Primes and Their Connection to Certain Polyhedral Number Sequences

Hindawi Journal of Mathematics Volume 2021, Article ID 9956343, 11 pages https://doi.org/10.1155/2021/9956343

Research Article Primes and Their Connection to Certain Polyhedral Number Sequences

Benjamin Lee Warren

Department of Mathematics, Rensselaer Polytechnic Institute, Troy 12180, New York, USA

Correspondence should be addressed to Benjamin Lee Warren; [email protected]

Received 4 April 2021; Accepted 23 June 2021; Published 9 August 2021

Academic Editor: Marco Fontana

Copyright © 2021 Benjamin Lee Warren. *is is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. A collection of results is given regarding whether a prime can be the sum or diﬀerence of two polyhedral numbers, as well as some primality restrictions on several sequences.

1. Introduction In this paper, another question will be examined and, very fortunately, answered for these types of sequences in *e sheer amount of various polyhedral figurate number se- general, which is as follows: when might a prime occur as the quences [1] makes studying them all a daunting task. Fortunately, difference between two elements from some chosen figurate some universal questions can be asked about them, though not all sequence? *e answer, along with some other results, is the answers are known. For instance, it is due to Euler [2] that we presented.. can find all numbers that are simultaneously triangular and square, which marks the general question of which numbers exist Theorem 1. Consider the tetrahedral numbers [4] in more than one figurate number sequence at a time (for ex- Tn � (1/6)n(n + 1)(n + 2). +en, the only primes which are ample, 36 is both triangular and square). Another significant the sum of two tetrahedral numbers are result is the general question suggested by Fermat, Lagrange, 2 � T1 + T1, 5 � T1 + T2, and 11 � T1 + T3. Waring, Gauss, and Cauchy, which is to determine the minimal number of elements required from a given figurate sequence in Proof. Using the closed form of Tn along with the factor- order to sum to all natural numbers [3]. *ough these are elegant ization n(n + 1)(n + 2) + m(m + 1)(m + 2) � (n + m + 2) and natural questions, few answers have so far been determined. (n2 + m2 + n + m − nm) gives the identities

2 2 T6n + T6m � 2(3n + 3m + 1)�6n + 6m + n + m − 6nm�,

2 2 T6n + T6m+1 �(2n + 2m + 1)�18n + 18m − 18nm + 9m + 1�,

2 2 T6n + T6m+2 � 2(3n + 3m + 2)�6n + 6m − 6nm − n + 5m + 1�,

2 2 T6n + T6m+3 �(6n + 6m + 5)�6n + 6m − 6nm − 2n + 7m + 2�,

2 2 T6n + T6m+4 � 2(n + m + 1)�18n + 18m − 18nm − 9n + 27m + 10�, 2 Journal of Mathematics

2 2 T6n + T6m+5 �(6n + 6m + 7)�6n + 6m − 6nm − 4n + 11m + 5�, 2 2 T6n+1 + T6m+1 �(3n + 3m + 2)�12n + 12m − 12nm + 4n + 4m + 1�, 2 2 T6n+1 + T6m+2 �(6n + 6m + 5)�6n + 6m − 6nm + n + 4m + 1�, 2 2 T6n+1 + T6m+3 �(n + m + 1)�36n + 36m − 36nm + 36m + 11�, 2 2 T6n+1 + T6m+4 �(6n + 6m + 7)�6n + 6m − 6nm − n + 8m + 3�, 2 2 T6n+1 + T6m+5 �(3n + 3m + 4)�12n + 12m − 12nm − 4n + 20m + 9�, 2 2 T6n+2 + T6m+2 � 2(n + m + 1)�18n + 18m − 18nm + 9n + 9m + 4�, 2 2 T n+ + T m+ �(6n + 6m + 7)�6n + 6m − 6nm + 2n + 5m + 2�, 6 2 6 3 ( ) 2 2 1 T6n+2 + T6m+4 � 2(3n + 3m + 4)�6n + 6m − 6nm + n + 7m + 3�, 2 2 T6n+2 + T6m+5 �(2n + 2m + 3)�18n + 18m − 18nm + 27m + 13�, 2 2 T6n+3 + T6m+3 �(3n + 3m + 4)�12n + 12m − 12nm + 8n + 8m + 5�, 2 2 T6n+3 + T6m+4 �(2n + 2m + 3)�18n + 18m − 18nm + 9n + 18m + 10�, 2 2 T6n+3 + T6m+5 �(3n + 3m + 5)�12n + 12m − 12nm + 4n + 16m + 9�, 2 2 T6n+4 + T6m+4 � 2(3n + 3m + 5)�6n + 6m − 6nm + 5n + 5m + 4�, 2 2 T6n+4 + T6m+5 �(6n + 6m + 11)�6n + 6m − 6nm + 4n + 7m + 5�, 2 2 T6n+5 + T6m+5 �(n + m + 2)�36n + 36m − 36nm + 36n + 36m + 35�. □ Theorem 2. Consider the pyramidal numbers Proof. Using the closed form of Pn along with the fac- n 2 Pn � �k�1 k � (1/6)n(n + 1)(2n + 1). +en, the only primes torization n(n + 1)(2n + 1) + m(m + 1)(2m + 1) � (n + m + which are the sum of two positive pyramidal numbers are 1)(2n2 + 2m2 − 2nm + n + m) gives the identities 2 � P1 + P1, 19 � P2 + P3, and 31 � P1 + P4.

