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On a Conjecture Involving Twin Practical Numbers

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22 December 2025

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23 December 2025

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Abstract
In this paper, we state a conjecture that for any positive integer $p$, there are infinitely many practical numbers $q$ such that both $q^p$ and $q^p + 2$ are practical numbers. We prove the conjecture when $p$ is a prime number between $7$ and $100$, refining a result involving twin practical numbers of Wang and Sun. We also prove several results related to this conjecture.
Keywords: 
;  ;  

1. Introduction

A positive integer m is called a practical number if each n = 1 , , m can be written as the sum of some distinct divisors of m. Clearly all practical numbers are even except 1, and all even perfect numbers and powers of 2 are practical. Let P ( x ) denote the number of practical numbers less than x. In 1984, Hausman and Shapiro [1] showed that
P ( x ) x ( log x ) β
for any β < 1 2 ( 1 1 log 2 ) 2 0.0979 . Margenstern [2] showed in 1991 that
P ( x ) x exp 1 2 log 2 ( log log x ) 2 + 3 log log x .
Tenenbaum [3,4] improved the bounds (1) and (2) to
x log x ( log log x ) 5 3 ε P ( x ) x log x log log x log log log x .
Using a sieve method, Saias [5] improved Tenenbaum’s result to
x log x P ( x ) x log x .
Margenstern conjectured that
P ( x ) λ 1 x log x
with λ 1 1.341 .
Like the famous twin prime conjecture, Margenstern [2] established that there are infinitely many twin practical numbers m and m + 2 , and another detailed proof was given by Melfi [6] in 1996. Let P 2 ( x ) denote the number of practical number pairs ( m , m + 2 ) less than x. Margenstern’s proof leads to a lower bound P 2 ( x ) log log x , and Melfi’s proof leads to a better lower bound P 2 ( x ) log x . In 2002, Melfi [7] improved the lower bound to
P 2 ( x ) x exp k ( log x ) 1 2
with k > 2 + log 3 2 . Margenstern also conjectured that
P 2 ( x ) λ 2 x log x
with λ 1 1.436 .
In 2022, Wang and Sun [8] generalized Margenstern’s result and showed that there are infinitely many practical numbers q such that q 4 and q 4 + 2 are also practical numbers. They showed their result by modifying Melfi’s cyclotomic method.
In this paper, we further state a conjecture and show some new results. For any positive integer C, there are infinitely many practical numbers q such that q C and q C + 2 are also practical numbers. Clearly the results of Margenstern [2] and Melfi [6,7] imply the case C = 1 of Conjecture Section 1, and Wang and Sun [8] solved the cases C = 2 and C = 4 . Wang and Sun [8] solved the cases C = 2 and C = 4 by proving that the sequence 2 35 × 3 k + 1 + 2 are practical for every integer k 0 . In this paper, we find four new sequences that generate infinitely many practical numbers.
Theorem 1. 
2 135 × 7 k + 1 + 2 , 2 165 × 3 k + 1 + 2 , 2 175 × 3 k + 1 + 2 and 2 189 × 5 k + 1 + 2 are practical for every integer k 0 .
Using Theorem 1 we can easily deduce the following three Theorems.
Theorem 2. 
Let p denotes a prime number, 7 p < 100 . Then there are infinitely many practical numbers q such that q p and q p + 2 are also practical numbers.
Theorem 3. 
There are infinitely many practical numbers q such that:
(1). q 3 and q 3 + 8 are also practical numbers.
(2). q 5 and q 5 + 32 are also practical numbers.
Theorem 4. 
For any positive integer C, there exists an integer h such that there are infinitely many practical numbers q such that q C and q C + 2 h are also practical numbers.

