Kyriakos Kefalas 2025
Hidden Integer Factors in a Radical Expression
Contents 1.0 Notation 1.1 The algebraic expression 1.2 Linear recurrences 1.3 Conclusion Kyriakos Kefalas - 4 Oct 2026
Abstract : We consider algebraic expressions whose values are integers in a specified domain, that can be factored into algebraic factors whose values are also integers. The results are integer divisors of the original integer, although not necessarily prime divisors.
1.0 Notation
$$\begin{array}{cc}\mathbb{N} & \text{non}\text{–} \text{negative integers} \\ {\mathbb{N}}_{1} & \text{odd positive integers} \\ {\mathbb{N}}_{2} & \text{even non}\text{–} \text{negative integers} \\ {\mathbb{N}}_{2}^{\ast } & \text{even positive integers} \\\end{array}$$
1.1 The algebraic expression
We consider the algebraic expression
$${y}_{n}=\frac{1}{4}\left( 2+\frac{{(1+\sqrt{2})}^{2n+1}-(17+12\sqrt{2}){(1-\sqrt{2})}^{2n+1}}{4+3\sqrt{2}} \right),\,\qquad n\in \mathbb{N} \tag{1}$$
Why does this expression always give an integer—and why is every term with, \(n\geq 3\), composite?
Question : Why does this expression always give an integer—and why is every term with, \(n\geq 3\), composite?
Indeed every term is a positive integer and admits the following factorization into positive integer-valued algebraic expressions
$$\boxed{{y}_{n}=\left\{\begin{array}{lll}\frac{{a}^{n-1}+{b}^{n-1}}{2}\frac{{a}^{n}+{b}^{n}}{a+b} & \, & n\in {\mathbb{N}}_{1} \\ \, & \, & \, \\ \frac{{a}^{n-1}+{b}^{n-1}}{a+b}\frac{{a}^{n}+{b}^{n}}{2} & \, & n\in {\mathbb{N}}_{2} \\\end{array}\right.},\,\qquad \begin{array}{c}a=1+\sqrt{2} \\ b=\sqrt{2}-1 \\\end{array} \tag{2}$$
Proof
1. Let, \(a=1+\sqrt{2}\), \(b=\sqrt{2}-1\), then the terms of this sequence can be written as
$${y}_{n}=\frac{{a}^{2n-1}+{b}^{2n-1}+a+b}{2(a+b)},\,\qquad n\in \mathbb{N} \tag{3}$$
since, \(a\,b=1\), \(a+b=2\sqrt{2}\), \({a}^{4}=17+12\sqrt{2}\), \({a}^{2} = \frac{4+3\sqrt{2}}{\sqrt{2}}\). This expression factors as
$${y}_{n}=\frac{({a}^{n-1}+{b}^{n-1})({a}^{n}+{b}^{n})}{2(a+b)} \tag{4}$$
where the two integer\(-\)valued algebraic factors are
$${y}_{n}=\left\{\begin{array}{lll}\frac{{a}^{n-1}+{b}^{n-1}}{2}\frac{{a}^{n}+{b}^{n}}{a+b} & \, & n\in {\mathbb{N}}_{1} \\ \, & \, & \, \\ \frac{{a}^{n-1}+{b}^{n-1}}{a+b}\frac{{a}^{n}+{b}^{n}}{2} & \, & n\in {\mathbb{N}}_{2} \\\end{array}\right. \tag{5}$$
and in particular
$$\frac{{a}^{-1}+{b}^{-1}}{a+b}=1,\,\quad \frac{{a}^{0}+{b}^{0}}{2}=1 \tag{6}$$
2. We show that the algebraic factors in (5 ) are positive integers. From the binomial expansion we have
$${a}^{k}={A}_{k}+\sqrt{2}{B}_{k},\,\qquad k,\,{A}_{k},\,{B}_{k}\in \mathbb{N} \tag{7}$$
$$\text{Multiplication by a},\,\text{gives}$$
$${A}_{0}=1,\,\quad {B}_{0}=0,\,\quad {A}_{k+1}={A}_{k}+2{B}_{k},\,\quad {B}_{k+1}={A}_{k}+{B}_{k} \tag{8}$$
By conjugation of (7 ) we have
$${b}^{k}={(-1)}^{k}({A}_{k}-\sqrt{2}{B}_{k}) \tag{9}$$
$$\text{hence}$$
$${a}^{k}+{b}^{k}=\left\{\begin{array}{ll}(a+b){B}_{k} & k\in {\mathbb{N}}_{1} \\ 2{A}_{k} & k\in {\mathbb{N}}_{2} \\\end{array}\right. \tag{10}$$
Then from (4 ) we get
$$\boxed{{y}_{n}=\left\{\begin{array}{lll}1 & \, & n=0 \\ {A}_{n-1}{B}_{n} & \, & n\in {\mathbb{N}}_{1} \\ {A}_{n}{B}_{n-1} & \, & n\in {\mathbb{N}}_{2}^{\ast } \\\end{array}\right.} \tag{11}$$
Every factor displayed is a positive integer. This proves that all sequence members are positive integers. ■■
Recurrence (8 ) shows that both sequences, \({A}_{n},\,{B}_{n}\), are non-decreasing. Since, \({A}_{2}=3,\,{B}_{3}=5\), the factor involving \(A\), in (11 ) is at least 3, and the factor involving \(B\), is at least 5, whenever, \(n\geq 3\). Hence every such term \({y}_{n}\), is composite. The first few terms of the sequence are
$${y}_{n}=1,\,1,\,3,\,15,\,85,\,493,\,2871,\,16731,\,97513,\,568345,\,3312555\ldots ,\,\qquad n\in \mathbb{N} \tag{12}$$
This sequence (without the first term), is given in \(\text{A011900}\), in shifted form.
1.2 Linear recurrences
The algebraic expression in (1 ) satisfies the non–homogeneous linear recurrence
$${y}_{n+2}=6{y}_{n+1}-{y}_{n}-2,\,\quad {y}_{0}={y}_{1}=1,\,\quad n\in \mathbb{N} \tag{13}$$
and the homogeneous linear recurrence
$${y}_{n+3}=7{y}_{n+2}-7{y}_{n+1}+{y}_{n},\,\quad \left\{ \begin{array}{c}{y}_{0}={y}_{1}=1 \\ {y}_{2}=3 \\\end{array} \right\},\,\quad n\in \mathbb{N} \tag{14}$$
Moreover the algebraic factors satisfy the relations
$${A}_{n}={B}_{n+1}-{B}_{n},\,\quad {A}_{n}^{2}-2{B}_{n}^{2}={(-1)}^{n} \tag{15}$$
1.3 Conclusion
Conclusion 1 ) yields positive integers and admits an explicit integer-valued algebraic factorization. For every, \(n\geq 3\), both factors exceed 1, proving that \({y}_{n}\), is composite.