2 2 P6n + P6m �(6n + 6m + 1)�12n + 12m − 12nm + n + m�, 2 2 P6n + P6m+1 �(3n + 3m + 1)�24n + 24m − 24nm − 2n + 10m + 1�, 2 2 P6n + P6m+2 �(2n + 2m + 1)�36n + 36m − 36nm − 9n + 27m + 5�, 2 2 P6n + P6m+3 �(3n + 3m + 2)�24n + 24m − 24nm − 10n + 26m + 7�, 2 2 P6n + P6m+4 �(6n + 6m + 5)�12n + 12m − 12nm − 7n + 17m + 6�, 2 2 P6n + P6m+5 �(n + m + 1)�72n + 72m − 72nm − 54n + 126m + 55�, 2 2 P6n+1 + P6m+1 �(2n + 2m + 1)�36n + 36m − 36nm + 9n + 9m + 2�, 2 2 P6n+1 + P6m+2 �(3n + 3m + 2)�24n + 24m − 24nm + 2n + 14m + 3�, 2 2 P6n+1 + P6m+3 �(6n + 6m + 5)�12n + 12m − 12nm − n + 11m + 3�, 2 2 P6n+1 + P6m+4 �(n + m + 1)�72n + 72m − 72nm − 18n + 90m + 31�, 2 2 P6n+1 + P6m+5 �(6n + 6m + 7)�12n + 12m − 12nm − 5n + 19m + 8�, 2 2 P6n+2 + P6m+2 �(6n + 6m + 5)�12n + 12m − 12nm + 5n + 5m + 2�, 2 2 P6n+2 + P6m+3 �(n + m + 1)�72n + 72m − 72nm + 18n + 54m + 19�, 2 2 P6n+2 + P6m+4 �(6n + 6m + 7)�12n + 12m − 12nm + n + 13m + 5�, 2 2 P6n+2 + P6m+5 �(3n + 3m + 4)�24n + 24m − 24nm − 2n + 34m + 15�, 2 2 P6n+3 + P6m+3 �(6n + 6m + 7)�12n + 12m − 12nm + 7n + 7m + 4�, Journal of Mathematics 3

2 2 P6n+3 + P6m+4 � (3n + 3m + 4)�24n + 24m − 24nm + 10n + 22m + 11�, 2 2 P6n+3 + P6m+5 � (2n + 2m + 3)�36n + 36m − 36nm + 9n + 45m + 23�, 2 2 P6n+4 + P6m+4 � (2n + 2m + 3)�36n + 36m − 36nm + 27n + 27m + 20�, (2) 2 2 P6n+4 + P6m+5 � (3n + 3m + 5)�24n + 24m − 24nm + 14n + 26m + 17�, 2 2 P6n+5 + P6m+5 � (6n + 6m + 11)�12n + 12m − 12nm + 11n + 11m + 10�. □ Lemma 1. Consider the pyramidal numbers sum of two positive octahedral numbers are 2 � O1 + O1 and Pn � (1/6)n(n + 1)(2n + 1). +en, for all n ∈ N, the following 7 � O1 + O2. identity follows by elementary algebra:

Pn(2n+3) + P(n+1)(2n− 1) � 2P2n(n+1). (3) Proof. Using the closed form of On along with the factorization n(2n2 + 1) + m(2m2 + 1) � (n + m)(2n2 + 2m2 − 2nm + 1) gives the identities Theorem 3. Consider the octahedral numbers 2 On � (1/3)n(2n + 1). +en, the only primes which are the

2 2 O3n + O3m �(n + m)�18n + 18m − 18nm + 1�, 2 2 O3n + O3m+1 �(3n + 3m + 1)�6n + 6m − 6nm − 2n + 4m + 1�, 2 2 O3n + O3m+2 �(3n + 3m + 2)�6n + 6m − 6nm − 4n + 8m + 3�, (4) 2 2 O3n+1 + O3m+1 �(3n + 3m + 2)�6n + 6m − 6nm + 2n + 2m + 1�, 2 2 O3n+1 + O3m+2 �(n + m + 1)�18n + 18m − 18nm + 18m + 7�, 2 2 O3n+2 + O3m+2 �(3n + 3m + 4)�6n + 6m − 6nm + 4n + 4m + 3�. □ Theorem 4. Consider the dodecahedral numbers Proof. Using the closed form of Dn along with the fac- Dn � (1/2)n(3n − 1)(3n − 2). +en, the only prime which is torization n(3n − 1)(3n − 2) + m(3m − 1)(3m − 2) � (3n + 2 2 the sum of two dodecahedral numbers is 2 � D1 + D1. 3m − 2)(3n + 3m − 3nm − n − m) gives the identities

2 2 D2n + D2m � 2(3n + 3m − 1)�6n + 6m − 6nm − n − m�, 2 2 D2n + D2m+1 � (6n + 6m + 1)�6n + 6m − 6nm − 4n + 5m + 1�, (5) 2 2 D2n+1 + D2m+1 � (3n + 3m + 2)�12n + 12m − 12nm + 4n + 4m + 1�. □ Theorem 5. Consider the icosahedral numbers In � (1/2) Proof. Using the closed form of In along with the factor- n(5n2 − 5n + 2). +en, primes which are the diﬀerence between ization n(5n2 − 5n + 2) − m(5m2 − 5m + 2) � (n − m)(5n2 2 two icosahedral numbers p � Ia − Ib occur only if a − b ≤ 2. +5m + 5nm − 5n − 5m + 2) gives the identities 4 Journal of Mathematics

2 2 I2n − I2m � 2(n − m)�10n + 10m + 10nm − 5n − 5m + 1�, 2 2 I2n − I2m+1 �(2n − 2m − 1)�10n + 10m + 10nm + 5m + 1�, (6) 2 2 I2n+1 − I2m+1 �(n − m)�20n + 20m + 20nm + 20n + 20m + 7�.