2. Proof of Theorem 1.2

In order to prove our Theorems, we need the following structure theorem:
Lemma 1 
([6], Lemma 1). Let m be any practical number. Then m n is practical for every n = 1 , , σ ( m ) + 1 . In particular, m n is practical for every 1 n 2 m .
Let Φ m ( x ) denotes the mth cyclotomic polynomial, then we have
Lemma 2. 
x n 1 = d n Φ d ( x ) for all n = 1 , 2 , 3 ,
Now we start our proof of Theorem 1. Note that we may use m k to represent different sequences in different subsection.

2.1. Sequence 1

We shall prove that 2 135 × 7 k + 1 + 2 are practical for every k 0 .
Write m k = 2 135 × 7 k + 1 + 2 for k = 0 , 1 , 2 , , then we only need to show that m k is practical for every k = 0 , 1 , 2 , . By a computer calculation we find that
m 0 = 87112285931760246646623899502532662132738 = 2 × 3 4 × 11 × 19 × 331 × 811 × 15121 × 87211 × 18837001 × 385838642647891 .
By Lemma 1, we know that m 0 is practical. Let x = 2 7 k , then we have m k = 2 ( x 135 + 1 ) and m k + 1 = 2 ( x 945 + 1 ) . Since k 0 , we have x 2 . By Lemma 2, we have
x 1890 + 1 x 945 + 1 = x 270 + 1 x 135 + 1 Φ 14 ( x ) Φ 42 ( x ) Φ 70 ( x ) Φ 126 ( x ) Φ 210 ( x ) Φ 378 ( x ) Φ 630 ( x ) Φ 1890 ( x ) .
Assume that 2 ( x 135 + 1 ) is practical, then we only need to show that
2 ( x 135 + 1 ) Φ 14 ( x ) Φ 42 ( x ) Φ 70 ( x ) Φ 126 ( x ) Φ 210 ( x ) Φ 378 ( x ) Φ 630 ( x ) Φ 1890 ( x )
is practical.
Note that
Φ 14 ( x ) = x 6 x 5 + x 4 x 3 + x 2 x + 1 , Φ 42 ( x ) = x 12 + x 11 x 9 x 8 + x 6 x 4 x 3 + x + 1 , Φ 70 ( x ) = x 24 + x 23 x 19 x 18 x 17 x 16 + x 14 + x 13 + x 12 + x 11 + x 10 x 8 x 7 x 6 x 5 + x + 1 , Φ 126 ( x ) = x 36 + x 33 x 27 x 24 + x 18 x 12 x 9 + x 3 + 1 , Φ 210 ( x ) = x 48 x 47 + x 46 + x 43 x 42 + 2 x 41 x 40 + x 39 + x 36 x 35 + x 34 x 33 + x 32 x 31 x 28 x 26 x 24 x 22 x 20 x 17 + x 16 x 15 + x 14 x 13 + x 12 + x 9 x 8 + 2 x 7 x 6 + x 5 + x 2 x + 1 , Φ 378 ( x ) = x 108 + x 99 x 81 x 72 + x 54 x 36 x 27 + x 