# Primes and Their Connection to Certain Polyhedral Number Sequences

Authors: Benjamin Lee Warren
Journal: Journal of Mathematics
 (2021)

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## Abstract

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.


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## Body

## 1. Introduction

The sheer amount of various polyhedral figurate number sequences [1] makes studying them all a daunting task. Fortunately, some universal questions can be asked about them, though not all the answers are known. For instance, it is due to Euler [2] that we can find all numbers that are simultaneously triangular and square, which marks the general question of which numbers exist in more than one figurate number sequence at a time (for example, 36 is both triangular and square). Another significant result is the general question suggested by Fermat, Lagrange, Waring, Gauss, and Cauchy, which is to determine the minimal number of elements required from a given figurate sequence in order to sum to all natural numbers [3]. Though these are elegant and natural questions, few answers have so far been determined.In this paper, another question will be examined and, very fortunately, answered for these types of sequences in general, which is as follows: when might a prime occur as the difference between two elements from some chosen figurate sequence? The answer, along with some other results, is presented..Theorem 1.
Consider the tetrahedral numbers [4] Tn=1/6nn+1n+2. Then, the only primes which are the sum of two tetrahedral numbers are 2=T1+T1,5=T1+T2, and 11=T1+T3.Proof.
Using the closed form ofTn along with the factorization nn+1n+2+mm+1m+2=n+m+2n2+m2+n+m−nm gives the identities(1)T6n+T6m=23n+3m+16n2+6m2+n+m−6nm,T6n+T6m+1=2n+2m+118n2+18m2−18nm+9m+1,T6n+T6m+2=23n+3m+26n2+6m2−6nm−n+5m+1,T6n+T6m+3=6n+6m+56n2+6m2−6nm−2n+7m+2,T6n+T6m+4=2n+m+118n2+18m2−18nm−9n+27m+10,T6n+T6m+5=6n+6m+76n2+6m2−6nm−4n+11m+5,T6n+1+T6m+1=3n+3m+212n2+12m2−12nm+4n+4m+1,T6n+1+T6m+2=6n+6m+56n2+6m2−6nm+n+4m+1,T6n+1+T6m+3=n+m+136n2+36m2−36nm+36m+11,T6n+1+T6m+4=6n+6m+76n2+6m2−6nm−n+8m+3,T6n+1+T6m+5=3n+3m+412n2+12m2−12nm−4n+20m+9,T6n+2+T6m+2=2n+m+118n2+18m2−18nm+9n+9m+4,T6n+2+T6m+3=6n+6m+76n2+6m2−6nm+2n+5m+2,T6n+2+T6m+4=23n+3m+46n2+6m2−6nm+n+7m+3,T6n+2+T6m+5=2n+2m+318n2+18m2−18nm+27m+13,T6n+3+T6m+3=3n+3m+412n2+12m2−12nm+8n+8m+5,T6n+3+T6m+4=2n+2m+318n2+18m2−18nm+9n+18m+10,T6n+3+T6m+5=3n+3m+512n2+12m2−12nm+4n+16m+9,T6n+4+T6m+4=23n+3m+56n2+6m2−6nm+5n+5m+4,T6n+4+T6m+5=6n+6m+116n2+6m2−6nm+4n+7m+5,T6n+5+T6m+5=n+m+236n2+36m2−36nm+36n+36m+35.Theorem 2.
Consider the pyramidal numbersPn=∑k=1nk2=1/6nn+12n+1. Then, the only primes which are the sum of two positive pyramidal numbers are 2=P1+P1,19=P2+P3, and 31=P1+P4.Proof.
