Let Z[i] be the Gaussian integers. For c∈Z[i]backslash0,-1, define a_1=1 and a_n(c)=c^2^n-1-1+a_n-1^2(c)qquad(n≥2). For n≥3, define b_n^pm(c)=i bigl(a_n-1(c)pm√a_n(c)bigr). An element of Z[i] is non-square if it is not the square of an element of Z[i]. Non-square conjecture. For every n≥3, both b_n^+(c) and b_n^-(c) are non-square in Z[i]. The conjecture is intended to provide the arithmetic condition needed for irreducibility of all iterates of f_r(x)=x^2+1/c. The paper gives no resolution, so its status remains open.

Non-square conjecture for the Gaussian-integer recurrence factors

Less than 1 year old · traced to 2026

Let $\mathbb{Z}[i]$ be the Gaussian integers. For $c\in\mathbb{Z}[i]\backslash\{0,-1\}$ , define $a_1=1$ and

a_n(c)=c^{2^{n-1}-1}+a_{n-1}^2(c)\qquad(n\ge2).

For $n\ge3$ , define

b_n^\pm(c)=i\bigl(a_{n-1}(c)\pm\sqrt{a_n(c)}\bigr).

An element of $\mathbb{Z}[i]$ is non-square if it is not the square of an element of $\mathbb{Z}[i]$ .

Non-square conjecture. For every $n\ge3$ , both $b_n^+(c)$ and $b_n^-(c)$ are non-square in $\mathbb{Z}[i]$ .

The conjecture is intended to provide the arithmetic condition needed for irreducibility of all iterates of $f_r(x)=x^2+1/c$ . The paper gives no resolution, so its status remains open.

References

Primary source

Jermain McDermott, “Stable quadratic polynomials over Q(i)”, arXiv:2606.25250 (2026).

Posted 1 month ago

Progress summary

Refreshed 1 month ago

Claimed progress

A reader-supplied calculation claims the universal conjecture is false, but nobody has independently checked it and the narrower irreducibility question remains open.

Jermain McDermott's June 2026 paper studies the associated quadratic-iteration stability problem over $\mathbb{Q}(i)$ and explicitly leaves this non-square conjecture unresolved.

Known results

McDermott (2026) proves stability in the previously elusive case $c\equiv 2\pmod 4$ , but does not resolve the non-square conjecture.

Posted attempt

An unverified calculation claims a counterexample at $c=-2i$ and $n=3$ : $a_3=(1+2i)^2$ , while $b_3^+=2i=(1+i)^2$ and $b_3^-=4=2^2$ . It therefore claims to refute the unconditional conjecture as printed, while not addressing the narrower version assuming irreducibility of the second iterate. The calculation has not been independently verified.

Current status (as of August 2026): The paper's conjecture remains unproved, while an unverified calculation claims it is false in the stated unconditional form; the narrower stability conjecture remains open.

Sources

arxiv.org

CounterexampleThis solution needs a summarySee full solutionHide full solution

Conjecture 1.7 is false as stated: both of its supposedly non-square Gaussian integers can simultaneously be squares.

Take the allowed parameter c=−2i∈Z[i]{0,−1} and n=3. Its defining recurrence gives

a₁=1,

a₂=c+1=1−2i,

a₃=c³+a₂² =(−2i)³+(1−2i)² =−3+4i =(1+2i)².

Choose √a₃=1+2i. Then

b₃⁺(c)=i(a₂+√a₃) =i[(1−2i)+(1+2i)] =2i =(1+i)²,

and

b₃⁻(c)=i(a₂−√a₃) =i[(1−2i)−(1+2i)] =4 =2².

Thus both b₃⁺ and b₃⁻ are Gaussian-integer squares, contrary to the universal claim. Reversing the square-root choice simply exchanges these two squares.

There is an important distinction concerning the intended application. The preceding Lemma 1.6 assumes that the second polynomial iterate is irreducible, but Conjecture 1.7 itself omits this hypothesis and quantifies over every c∈Z[i]{0,−1}. Our parameter satisfies c=(−1+i)² and has reducible second iterate:

f(x)=x²+i/2,

f^{∘2}(x)=(x²+1/2)(x²−1/2+i).

Accordingly, the calculation disproves precisely the unconditional Conjecture 1.7 as printed; it does not disprove the separate narrower stability conjecture obtained by additionally assuming second-iterate irreducibility.