9 + 1 , Φ 630 ( x ) = x 144 x 141 + x 138 + x 129 x 126 + 2 x 123 x 120 + x 117 + x 108 x 105 + x 102 x 99 + x 96 x 93 x 84 x 78 x 72 x 66 x 60 x 51 + x 48 x 45 + x 42 x 39 + x 36 + x 27 x 24 + 2 x 21 x 18 + x 15 + x 6 x 3 + 1 , Φ 1890 ( x ) = x 432 x 423 + x 414 + x 387 x 378 + 2 x 369 x 360 + x 351 + x 324 x 315 + x 306 x 297 + x 288 x 279 x 252 x 234 x 216 x 198 x 180 x 153 + x 144 x 135 + x 126 x 117 + x 108 + x 81 x 72 + 2 x 63 x 54 + x 45 + x 18 x 9 + 1 .
Since x 2 , we have
1 2 x 6 Φ 14 ( x ) x 6 ,
x 12 Φ 42 ( x ) 2 x 12 ,
x 24 Φ 70 ( x ) 2 x 24 ,
x 36 Φ 126 ( x ) 2 x 36 ,
1 2 x 48 Φ 210 ( x ) x 48 ,
x 108 Φ 378 ( x ) 2 x 108 ,
1 2 x 144 Φ 630 ( x ) x 144 ,
1 2 x 432 Φ 1890 ( x ) x 432 .
By (9)–(13), we have
Φ 14 ( x ) Φ 42 ( x ) Φ 70 ( x ) Φ 126 ( x ) Φ 210 ( x ) 8 x 126 x 129 2 ( x 135 + 1 )
and
2 ( x 135 + 1 ) Φ 14 ( x ) Φ 42 ( x ) Φ 70 ( x ) Φ 126 ( x ) Φ 210 ( x )
is practical by Lemma 1.
By (9)–(13), we have
2 ( x 135 + 1 ) Φ 14 ( x ) Φ 42 ( x ) Φ 70 ( x ) Φ 126 ( x ) Φ 210 ( x ) 1 2 x 261 x 260 .
By (14)–(15), we have
Φ 378 ( x ) Φ 630 ( x ) 2 x 252 x 253 .
By (18)–(19), we have
Φ 378 ( x ) Φ 630 ( x ) x 253 x 260 2 ( x 135 + 1 ) Φ 14 ( x ) Φ 42 ( x ) Φ 70 ( x ) Φ 126 ( x ) Φ 210 ( x )
and
2 ( x 135 + 1 ) Φ 14 ( x ) Φ 42 ( x ) Φ 70 ( x ) Φ 126 ( x ) Φ 210 ( x ) Φ 378 ( x ) Φ 630 ( x )
is practical by Lemma 1.
By (9)–(15), we have
2 ( x 135 + 1 ) Φ 14 ( x ) Φ 42 ( x ) Φ 70 ( x ) Φ 126 ( x ) Φ 210 ( x ) Φ 378 ( x ) Φ 630 ( x ) 1 4 x 513 x 511 .
By (16) and (21), we have
Φ 1890 ( x ) x 432 x 511 2 ( x 135 + 1 ) Φ 14 ( x ) Φ 42 ( x ) Φ 70 ( x ) Φ 126 ( x ) Φ 210 ( x ) Φ 378 ( x ) Φ 630 ( x )
and
2 ( x 135 + 1 ) Φ 14 ( x ) Φ 42 ( x ) Φ 70 ( x ) Φ 126 ( x ) Φ 210 ( x ) Φ 378 ( x ) Φ 630 ( x ) Φ 1890 ( x )
is practical by Lemma 1.
Since m 0 is practical, we can show that m k is practical for every k = 0 , 1 , 2 , by an inductive process. Now the proof of sequence 2 135 × 7 k + 1 + 2 is completed.