*e first several primes which are the difference between difference between two truncated tetrahedral numbers p � two icosahedral numbers are 11 � I2 − I1, 47 � I3 − I1, τa − τb occur only if a − b ≤ 6. 131 � I5 − I4, etc. □ Proof. Using the closed form of τn along with the factorization n(23n2 − 27n + 10) − m(23m2 − 27m + 10) � (n− Theorem 6. Consider the truncated tetrahedral numbers m)(23n2 + 23m2 + 23nm − 27n − 27m + 10) gives the 2 τn � (1/6)n(23n − 27n + 10). +en, primes which are the identities

2 2 τ6n − τ6m � 2(n − m)�414n + 414m + 414nm − 81n − 81m + 5�,

2 2 τ6n − τ6m+1 �(6n − 6m − 1)�138n + 138m + 138nm − 4n + 19m + 1�,

2 2 τ6n − τ6m+2 �(6n − 6m − 2)�138n + 138m + 138nm + 19n + 65m + 8�,

2 2 τ6n − τ6m+3 �(2n − 2m − 1)�414n + 414m + 414nm + 126n + 333m + 68�,

2 2 τ6n − τ6m+4 � 2(3n − 3m − 2)�138n + 138m + 138nm + 65n + 157m + 45�,

2 2 τ6n − τ6m+5 �(6n − 6m − 5)�138n + 138m + 138nm + 88n + 203m + 75�,

2 2 τ6n+1 − τ6m+1 �(n − m)�828n + 828m + 828nm + 252n + 252m + 25�,

2 2 τ6n+1 − τ6m+2 �(6n − 6m − 1)�138n + 138m + 138nm + 65n + 88m + 15�,

2 2 τ6n+1 − τ6m+3 �(3n − 3m − 1)�276n + 276m + 276nm + 176n + 268m + 67�,

2 2 τ6n+1 − τ6m+4 �(2n − 2m − 1)�414n + 414m + 414nm + 333n + 540m + 179�,

2 2 τ6n+1 − τ6m+5 �(3n − 3m − 2)�276n + 276m + 276nm + 268n + 452m + 187�, (7)

2 2 τ6n+2 − τ6m+2 � 2(n − m)�414n + 414m + 414nm + 333n + 333m + 89�,

2 2 τ6n+2 − τ6m+3 �(6n − 6m − 1)�138n + 138m + 138nm + 134n + 157m + 52�,

2 2 τ6n+2 − τ6m+4 � 2(3n − 3m − 1)�138n + 138m + 138nm + 157n + 203m + 82�,

2 2 τ6n+2 − τ6m+5 �(2n − 2m − 1)�414n + 414m + 414nm + 540n + 747m + 359�,

2 2 τ6n+3 − τ6m+3 �(n − m)�828n + 828m + 828nm + 1080n + 1080m + 469�,

2 2 τ6n+3 − τ6m+4 �(6n − 6m − 1)�138n + 138m + 138nm + 203n + 226m + 112�,

2 2 τ6n+3 − τ6m+5 �(3n − 3m − 1)�276n + 276m + 276nm + 452n + 544m + 307�,

2 2 τ6n+4 − τ6m+4 � 2(n − m)�414n + 414m + 414nm + 747n + 747m + 449�,

2 2 τ6n+4 − τ6m+5 �(6n − 6m − 1)�138n + 138m + 138nm + 272n + 295m + 195�,

2 2 τ6n+5 − τ6m+5 �(n − m)�828n + 828m + 828nm + 1908n + 1908m + 1465�. Journal of Mathematics 5

*e first several primes which are the difference between the difference between two truncated cube numbers p � Ka − two truncated tetrahedral numbers are 67 � τ3 − τ1, Kb occur only if a − b ≤ 3. 179 � τ4 − τ1, 359 � τ5 − τ2, etc. □ Proof. Using the closed form of Kn along with the factorization 77n3 + 69n2 + 19n − 77m3 − 69m2 − 19m � (n− Theorem 7. Consider the truncated cube numbers m)(77n2 + 77m2 + 77nm + 69n + 69m + 19) gives the K � (1/3)(77n3 + 69n2 + 19n + 3). +en, primes which are n identities

2 2 K3n − K3m �(n − m)�693n + 693m + 693nm + 207n + 207m + 19�, 2 2 K3n − K3m+1 �(3n − 3m − 1)�231n + 231m + 231nm + 146n + 223m + 55�, 2 2 K3n − K3m+2 �(3n − 3m − 2)�231n + 231m + 231nm + 223n + 377m + 155�, (8) 2 2 K3n+1 − K3m+1 �(n − m)�693n + 693m + 693nm + 900n + 900m + 388�, 2 2 K3n+1 − K3m+2 �(3n − 3m − 1)�231n + 231m + 231nm + 377n + 454m + 255�, 2 2 K3n+2 − K3m+2 �(n − m)�693n + 693m + 693nm + 1593n + 1593m + 1219�.

