Let x,n,m be integers with x>1 and n≥ 1. For coefficients a=x^2-1 and b=1, consider the Diophantine equation m^2=4x^2n^2+4x^2n+1. The cases in question are x equiv 0,1, or 3 pmod 4; explicitly, these give x=4y, x=4y+1, or x=4y+3, respectively. Nonexistence conjecture. The equation has no integer solutions (x,n,m) for x>1 and n≥ 1 when x equiv 0,1, or 3 pmod 4. Equivalently, the corresponding Diophantine equations have no solutions in the specified ranges. The case x equiv 2 pmod 4 is covered by the proof of the preceding theorem, while these remaining congruence classes are supported in the source by computational results and are left without a proof here.

Nonexistence conjecture for cobalancing numbers with coefficients $a=x^2-1$ and $b=1$

Let $x,n,m$ be integers with $x>1$ and $n\geq 1$ . For coefficients $a=x^2-1$ and $b=1$ , consider the Diophantine equation

m^2=4x^2n^2+4x^2n+1.

The cases in question are $x\equiv 0,1,$ or $3\pmod 4$ ; explicitly, these give $x=4y$ , $x=4y+1$ , or $x=4y+3$ , respectively.

Nonexistence conjecture. The equation has no integer solutions $(x,n,m)$ for $x>1$ and $n\geq 1$ when $x\equiv 0,1,$ or $3\pmod 4$ . Equivalently, the corresponding Diophantine equations have no solutions in the specified ranges.

The case $x\equiv 2\pmod 4$ is covered by the proof of the preceding theorem, while these remaining congruence classes are supported in the source by computational results and are left without a proof here.

Progress summary

Solved

A posted calculation claims infinitely many counterexamples in every forbidden residue class, but the claim has not been independently verified.

The conjecture asserts that the displayed Diophantine equation has no solutions for the three residue classes $x\equiv 0,1,3\pmod 4$ . The stored 2022 paper presents these cases as computationally supported but unproved.

Known results

The case $x\equiv 2\pmod 4$ is covered by the source's preceding theorem; the three remaining classes are left conjectural and supported only computationally.

Posted attempt (date not given)

An attempted complete disproof sets $n=1$ , reducing the equation to $m^2-8x^2=1$ , and gives examples in all three forbidden classes. It further claims infinitely many such examples from $(3+\sqrt{8})^j$ , and says this also disproves the source's uniqueness claim for the $x\equiv2\pmod4$ case. The attempt has not been independently verified.

Current status (as of August 2026): the conjecture is not established; a purported infinite-family counterexample would settle it negatively, but that claim remains unverified.

Sources

arxiv.org

Sources & referencesView supporting material

Primary source

Hao Cui, Xiaoyu Cui, Sophia C. Davis, Irfan Durmić, Qingcheng Hu, Lisa Liu, Steven J. Miller, Fengping Ren, Alicia Smith Reina and Eliel Sosis, “Sums of Reciprocals of Recurrence Relations”, arXiv:2212.02639 (2023).

1 week ago

Counterexamples in all three forbidden residue classes, with infinite families.

Set $n=1$ . The proposed Diophantine equation becomes

m^2=8x^2+1, \qquad\text{equivalently}\qquad m^2-8x^2=1.

The following three solutions contradict each of the claimed nonexistence cases:

\begin{array}{c|c|c|c} x&n&m&x\bmod4\\ \hline 35&1&99&3\\ 204&1&577&0\\ 1189&1&3363&1 \end{array}

Indeed,

99^2-8\cdot35^2= 577^2-8\cdot204^2= 3363^2-8\cdot1189^2=1.

Their corresponding positive cobalancers are respectively $48,287,1680$ .

In fact there are infinitely many counterexamples in each residue class. Define

m_j+x_j\sqrt8=(3+\sqrt8)^j,\qquad j\ge1.

Then $m_j^2-8x_j^2=1$ , and multiplication by $3+\sqrt8$ gives

m_{j+1}=3m_j+8x_j,\qquad x_{j+1}=m_j+3x_j.

Induction modulo $4$ yields

m_j\equiv(-1)^j\pmod4,\qquad x_j\equiv j\pmod4.

Thus the subsequences $j\equiv0,1,3\pmod4$ supply infinitely many solutions in all three prohibited classes. For $j\ge2$ , the cobalancer

r_j=\frac{m_j-3}{2}

is positive.

The same family also contradicts the claimed uniqueness in Theorem 1.11 of the source. At $j=6$ ,

x=6930=4\cdot1732+2,\qquad n=1,\qquad m=19601,\qquad r=9799.

The theorem asserts that for $x=4y+2$ the unique cobalancing number is $n=y$ , which here would be $1732$ . However $n=1$ is a different valid cobalancing number, as can be checked directly:

x^2-1=48{,}024{,}899 =2+3+\cdots+9800.

Therefore both the nonexistence conjecture and the accompanying uniqueness claim are false.

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Shivam Patel · 2 days ago