2.2. Sequence 2

We shall prove that 2 165 × 3 k + 1 + 2 are practical for every k 0 .
Write m k = 2 165 × 3 k + 1 + 2 for k = 0 , 1 , 2 , , then we only need to show that m k is practical for every k = 0 , 1 , 2 , . By a computer calculation we find that
m 0 = 93536104789177786765035829293842113257979682750466 = 2 × 3 2 × 11 2 × 67 × 331 × 683 × 2971 × 20857 × 48912491 × 415365721 × 2252127523412251 .
By Lemma 1, we know that m 0 is practical. Let x = 2 3 k , then we have m k = 2 ( x 165 + 1 ) and m k + 1 = 2 ( x 495 + 1 ) . Since k 0 , we have x 2 . By Lemma 2, we have
x 990 + 1 x 495 + 1 = x 330 + 1 x 165 + 1 Φ 18 ( x ) Φ 90 ( x ) Φ 198 ( x ) Φ 990 ( x ) .
Assume that 2 ( x 165 + 1 ) is practical, then we only need to show that
2 ( x 165 + 1 ) Φ 18 ( x ) Φ 90 ( x ) Φ 198 ( x ) Φ 990 ( x )
is practical.
Note that
Φ 18 ( x ) = x 6 x 3 + 1 , Φ 90 ( x ) = x 24 + x 21 x 15 x 12 x 9 + x 3 + 1 , Φ 198 ( x ) = x 60 + x 57 x 51 x 48 + x 42 + x 39 x 33 x 30 x 27 + x 21 + x 18 x 12 x 9 + x 3 + 1 , Φ 990 ( x ) = x 240 x 237 + x 234 + x 225 x 222 + x 219 + x 207 x 204 + x 201 x 195 + 2 x 192 2 x 189 + x 186 x 180 + x 177 x 174 x 162 + x 159 x 156 + x 150 2 x 147 + 2 x 144 2 x 141 + x 138 x 132 + x 129 x 126 + x 123 x 120 + x 117 x 114 + x 111 x 108 + x 102 2 x 99 + 2 x 96 2 x 93 + x 90 x 84 + x 81 x 78 x 66 + x 63 x 60 + x 54 2 x 51 + 2 x 48 x 45 + x 39 x 36 + x 33 + x 21 x 18 + x 15 + x 6 x 3 + 1 .
Since x 2 , we have
1 2 x 6 Φ 18 ( x ) x 6 ,
x 24 Φ 90 ( x ) 2 x 24 ,
x 60 Φ 198 ( x ) 2 x 60 ,
1 2 x 240 Φ 990 ( x ) x 240 .
By (24)–(26), we have
Φ 18 ( x ) Φ 90 ( x ) Φ 198 ( x ) 4 x 90 x 92 2 ( x 165 + 1 )
and
2 ( x 165 + 1 ) Φ 18 ( x ) Φ 90 ( x ) Φ 198 ( x )
is practical by Lemma 1.
By (24)–(26), we have
2 ( x 165 + 1 ) Φ 18 ( x ) Φ 90 ( x ) Φ 198 ( x ) x 255 .
By (27) and (29), we have
Φ 990 ( x ) x 240 x 255 2 ( x 165 + 1 ) Φ 18 ( x ) Φ 90 ( x ) Φ 198 ( x )
and
2 ( x 165 + 1 ) Φ 18 ( x ) Φ 90 ( x ) Φ 198 ( x ) Φ 990 ( x )
is practical by Lemma 1.
Since m 0 is practical, we can show that m k is practical for every k = 0 , 1 , 2 , by an inductive process. Now the proof of sequence 2 165 × 3 k + 1 + 2 is completed.