Using the closed form ofPn along with the factorization nn+12n+1+mm+12m+1=n+m+12n2+2m2−2nm+n+m gives the identities(2)P6n+P6m=6n+6m+112n2+12m2−12nm+n+m,P6n+P6m+1=3n+3m+124n2+24m2−24nm−2n+10m+1,P6n+P6m+2=2n+2m+136n2+36m2−36nm−9n+27m+5,P6n+P6m+3=3n+3m+224n2+24m2−24nm−10n+26m+7,P6n+P6m+4=6n+6m+512n2+12m2−12nm−7n+17m+6,P6n+P6m+5=n+m+172n2+72m2−72nm−54n+126m+55,P6n+1+P6m+1=2n+2m+136n2+36m2−36nm+9n+9m+2,P6n+1+P6m+2=3n+3m+224n2+24m2−24nm+2n+14m+3,P6n+1+P6m+3=6n+6m+512n2+12m2−12nm−n+11m+3,P6n+1+P6m+4=n+m+172n2+72m2−72nm−18n+90m+31,P6n+1+P6m+5=6n+6m+712n2+12m2−12nm−5n+19m+8,P6n+2+P6m+2=6n+6m+512n2+12m2−12nm+5n+5m+2,P6n+2+P6m+3=n+m+172n2+72m2−72nm+18n+54m+19,P6n+2+P6m+4=6n+6m+712n2+12m2−12nm+n+13m+5,P6n+2+P6m+5=3n+3m+424n2+24m2−24nm−2n+34m+15,P6n+3+P6m+3=6n+6m+712n2+12m2−12nm+7n+7m+4,P6n+3+P6m+4=3n+3m+424n2+24m2−24nm+10n+22m+11,P6n+3+P6m+5=2n+2m+336n2+36m2−36nm+9n+45m+23,P6n+4+P6m+4=2n+2m+336n2+36m2−36nm+27n+27m+20,P6n+4+P6m+5=3n+3m+524n2+24m2−24nm+14n+26m+17,P6n+5+P6m+5=6n+6m+1112n2+12m2−12nm+11n+11m+10.Lemma 1.
Consider the pyramidal numbersPn=1/6nn+12n+1. Then, for all n∈ℕ, the following identity follows by elementary algebra:(3)Pn2n+3+Pn+12n−1=2P2nn+1.Theorem 3.
Consider the octahedral numbersOn=1/3n2n2+1. Then, the only primes which are the sum of two positive octahedral numbers are 2=O1+O1 and 7=O1+O2.Proof.
Using the closed form ofOn along with the factorization n2n2+1+m2m2+1=n+m2n2+2m2−2nm+1 gives the identities(4)O3n+O3m=n+m18n2+18m2−18nm+1,O3n+O3m+1=3n+3m+16n2+6m2−6nm−2n+4m+1,O3n+O3m+2=3n+3m+26n2+6m2−6nm−4n+8m+3,O3n+1+O3m+1=3n+3m+26n2+6m2−6nm+2n+2m+1,O3n+1+O3m+2=n+m+118n2+18m2−18nm+18m+7,O3n+2+O3m+2=3n+3m+46n2+6m2−6nm+4n+4m+3.Theorem 4.
Consider the dodecahedral numbersDn=1/2n3n−13n−2. Then, the only prime which is the sum of two dodecahedral numbers is 2=D1+D1.Proof.
Using the closed form ofDn along with the factorization n3n−13n−2+m3m−13m−2=3n+3m−23n2+3m2−3nm−n−m gives the identities(5)D2n+D2m=23n+3m−16n2+6m2−6nm−n−m,D2n+D2m+1=6n+6m+16n2+6m2−6nm−4n+5m+1,D2n+1+D2m+1=3n+3m+212n2+12m2−12nm+4n+4m+1.Theorem 5.
Consider the icosahedral numbersIn=1/2n5n2−5n+2. Then, primes which are the difference between two icosahedral numbers p=Ia−Ib occur only if a−b≤2.Proof.
Using the closed form ofIn along with the factorization n5n2−5n+2−m5m2−5m+2=n−m5n2+5m2+5nm−5n−5m+2 gives the identities(6)I2n−I2m=2n−m10n2+10m2+10nm−5n−5m+1,I2n−I2m+1=2n−2m−110n2+10m2+10nm+5m+1,I2n+1−I2m+1=n−m20n2+20m2+20nm+20n+20m+7.
The first several primes which are the difference between two icosahedral numbers are11=I2−I1, 47=I3−I1, 131=I5−I4, etc.Theorem 6.
Consider the truncated tetrahedral numbersτn=1/6n23n2−27n+10. Then, primes which are the difference between two truncated tetrahedral numbers p=τa−τb occur only if a−b≤6.Proof.
Using the closed form ofτn along with the factorization n23n2−27n+10−m23m2−27m+10=n−m23n2+23m2+23nm−27n−27m+10 gives the identities(7)τ6n−τ6m=2n−m414n2+414m2+414nm−81n−81m+5,τ6n−τ6m+1=6n−6m−1138n2+138m2+138nm−4n+19m+1,τ6n−τ6m+2=6n−6m−2138n2+138m2+138nm+19n+65m+8,τ6n−τ6m+3=2n−2m−1414n2+414m2+414nm+126n+333m+68,τ6n−τ6m+4=23n−3m−2138n2+138m2+138nm+65n+157m+45,τ6n−τ6m+5=6n−6m−5138n2+138m2+138nm+88n+203m+75,τ6n+1−τ6m+1=n−m828n2+828m2+828nm+252n+252m+25,τ6n+1−τ6m+2=6n−6m−1138n2+138m2+138nm+65n+88m+15,τ6n+1−τ6m+3=3n−3m−1276n2+276m2+276nm+176n+268m+67,τ6n+1−τ6m+4=2n−2m−1414n2+414m2+414nm+333n+540m+179,