2 *e ﬁrst several primes which are the diﬀerence between Proof. Using πn � (1/2)n(3n − 3n + 2) along with the 2 2 two truncated cube numbers are 919 � K4 − K1, factorization n(3n − 3n + 2) − m(3m − 3m + 2) � (n− 2 2 1117 � K5 − K4, 14221 � K10 − K7, etc. □ m)(3n + 3m + 3nm − 3n − 3m + 2) gives the identities

Theorem 8. Consider the triakis tetrahedral numbers 2 πn � (1/2)n(3n − 3n + 2). +en, πa − πb is prime only if a − b ≤ 2.

2 2 π2n − π2m � 2(n − m)�6n + 6m + 6nm − 3n − 3m + 1�, 2 2 π2n − π2m+1 �(2n − 2m − 1)�6n + 6m + 6nm + 3m + 1�, (9) 2 2 π2n+1 − π2m+1 �(n − m)�12n + 12m + 12nm + 12n + 12m + 5�.

2 *e ﬁrst several primes which are the diﬀerence between Proof. Using σn � (1/2)n(7n − 9n + 4) along with the 2 2 two triakis tetrahedral numbers are 7 � π2 − π1, factorization n(7n − 9n + 4) − m(7m − 9m + 4) � (n − m) 2 2 29 � π3 − π1, 79 � π5 − π4, etc. □ (7n + 7m + 7nm − 9n − 9m + 4) gives the identities

Theorem 9. Consider the triakis octahedral numbers 2 σn � (1/2)n(7n − 9n + 4). +en, σa − σb is prime only if a − b ≤ 2.

2 2 σ2n − σ2m � 2(n − m)�14n + 14m + 14nm − 9n − 9m + 2�, 2 2 σ2n − σ2m+1 �(2n − 2m − 1)�14n + 14m + 14nm − 2n + 5m + 1�, (10) 2 2 σ2n+1 − σ2m+1 �(n − m)�28n + 28m + 28nm + 24n + 24m + 7�.

*e ﬁrst several primes which are the diﬀerence between Theorem 10. Consider the tetrakis hexahedral numbers 2 two triakis octahedral numbers are 13 � σ2 − σ1, θn � (1/3)n(10n − 12n + 5). +en, θa − θb is prime only if 59 � σ3 − σ1, 271 � σ5 − σ4, etc. □ a − b ≤ 3. 6 Journal of Mathematics

2 2 2 Proof. Using θn � (1/3)n(10n − 12n + 5) along with the (n − m)(10n + 10m + 10nm − 12n − 12m + 5) gives the factorization n(10n2 − 12n + 5) − m(10m2 − 12m + 5) � identities

2 2 θ3n − θ3m �(n − m)�90n + 90m + 90nm − 36n − 36m + 5�,

2 2 θ3n − θ3m+1 �(3n − 3m − 1)�30n + 30m + 30nm − 2n + 8m + 1�,

2 2 θ3n − θ3m+2 �(3n − 3m − 2)�30n + 30m + 30nm + 8n + 28m + 7�, (11) 2 2 θ3n+1 − θ3m+1 �(n − m)�90n + 90m + 90nm + 54n + 54m + 11�,

2 2 θ3n+1 − θ3m+2 �(3n − 3m − 1)�30n + 30m + 30nm + 28n + 38m + 13�,

2 2 θ3n+2 − θ3m+2 �(n − m)�90n + 90m + 90nm + 144n + 144m + 77�.

*e first several primes which are the difference between *e first several primes which are the difference between two tetrakis hexahedral numbers are 13 � θ2 − θ1, two triangular numbers are 2 � Δ2 − Δ1, 5 � Δ3 − Δ1, 97 � θ4 − θ3, 311 � θ5 − θ2, etc. □ 7 � Δ4 − Δ2, etc. □

Corollary 1. �a n is prime only if a − b � 1. Theorem 11. Consider the triangular numbers n�b Δn � (1/2)n(n + 1) with n ∈ N. +en, primes which are the n Proof. Using the well-known result �m�0 m � diﬀerence between two triangular numbers p � Δa − Δb occur a (1/2)n(n + 1) � Δn, it follows that �n�b n � Δa − Δb− 1, so by only if a − b ≤ 2. a the previous theorem, a prime of the form p � �n�b n exists only if a − (b − 1) ≤ 2⇒a − b � 1. □ Proof. Using Δn � (1/2)n(n + 1) along with the identity n(n + 1) − m(m + 1) � (n − m)(n + m + 1) gives the fol- Theorem 12. Consider the triakis icosahedral numbers 2 lowing three equations: cn � (1/2)n(19n − 27n + 10). +en, ca − cb is prime only if a − b ≤ 2. Δ2n − Δ2m �(n − m)(2n + 2m + 1), 2 Δ2n − Δ2m+1 �(n + m + 1)(2n − 2m − 1), (12) Proof. Using cn � (1/2)n(19n − 27n + 10) along with the factorization n(19n2 − 27n + 10) − m(19m2 − 27m + 10) Δ − Δ �(n − m)(2n + 2m + 3). 2n+1 2m+1 � (n − m)(19n2 + 19m2 + 19nm − 27n − 27m + 10) gives the identities

2 2 c2n − c2m � 2(n − m)�38n + 38m + 38nm − 27n − 27m + 5�, 2 2 c2n − c2m+1 �(2n − 2m − 1)�38n + 38m + 38nm − 8n + 11m + 1�, (13) 2 2 c2n+1 − c2m+1 �(n − m)�76n + 76m + 76nm + 60n + 60m + 13�.