2.3. Sequence 3

We shall prove that 2 175 × 3 k + 1 + 2 are practical for every k 0 .
Write m k = 2 175 × 3 k + 1 + 2 for k = 0 , 1 , 2 , , then we only need to show that m k is practical for every k = 0 , 1 , 2 , . By a computer calculation we find that
m 0 = 95780971304118053647396689196894323976171195136475138 = 2 × 3 × 11 × 43 × 251 × 281 × 1051 × 4051 × 86171 × 110251 × 347833278451 × 34010032331525251 .
By Lemma 1, we know that m 0 is practical. Let x = 2 3 k , then we have m k = 2 ( x 175 + 1 ) and m k + 1 = 2 ( x 525 + 1 ) . Since k 0 , we have x 2 . By Lemma 2, we have
x 1050 + 1 x 525 + 1 = x 350 + 1 x 175 + 1 Φ 6 ( x ) Φ 30 ( x ) Φ 42 ( x ) Φ 150 ( x ) Φ 210 ( x ) Φ 1050 ( x ) .
Assume that 2 ( x 175 + 1 ) is practical, then we only need to show that
2 ( x 175 + 1 ) Φ 6 ( x ) Φ 30 ( x ) Φ 42 ( x ) Φ 150 ( x ) Φ 210 ( x ) Φ 1050 ( x )
is practical.
Note that
Φ 6 ( x ) = x 2 x + 1 , Φ 30 ( x ) = x 8 + x 7 x 5 x 4 x 3 + x + 1 , Φ 42 ( x ) = x 12 + x 11 x 9 x 8 + x 6 x 4 x 3 + x + 1 , Φ 150 ( x ) = x 40 + x 35 x 25 x 20 x 15 + x 5 + 1 , Φ 210 ( x ) = x 48 x 47 + x 46 + x 43 x 42 + 2 x 41 x 40 + x 39 + x 36 x 35 + x 34 x 33 + x 32 x 31 x 28 x 26 x 24 x 22 x 20 x 17 + x 16 x 15 + x 14 x 13 + x 12 + x 9 x 8 + 2 x 7 x 6 + x 5 + x 2 x + 1 , Φ 1050 ( x ) = x 240 x 235 + x 230 + x 215 x 210 + 2 x 205 x 200 + x 195 + x 180 x 175 + x 170 x 165 + x 160 x 155 x 140 x 130 x 120 x 110 x 100 x 85 + x 80 x 75 + x 70 x 65 + x 60 + x 45 x 40 + 2 x 35 x 30 + x 25 + x 10 x 5 + 1 .
Since x 2 , we have
x Φ 6 ( x ) x 2 ,
x 8 Φ 30 ( x ) 2 x 8 ,
x 12 Φ 42 ( x ) 2 x 12 ,
x 40 Φ 150 ( x ) 2 x 40 ,
1 2 x 48 Φ 210 ( x ) x 48 ,
1 2 x 240 Φ 1050 ( x ) x 240 .
By (32)–(36), we have
Φ 6 ( x ) Φ 30 ( x ) Φ 42 ( x ) Φ 150 ( x ) Φ 210 ( x ) 8 x 110 x 113 2 ( x 175 + 1 )
and
2 ( x 175 + 1 ) Φ 6 ( x ) Φ 30 ( x ) Φ 42 ( x ) Φ 150 ( x ) Φ 210 ( x )
is practical by Lemma 1.
By (32)–(36), we have
2 ( x 175 + 1 ) Φ 6 ( x ) Φ 30 ( x ) Φ 42 ( x ) Φ 150 ( x ) Φ 210 ( x ) x 284 .
By (37) and (39), we have
Φ 1050 ( x ) x 240 x 284 2 ( x 175 + 1 ) Φ 6 ( x ) Φ 30 ( x ) Φ 42 ( x ) Φ 150 ( x ) Φ 210 ( x )
and
2 ( x 175 + 1 ) Φ 6 ( x ) Φ 30 ( x ) Φ 42 ( x ) Φ 150 ( x ) Φ 210 ( x ) Φ 1050 ( x )
is practical by Lemma 1.
Since m 0 is practical, we can show that m k is practical for every k = 0 , 1 , 2 , by an inductive process. Now the proof of sequence 2 175 × 3 k + 1 + 2 is completed.

