ was chosen arbitrarily, this inequality holds for any $t^{2}\in T_{j}$.

Choose any $u^{2}\in T_{j+1}$. We have $c_{j}\pm u\in T_{j}$ by definition. So

[TABLE]

Since $c_{j}+u$ and $c_{j}-u$ share a common power of $p$, their sum, $(c_{j}+u)+(c_{j}-u)=2c_{j}$ must share the same common power. That is,

[TABLE]

And since $p\neq 2$,

[TABLE]

∎

By proposition 2.3, later coefficients “pick up” many primes as divisors. By lemma 3.1, once a coefficient acquires a divisor $p\equiv 3\,(\mathrm{mod\ }4)$, each later coefficient is divisible many times by the same prime. Define

[TABLE]

Together, proposition 2.3 and lemma 3.1 imply:

Proposition 3.2**.**

Suppose $P(x)=(\cdots(x^{2}-c_{1})^{2}-\cdots)^{2}-c_{k}\in\mathbb{Z}[x]$ crumbles over $\mathbb{Z}$. Let

[TABLE]

For each $j$, $D_{j}$ divides $c_{j}$. If $c_{k}\neq 0$, then for some absolute constant $\lambda>0$ and each $j\geq 3$,

[TABLE]

Proof.

For primes $p\leq 2^{j-1}$:

If $p\not\equiv 3\,(\mathrm{mod\ }4)$ then $\nu_{p}(D_{j})=1$.

If $p\equiv 3\,(\mathrm{mod\ }4)$ then

[TABLE]

Thus by corollary 2.1 and lemma 3.1, $D_{j}$ divides $c_{j}$.

Corollaries of the Siegel-Walfisz theorem give $\ln(x\#_{3:4})\sim\frac{1}{2}x$, c.f. Montgomery and Vaughan (2006, corollaries 11.15, 11.20). So there must be some constant $\lambda_{3:4}>0$ satisfying $\lambda_{3:4}x<\ln(x\#_{3:4})$ for all $x\geq 3$. Similarly, by the prime number theorem, $\ln(x\#)\sim x$, and thus there is a constant $\lambda_{1}$ so that $\lambda_{1}x<\ln(x\#)$ for all $x\geq 2$. Let $4\lambda=\min(\lambda_{1},\lambda_{3:4})$. Then for $j\geq 3$

[TABLE]

If $c_{k}\neq 0$, then $c_{j}>0$ as argued in proposition 2.3. Since $D_{j}$ divides $c_{j}$, positivity of $c_{j}$ forces $D_{j}\leq c_{j}$. And so $\lambda j2^{j}<\ln D_{j}\leq\ln c_{j}$. ∎

We can sharpen the closed form estimate slightly:

Corollary 3.3**.**

Suppose $P(x)=(\cdots(x^{2}-c_{1})^{2}-\cdots)^{2}-c_{k}\in\mathbb{Z}[x]$ crumbles over $\mathbb{Z}$, $k\geq 3$, and $c_{k}>0$. Then there exists an absolute constant $\lambda>0$ so that for all $j$,

[TABLE]

Proof.

We observed in the proof of proposition 2.3 that $c_{j}\geq\sqrt[2^{k-j}]{c_{k}}$ so,

[TABLE]

∎

It is interesting to note that this statement is enough to show the length of the chain influences the size of $c_{1}$, as it shows that $\ln c_{1}\geq 2\lambda k$.

4. Discussion and related work

To factor a product of two 500-bit primes using the algorithm described in section 1, we would need to start by constructing a fundamental crumbling square chain of length not much smaller than 500. According to proposition 2.3, the coefficient $c_{400}$ of such a chain would be at least $\lg e^{2^{398}}\approx 9.3\times{10}^{119}$ bits in length. By way of comparison, estimates place the total digital storage capacity of the world at approximately ${10}^{22}$ bits as of the year 2019. Even if we knew how to construct such a chain, precalculating the coefficients of such a chain would clearly be infeasible.

Finding fundamental crumbling square chains has proven difficult. While Dilcher (2000) provides a characterization of length 3 fundamental crumbling square chains, and Bremner (2008) describes two infinite families of length 4 fundamental crumbling square chains, no fundamental crumbling square chains of length 5 are known. Indeed, Borchert et al. (2013) points out that a crumbling square chain of length 5 with distinct roots would advance understanding of a historied question known as the Prouhet-Tarry-Escott problem.

Borchert et al. (2013) discuss a more general family of polynomials, which they term gems. By construction, their gems are polynomials which are efficiently computable, crumble over $\mathbb{Z}$, and have distinct roots. While the highest known degree of a square chain that crumbles over $\mathbb{Z}$ is 16, the authors of that article describe gems of degrees up to 55.