*e ﬁrst several primes which are the diﬀerence between Theorem 13. Consider the pentakis dodecahedral numbers 2 two triakis icosahedral numbers are 31 � c2 − c1, ρn � (1/6)n(55n − 75n + 26). +en, ρa − ρb is prime only if 149 � c3 − c1, 463 � c5 − c4, etc. □ a − b ≤ 6. Journal of Mathematics 7

2 2 2 Proof. Using ρn � (1/6)n(55n − 75n + 26) along with the � (n − m)(55n + 55m + 55nm − 75n − 75m + 26) gives the factorization n(55n2 − 75n + 26) − m(55m2 − 75m + 26) identities

2 2 ρ6n − ρ6m � 2(n − m)�990n + 990m + 990nm − 225n − 225m + 13�, 2 2 ρ6n − ρ6m+1 �(6n − 6m − 1)�330n + 330m + 330nm − 20n + 35m + 1�, 2 2 ρ6n − ρ6m+2 � 2(3n − 3m − 1)�330n + 330m + 330nm + 35n + 145m + 16�, 2 2 ρ6n − ρ6m+3 �(2n − 2m − 1)�990n + 990m + 990nm + 270n + 765m + 148�, 2 2 ρ6n − ρ6m+4 � 2(3n − 3m − 2)�330n + 330m + 330nm + 145n + 365m + 101�, 2 2 ρ6n − ρ6m+5 �(6n − 6m − 5)�330n + 330m + 330nm + 200n + 475m + 171�, 2 2 ρ6n+1 − ρ6m+1 �(n − m)�1980n + 1980m + 1980nm + 540n + 540m + 41�, 2 2 ρ6n+1 − ρ6m+2 �(6n − 6m − 1)�330n + 330m + 330nm + 145n + 200m + 31�, 2 2 ρ6n+1 − ρ6m+3 �(3n − 3m − 1)�660n + 660m + 660nm + 400n + 620m + 147�, 2 2 ρ6n+1 − ρ6m+4 �(2n − 2m − 1)�990n + 990m + 990nm + 765n + 1260m + 403�, 2 2 ρ6n+1 − ρ6m+5 �(3n − 3m − 2)�660n + 660m + 660nm + 620n + 1060m + 427�, (14) 2 2 ρ6n+2 − ρ6m+2 � 2(n − m)�990n + 990m + 990nm + 765n + 765m + 193�, 2 2 ρ6n+2 − ρ6m+3 �(6n − 6m − 1)�330n + 330m + 330nm + 310n + 365m + 116�, 2 2 ρ6n+2 − ρ6m+4 � 2(3n − 3m − 1)�330n + 330m + 330nm + 365n + 475m + 186�, 2 2 ρ6n+2 − ρ6m+5 �(2n − 2m − 1)�990n + 990m + 990nm + 1260n + 1755m + 823�, 2 2 ρ6n+3 − ρ6m+3 �(n − m)�1980n + 1980m + 1980nm + 2520n + 2520m + 1061�, 2 2 ρ6n+3 − ρ6m+4 �(6n − 6m − 1)�330n + 330m + 330nm + 475n + 530m + 256�, 2 2 ρ6n+3 − ρ6m+5 �(3n − 3m − 1)�660n + 660m + 660nm + 1060n + 1280m + 707�, 2 2 ρ6n+4 − ρ6m+4 � 2(n − m)�990n + 990m + 990nm + 1755n + 1755m + 1033�, 2 2 ρ6n+4 − ρ6m+5 �(6n − 6m − 1)�330n + 330m + 330nm + 640n + 695m + 451�, 2 2 ρ6n+5 − ρ6m+5 �(n − m)�1980n + 1980m + 1980nm + 4500n + 4500m + 3401�.

2 *e ﬁrst several primes which are the diﬀerence between Proof. Using Λn � (1/3)n(55n − 75n + 23) along with the 2 2 two pentakis dodecahedral numbers are 31 � ρ2 − ρ1, factorization n(55n − 75n + 23) − m(55m − 75m + 23) 2 2 701 � ρ6 − ρ5, 823 � ρ5 − ρ2, etc. □ � (n − m)(55n + 55m + 55nm − 75n − 75m + 23) gives the identities Theorem 14. Consider the disdyakis triacontahedral num- 2 bers Λn � (1/3)n(55n − 75n + 23). +en, Λa − Λb is prime only if a − b ≤ 3.

2 2 Λ3n − Λ3m �(n − m)�495n + 495m + 495nm − 225n − 225m + 23�, 2 2 Λ3n − Λ3m+1 �(3n − 3m − 1)�165n + 165m + 165nm − 20n + 35m + 1�, 2 2 Λ3n − Λ3m+2 �(3n − 3m − 2)�165n + 165m + 165nm + 35n + 145m + 31�, (15) 2 2 Λ3n+1 − Λ3m+1 �(n − m)�495n + 495m + 495nm + 270n + 270m + 38�, 2 2 Λ3n+1 − Λ3m+2 �(3n − 3m − 1)�165n + 165m + 165nm + 145n + 200m + 61�, 2 2 Λ3n+2 − Λ3m+2 �(n − m)�495n + 495m + 495nm + 765n + 765m + 383�. 8 Journal of Mathematics