2.4. Sequence 4

We shall prove that 2 189 × 5 k + 1 + 2 are practical for every k 0 .
Write m k = 2 189 × 5 k + 1 + 2 for k = 0 , 1 , 2 , , then we only need to show that m k is practical for every k = 0 , 1 , 2 , . By a computer calculation we find that
m 0 = 1569275433846670190958947355801916604025588861116008628226 = 2 × 3 4 × 19 × 43 × 379 × 5419 × 87211 × 119827 × 77158673929 × 127391413339 × 56202143607667 .
By Lemma 1, we know that m 0 is practical. Let x = 2 5 k , then we have m k = 2 ( x 189 + 1 ) and m k + 1 = 2 ( x 945 + 1 ) . Since k 0 , we have x 2 . By Lemma 2, we have
x 1890 + 1 x 945 + 1 = x 378 + 1 x 189 + 1 Φ 10 ( x ) Φ 30 ( x ) Φ 70 ( x ) Φ 90 ( x ) Φ 210 ( x ) Φ 270 ( x ) Φ 630 ( x ) Φ 1890 ( x ) .
Assume that 2 ( x 189 + 1 ) is practical, then we only need to show that
2 ( x 189 + 1 ) Φ 10 ( x ) Φ 30 ( x ) Φ 70 ( x ) Φ 90 ( x ) Φ 210 ( x ) Φ 270 ( x ) Φ 630 ( x ) Φ 1890 ( x )
is practical.
Note that
Φ 10 ( x ) = x 4 x 3 + x 2 x + 1 , Φ 30 ( x ) = x 8 + x 7 x 5 x 4 x 3 + x + 1 , Φ 70 ( x ) = x 24 + x 23 x 19 x 18 x 17 x 16 + x 14 + x 13 + x 12 + x 11 + x 10 x 8 x 7 x 6 x 5 + x + 1 , Φ 90 ( x ) = x 24 + x 21 x 15 x 12 x 9 + x 3 + 1 , Φ 210 ( x ) = x 48 x 47 + x 46 + x 43 x 42 + 2 x 41 x 40 + x 39 + x 36 x 35 + x 34 x 33 + x 32 x 31 x 28 x 26 x 24 x 22 x 20 x 17 + x 16 x 15 + x 14 x 13 + x 12 + x 9 x 8 + 2 x 7 x 6 + x 5 + x 2 x + 1 , Φ 270 ( x ) = x 72 + x 63 x 45 x 36 x 27 + x 9 + 1 , Φ 630 ( x ) = x 144 x 141 + x 138 + x 129 x 126 + 2 x 123 x 120 + x 117 + x 108 x 105 + x 102 x 99 + x 96 x 93 x 84 x 78 x 72 x 66 x 60 x 51 + x 48 x 45 + x 42 x 39 + x 36 + x 27 x 24 + 2 x 21 x 18 + x 15 + x 6 x 3 + 1 , Φ 1890 ( x ) = x 432 x 423 + x 414 + x 387 x 378 + 2 x 369 x 360 + x 351 + x 324 x 315 + x 306 x 297 + x 288 x 279 x 252 x 234 x 216 x 198 x 180 x 153 + x 144 x 135 + x 126 x 117 + x 108 + x 81 x 72 + 2 x 63 x 54 + x 45 + x 18 x 9 + 1 .
Since x 2 , we have
1 2 x 4 Φ 10 ( x ) x 4 ,
x 8 Φ 30 ( x ) 2 x 8 ,
x 24 Φ 70 ( x ) 2 x 24 ,
x 24 Φ 90 ( x ) 2 x 24 ,
1 2 x 48 Φ 210 ( x ) x 48 ,
x 72 Φ 270 ( x ) 2 x 72 ,
1 2 x 144 Φ 630 ( x ) x 144 ,
1 2 x 432 Φ 1890 ( x ) x 432 .
By (42)–(47), we have
Φ 10 ( x ) Φ 30 ( x ) Φ 70 ( x ) Φ 90 ( x ) Φ 210 ( x ) Φ 270 ( x ) 16 x 180 x 184 2 ( x 189 + 1 )
and
2 ( x 189 + 1 ) Φ 10 ( x ) Φ 30 ( x ) Φ 70 ( x ) Φ 90 ( x ) Φ 210 ( x ) Φ 270 ( x )
is practical by Lemma 1.
By (42)–(47), we have
2 ( x 189 + 1 ) Φ 10 ( x ) Φ 30 ( x ) Φ 70 ( x ) Φ 90 ( x ) Φ 210 ( x ) Φ 270 ( x ) 1 2 x 369 x 368 .
By (48) and (51), we have
Φ 630 ( x ) x 144 x 368 2 ( x 189 + 1 ) Φ 10 ( x ) Φ 30 ( x ) Φ 70 ( x ) Φ 90 ( x ) Φ 210 ( x ) Φ 270 ( x )
and
2 ( x 189 + 1 ) Φ 10 ( x ) Φ 30 ( x ) Φ 70 ( x ) Φ 90 ( x ) Φ 210 ( x ) Φ 270 ( x ) Φ 630 ( x )
is practical by Lemma 1.
By (42)–(48), we have
2 ( x 189 + 1 ) Φ 10 ( x ) Φ 30 ( x ) Φ 70 ( x ) Φ 90 ( x ) Φ 210 ( x ) Φ 270 ( x ) Φ 630 ( x ) 1 4 x 513 x 511 .
By (49) and (53), we have
Φ 1890 ( x ) x 432 x 511 2 ( x 189 + 1 ) Φ 10 ( x ) Φ 30 ( x ) Φ 70 ( x ) Φ 90 ( x ) Φ 210 ( x ) Φ 270 ( x ) Φ 630 ( x )
and
2 ( x 189 + 1 ) Φ 10 ( x ) Φ 30 ( x ) Φ 70 ( x ) Φ 90 ( x ) Φ 210 ( x ) Φ 270 ( x ) Φ 630 ( x ) Φ 1890 ( x )
is practical by Lemma 1.
Since m 0 is practical, we can show that m k is practical for every k = 0 , 1 , 2 , by an inductive process. Now the proof of sequence 2 189 × 5 k + 1 + 2 is completed.
Finally, we get Theorem 1 by combining the above four cases.