*e first several primes which are the difference between Case 2: z divides U or z divides V. If z divides U, then two disdyakis triacontahedral numbers are 61 � Λ2 − Λ1, we have Scenario 1. If z divides V, then the only values 2011 � Λ7 − Λ6, 6143 � Λ8 − Λ5, etc. □ of n and m which can make V/z equal to 1 correspond to n � m � 0 which indicates that a − b � x − y < z. Theorem 15. Consider the pentagonal numbers Case 3: some of the factors of z divide U and the rest of pn � (1/2)n(3n − 1). +en, pa − pb is prime only if a − b ≤ 2. the factors of z divide V. If this is true, then U will reduce into (z/gcd(z, x − y))(n − m)+ Proof (x − y/gcd(z, x − y)) which if set equal to 1 gives n � m + (1/z)(gcd(z, x − y) + y − x). *ese are integers only if (1/z)(gcd(z, x − y) + y − x) is an integer, and p2n − p2m �(n − m)(6n + 6m − 1), this can occur only if gcd(x, x − y) + y − x � 0. *is p2n − p2m+1 �(2n − 2m − 1)(3n + 3m + 1), (16) indicates that n � m, so a − b � x − y < z. Also, when V n � m � p2n+1 − p2m+1 �(n − m)(6n + 6m + 5). is reduced, then it can be 1 only if 0, so a − b � x − y < z. *e first several primes which are the difference between two pentagonal numbers are 7 � p3 − p2, 11 − p3 − p1, − p − p � p − p 11 3 117 4 2, etc. Scenario 3. x < y and n > m. If this is true, then there will be *ere is a universal generalization of some of these which the same three cases as in Scenario 2. is encompassed in the following several results. □ Case 1: this case was shown to be impossible.

Lemma 2. Let a0, a1, ... , aR ∈ R. +en, there exists the Case 2: this case is similar to case 2 in Scenario 2. following algebraic identity: Case 3: U will reduce into (z/gcd(z, y − x))(n − m) + R R R R (x − y/gcd(z, y − x)) which if set equal to 1 gives k k j ℓ− j n � akn − m � akm �(n − m) � n � aℓm . (17) n � m + (1/z)(gcd(z, x − y) + y − x). *ese are inte- k�0 k�0 j�0 ℓ�j gers only if (1/z)(gcd(z, y − x) + y − x) is an integer, Now, the universal result is ready to be given. and this can occur only if gcd(x, y − x) + y − x � 1. *is indicates that n � m + 1, so a − b � z + x − y < z. ∞ Also, when V is reduced, then it can be equal to 1 only if Theorem 16. Let �An �n�0 be a sequence of the form An � R k n � m � 0, but this will not happen since n > m. (1/z)(g + n �k�0 akn ) satisfying the following: All the polyhedra in this paper have associated poly- (1) z, g ∈ Z and a ∈ Z for all i � 0: R i hedral number sequences that are of the form referred to by z n R a nk (2) is the largest positive integer that divides �k�0 k the previous theorem. Many more R− dimensional polyhe- for all positive integers n dra not discussed here also have associated sequences that (3) An is a positive integer for all n ∈ N are of the type in *eorem 15. Many statements follow as an immediate result from +en, A − A is prime only if a − b ≤ z. a b n m *eorem 15, such as the statement � � − � � is prime k k Proof. Let x, y ∈ N such that 0 ≤ x < z and 0 ≤ y < z. *en, n p m p letting a � zn + x and b � zm + y in Lemma 2 gives only if n − m ≤ k!. Another one is that �k�0 k − �k�0 k is prime only if n − m ≤ Q, where Q is the reduced denomi- Aa − Ab � Azn+x − Azm+y � (1/z)UV, where U � zn − zm + R j R ℓ− j nator in the Faulhaber formula evaluated at a given positive x − y and V � �j�0 (zn + x) �ℓ�j aℓ(zm + y) . Now, there will be three scenarios. □ integer p. Another statement that follows from *eorem 15 is that if gn(k) denotes the nth polygonal number from a k g (k) � ( )n[nk − n − k + ] Scenario 1. x � y and n > m. If this is true, then z divides U polygon with sides n 1/2 2 4 , then g (k) − g (k) a − b � k a − as U/z � n − m which when set equal to 1 gives n � m + 1 a b is prime only if 1 if is even or b k which means a − b � z. ≤ 2 if is odd. One more great statement that follows from *eorem 15 is that if Pn(k) denotes the nth number from a polygonal pyramid from a polygon with k sides Scenario 2. x > y and n ≥ m. If this is true, there will be three P (k) � (1/6)n(kn2 − 2n2 + 3n − k + 5), then P (k) cases: n a − Pb(k) is prime only if a − b ≤ 6/gcd(3, k − 2, |5 − k|). Case 1: some of the factors of z divide U and some of *ere is also another result for a sequence which is a case the factors of z divide V but not all the factors of z of the sequences referred to in *eorem 15, but it has an- divide U or V. *is is impossible since it would imply other very striking property, as shown in the following that Aa − Ab is not an integer, but it clearly is an integer. theorem. Journal of Mathematics 9

∞ Theorem 17. Let �Bn �n�0 be a sequence of the form Bn � Case 1: z divides μ or z divides ]. If this is true, Ba + Bb R k (1/z)n �k�0 akn satisfying will factor into two positive polynomials either of which will be 1 only if n � m � 0, so there will be at most (1) z ∈ Z and ai ∈ Z for all i � 0: R finitely many primes of the form p � Ba + Bb. (2) z divides n �R a nk for all positive integers n k�0 k Case 2: some of the factors of z divide μ and the rest of (3) B− n � − Bn for all n ∈ Z the factors of z divide ]. If this is true, then like case 1, (4) Bn is a positive integer for all n ∈ N Ba + Bb will factor into two positive polynomials, either of them can be 1 only if n � m � 0, so there will be at +en, for each R ∈ N, there are at most finitely many most finitely many primes of the form p � Ba + Bb. primes of the form p � Ba + Bb. Case 3: some of the factors of z divide μ and some of the factors of z divide ] but not all the factors of z divide μ Proof. Replace m with − m in Lemma 2, and apply the R k or ]. *is case is impossible because it would imply that assumption that every exponent in n �k�0 akn is odd. R Ba + Bb is not an integer, but it clearly is an integer. *is will give the identity n �k�0 k R k R j R ℓ− j ℓ− j akn + m �k�0 akm � (n + m) �j�0 n �ℓ�j aℓ(− 1) m . Several more results include some other interesting So, now define positive integers x, y as 0 ≤ x < z and properties of some polyhedral numbers. 0 ≤ y < z, and let a � zn + x and b � zm + y. *en, *e following identity can be verified with elementary Ba + Bb � Bzn+x + Bzm+y � (1/z)μ], where μ � zn + zm + x algebra. □ R j R ℓ− j ℓ− j +y and ] � �j�0 (zn + x) �ℓ�j aℓ(− 1) (zm + y) . So, now there will be three cases: Lemma 3

a(a + b)(a + c) + d(c + d − b)(c + d) � (a + c + d)�a2 + d2 + ab + c d − a d − b d�. (18)

Using some cases of this lemma leads to the following a2 + b2 + ar + br − ab, then k is coprime to both Y and Z theorems. which means A(n, r, k) + A(m, r, k) is not an integer.

Theorem 18. Let C(n, r, k) � (1/k)n(n + r)(n + 2r) for Scenario 5. If there are some similar factors between k and positive integers n, r, and k. +en, for all fixed r and k, there a + b + 2r and the rest of the factors of k are similar to some are at most finitely many primes of the form factors of a2 + b2 + ar + br − ab, then (1/k)YZ would sim- p � C(n, r, k) + C(m, r, k). plify into the product of two polynomial factors, which means C(n, r, k) + C(m, r, k) runs through at most finitely Proof. First, note that from Lemma 3, it follows that n(n + many primes. r)(n + 2r) + m(m + r) 2 2 (m + 2r) � (n + m + 2r)(n + m + 2rm + nr − nm − rm). Scenario 6. If there are some similar factors between k and Now, for two positive integers a and b, let n � kn′ + a a + b + 2r or some similar factors between k and a2 + b2 + and m � km′ + b. ar + br − ab but not all the factors of k are similar to the *en, C(n, r, k) + C(m, r, k) � C(kn′ + a, r, k) + C(k factors of both a + b + 2r and a2 + b2 + ar + br − ab, then m′ + b, r, k) � (1/k)YZ, where Y � kn′ + km′ + a + b + 2r ( k)YZ ( k′)Y′Z′ k′ 2 2 2 2 2 2 1/ will reduce into 1/ , but is coprime to and Z � a +b + ar + br − ab + k n′ + k m′ + kn′r + km′r Y′ Z′ C(n, r, k) + C(m, r, k) 2 both and which means is not an +2kn′a + 2km′b − k n′m′ − kn′b − km′a. Now, there will be integer. two cases: From this, it follows that if C(n, r, k) + C(m, r, k) is an Case 1: k divides a + b + 2r or k divides integer for some fixed r and k, it must factor into two a2 + b2 + ar + br − ab. *en, from this, k divides either positive polynomials, so there can only be finitely many Y or Z, so C(n, r, k) + C(m, r, k) runs through at most primes of the form p � C(n, r, k) + C(m, r, k). finitely many primes.

Case 2: k does not divide a + b + 2r and k does not Corollary 2. Consider the numbers Fn � (1/4)n(n + 1)(n + 2 2 divide a + b + ar + br − ab. 2) for n ∈ N. +en, the only prime of the form p � Fn + Fm is 3 � F + F . In this case, there are three scenarios. □ 1 1

Scenario 4. If there are no similar factors between k and a + Proof. Letting r � 1 and k � 4 in *eorem 17 gives the b + 2r and there are no similar factors between k and identities 10 Journal of Mathematics

2 2 F4n + F4m � 2(2n + 2m + 1)�4n + 4m − 4nm + n + m�,

1 F + F � (4n + 4m + 3)�8n2 + 8m2 − 8nm + 6m + 1�, 4n 4m+1 2

2 2 F4n + F4m+2 �(n + m + 1)�16n + 16m − 16nm − 4n + 20m + 6�,

2 2 F4n + F4m+3 �(4n + 4m + 5)�4n + 4m − 4nm − 2n + 7m + 3�,

2 2 F4n+1 + F4m+1 �(n + m + 1)�16n + 16m − 16nm + 8n + 8m + 3�, (19) 1 F + F � (4n + 4m + 5)�8n2 + 8m2 − 8nm + 2n + 8m + 3�, 4n+1 4m+2 2 1 F + F � (2n + 2m + 3)�16n2 + 16m2 − 16nm + 24m + 11�, 4n+1 4m+3 2