3. Proof of Theorems 1.3, 1.4 and 1.5

In order to prove Theorem 2, we need to show that at least one of the following five sequences generate infinitely many multiples of p for all prime 7 p < 100 :
35 × 3 k + 1 , 135 × 7 k + 1 , 165 × 3 k + 1 , 175 × 3 k + 1 , 189 × 5 k + 1 .
By the following identities (where m is any positive integer)
165 × 3 5 + 6 m + 1 0 ( mod 7 ) , 35 × 3 3 + 10 m + 1 0 ( mod 11 ) , 135 × 7 3 + 12 m + 1 0 ( mod 13 ) , 35 × 3 8 + 16 m + 1 0 ( mod 17 ) , 35 × 3 17 + 18 m + 1 0 ( mod 19 ) , 135 × 7 20 + 22 m + 1 0 ( mod 23 ) , 35 × 3 24 + 28 m + 1 0 ( mod 29 ) , 35 × 3 27 + 30 m + 1 0 ( mod 31 ) , 189 × 5 32 + 36 m + 1 0 ( mod 37 ) , 135 × 7 7 + 40 m + 1 0 ( mod 41 ) , 35 × 3 3 + 42 m + 1 0 ( mod 43 ) , 35 × 3 11 + 46 m + 1 0 ( mod 47 ) , 35 × 3 1 + 52 m + 1 0 ( mod 53 ) , 165 × 3 21 + 58 m + 1 0 ( mod 59 ) , 135 × 7 50 + 60 m + 1 0 ( mod 61 ) , 135 × 7 33 + 66 m + 1 0 ( mod 67 ) , 35 × 3 11 + 70 m + 1 0 ( mod 71 ) , 189 × 5 57 + 72 m + 1 0 ( mod 73 ) , 35 × 3 2 + 78 m + 1 0 ( mod 79 ) , 35 × 3 24 + 82 m + 1 0 ( mod 83 ) , 35 × 3 69 + 88 m + 1 0 ( mod 89 ) , 35 × 3 40 + 96 m + 1 0 ( mod 97 ) ,
we complete the proof of Theorem 2. By combining those identities, we can also show that Conjecture Section 1 holds true for many other integers.
We can prove Theorem 3 by taking the sequences
4 × 2 35 × 3 k + 1 + 2 = 2 35 × 3 k + 3 + 8
and
16 × 2 35 × 3 k + 1 + 2 = 2 35 × 3 k + 5 + 32 .
Similarly, for Theorem 4 we can take the sequence
2 h 1 × 2 35 × 3 k + 1 + 2 = 2 35 × 3 k + h + 2 h
with a suitable h such that 35 × 3 k + h is a multiple of C.

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