2 2 F4n+2 + F4m+2 � 2(2n + 2m + 3)�4n + 4m − 4nm + 3n + 3m + 2�,

2 2 F4n+2 + F4m+3 �(4n + 4m + 7)�4n + 4m − 4nm + 2n + 5m + 3�,

2 2 F4n+3 + F4m+3 �(n + m + 2)�16n + 16m − 16nm + 16n + 16m + 15�. □ Theorem 19. Consider the triangular numbers Proof. Let G(n, m, k) � (n + k + 1)Δn + mΔm+k. Using Δn � Δn � n(n + 1)/2, and let n, m, and k be nonnegative integers. n(n + 1)/2 and the fact that n(n + 1)(n + k + 1)+ m(m + +en, the only primes that are of the form k)(m+ k + 1) � (n + m + k + 1)(n2 + m2 − nm + n + km), it p � (n + k + 1)Δn + mΔm+k are 2 � (1 + 0 + 1)Δ1 + 0Δ0+0 is easy to verify the following identities: and 3 � (1 + 1 + 1)Δ1 + 0Δ0+1 � (1 + 0 + 1)Δ1 + 1Δ1+0 � (0 + 1 + 1)Δ0 + 1Δ2.

G(2n, 2m, 2k) �(2n + 2m + 2k + 1)�2n2 + 2m2 − 2nm + 2mk + n�, G(2n, 2m, 2k + 1) � 2(n + m + k + 1)�2n2 + 2m2 − 2nm + 2mk + n + m�, G(2n, 2m + 1, 2k) �(n + m + k + 1)�4n2 + 4m2 − 4nm + 4mk + 4m + 2k + 1�, G(2n, 2m + 1, 2k + 1) �(2n + 2m + 2k + 3)�2n2 + 2m2 − 2nm + 2mk + 3m + k + 1�, (20) G(2n + 1, 2m, 2k) � 2(n + m + k + 1)�2n2 + 2m2 − 2nm + 2mk + 3n − m + 1�, G(2n + 1, 2m, 2k + 1) �(2n + 2m + 2k + 3)�2n2 + 2m2 − 2nm + 2mk + 3n + 1�, G(2n + 1, 2m + 1, 2k) �(2n + 2m + 2k + 3)�2n2 + 2m2 − 2nm + 2mk + 2n + m + k + 1�, G(2n + 1, 2m + 1, 2k + 1) �(n + m + k + 2)�4n2 + 4m2 − 4nm + 4mk + 4n + 4m + 2k + 3�. □ Lemma 4. Consider the tetrahedral numbers Theorem 20. Still using Tn � (1/6)n(n + 1)(n + 2), there Tn � (1/6)n(n + 1)(n + 2). +en, for all n ∈ N, the following exists the identity identity can be shown with elementary algebra: n n+m 1 � (− 1) Tm(m+ ) � Tn(n+ ). (22) Tn(n+ ) + T(n+ )(n+ ) � 2T(n+ )(n+ ). (21) 2 3 3 1 4 1 3 m�1 2 Journal of Mathematics 11

Proof. From Tn(n+3) + T(n− 1)(n+2) � 2Tn(n+2), we can as- Acknowledgments semble the list of equations: All funding invested in the article is provided by the author. Tn(n+3) + T(n− 1)(n+2) � 2Tn(n+2),

− T(n− 1)(n+2) − T(n− 2)(n+1) � − 2T(n− 1)(n+1), References

T(n− 2)(n+1) + T(n− 3)n � 2T(n− 2)n, (23) [1] E. Deza and M. Deza, Figurate Numbers, vol. 93, World Scientific Publishing, Singapore, 2012. − T(n− 3)n − T(n− 4)(n− 1) � − 2T(n− 3)(n− 1), [2] L. Euler, “Regula facilis problemata Diophantea per numeros (− 1)n+1T � 2(− 1)n+1T . integros expedite resolvendi (an easy rule for Diophantine (1)(4) (1)(3) problems which are to be resolved quickly by integral num- n n+m bers),” Memoires de l’Academie des Sciences de St.-Petersbourg, Adding up all these gives 2 �m�1 (− 1) Tm(m+2) � T vol. 4, no. 3, p. 17, 1813. n(n+3). □ [3] Nathanson and B. Melvyn, Additive Number +eory +e Classical Bases, Springer, vol. 164, Berlin, Germany, 1996. 2. Conclusion [4] W. Sierpinski, “On tetrahedral numbers (in Polish),” pp. 209–217, 1964. We now have a result to determine very easily whether the [5] T. Pappas, +e Five Platonic Solids, +e Joy of Mathematics, difference between two elements from a polyhedral number pp. 110-111, Wide World Publishing, San Carlos, CA, USA, sequence may be eligible to be a prime. We also see that the 1989. only platonic number sequence [5] in which it is unknown whether there are infinitely primes as the sum of two elements is the icosahedral number. *ere exist some additional following identities relating some of these sequences to each other. Let Δn � (1/2)n(n + 1) denote the nth triangular number, Tn � (1/6)n(n + 1) (n + 2) denote the nth tetrahedral number, 2 On � (1/3)n(2n + 1) denote the nth octahedral number, and Pn � (1/6)n(n + 1)(2n + 1) denote the nth pyramidal number.

Lemma 5 2n + 1 2n + 4 � � +(n + 1)4 �� + n4. (24) 4 4

Lemma 6

Tn+2 + Pn+1 � Tn + Pn+2. (25)

Lemma 7 4 Tn2− 2 + n � Tn2 . (26)

Lemma 8

T 2 + O � T 2 + O . ( ) (n+2) Δn (n+1) Δn+2 27

Data Availability *e mathematical data used to support the ﬁndings of this study are included within the article.

Conflicts of Interest *e author declares that there are no conﬂicts of interest.