τ6n+1−τ6m+5=3n−3m−2276n2+276m2+276nm+268n+452m+187,τ6n+2−τ6m+2=2n−m414n2+414m2+414nm+333n+333m+89,τ6n+2−τ6m+3=6n−6m−1138n2+138m2+138nm+134n+157m+52,τ6n+2−τ6m+4=23n−3m−1138n2+138m2+138nm+157n+203m+82,τ6n+2−τ6m+5=2n−2m−1414n2+414m2+414nm+540n+747m+359,τ6n+3−τ6m+3=n−m828n2+828m2+828nm+1080n+1080m+469,τ6n+3−τ6m+4=6n−6m−1138n2+138m2+138nm+203n+226m+112,τ6n+3−τ6m+5=3n−3m−1276n2+276m2+276nm+452n+544m+307,τ6n+4−τ6m+4=2n−m414n2+414m2+414nm+747n+747m+449,τ6n+4−τ6m+5=6n−6m−1138n2+138m2+138nm+272n+295m+195,τ6n+5−τ6m+5=n−m828n2+828m2+828nm+1908n+1908m+1465.
The first several primes which are the difference between two truncated tetrahedral numbers are67=τ3−τ1, 179=τ4−τ1, 359=τ5−τ2, etc.Theorem 7.
Consider the truncated cube numbersKn=1/377n3+69n2+19n+3. Then, primes which are the difference between two truncated cube numbers p=Ka−Kb occur only if a−b≤3.Proof.
Using the closed form ofKn along with the factorization 77n3+69n2+19n−77m3−69m2−19m=n−m77n2+77m2+77nm+69n+69m+19 gives the identities(8)K3n−K3m=n−m693n2+693m2+693nm+207n+207m+19,K3n−K3m+1=3n−3m−1231n2+231m2+231nm+146n+223m+55,K3n−K3m+2=3n−3m−2231n2+231m2+231nm+223n+377m+155,K3n+1−K3m+1=n−m693n2+693m2+693nm+900n+900m+388,K3n+1−K3m+2=3n−3m−1231n2+231m2+231nm+377n+454m+255,K3n+2−K3m+2=n−m693n2+693m2+693nm+1593n+1593m+1219.
The first several primes which are the difference between two truncated cube numbers are919=K4−K1, 1117=K5−K4, 14221=K10−K7, etc.Theorem 8.
Consider the triakis tetrahedral numbersπn=1/2n3n2−3n+2. Then, πa−πb is prime only if a−b≤2.Proof.
Usingπn=1/2n3n2−3n+2 along with the factorization n3n2−3n+2−m3m2−3m+2=n−m3n2+3m2+3nm−3n−3m+2 gives the identities(9)π2n−π2m=2n−m6n2+6m2+6nm−3n−3m+1,π2n−π2m+1=2n−2m−16n2+6m2+6nm+3m+1,π2n+1−π2m+1=n−m12n2+12m2+12nm+12n+12m+5.
The first several primes which are the difference between two triakis tetrahedral numbers are7=π2−π1, 29=π3−π1, 79=π5−π4, etc.Theorem 9.
Consider the triakis octahedral numbersσn=1/2n7n2−9n+4. Then, σa−σb is prime only if a−b≤2.Proof.
Usingσn=1/2n7n2−9n+4 along with the factorization n7n2−9n+4−m7m2−9m+4=n−m7n2+7m2+7nm−9n−9m+4 gives the identities(10)σ2n−σ2m=2n−m14n2+14m2+14nm−9n−9m+2,σ2n−σ2m+1=2n−2m−114n2+14m2+14nm−2n+5m+1,σ2n+1−σ2m+1=n−m28n2+28m2+28nm+24n+24m+7.
The first several primes which are the difference between two triakis octahedral numbers are13=σ2−σ1, 59=σ3−σ1, 271=σ5−σ4, etc.Theorem 10.
Consider the tetrakis hexahedral numbersθn=1/3n10n2−12n+5. Then, θa−θb is prime only if a−b≤3.Proof.
Usingθn=1/3n10n2−12n+5 along with the factorization n10n2−12n+5−m10m2−12m+5=n−m10n2+10m2+10nm−12n−12m+5 gives the identities(11)θ3n−θ3m=n−m90n2+90m2+90nm−36n−36m+5,θ3n−θ3m+1=3n−3m−130n2+30m2+30nm−2n+8m+1,θ3n−θ3m+2=3n−3m−230n2+30m2+30nm+8n+28m+7,θ3n+1−θ3m+1=n−m90n2+90m2+90nm+54n+54m+11,θ3n+1−θ3m+2=3n−3m−130n2+30m2+30nm+28n+38m+13,θ3n+2−θ3m+2=n−m90n2+90m2+90nm+144n+144m+77.
The first several primes which are the difference between two tetrakis hexahedral numbers are13=θ2−θ1, 97=θ4−θ3, 311=θ5−θ2, etc.Theorem 11.
Consider the triangular numbersΔn=1/2nn+1 with n∈ℕ. Then, primes which are the difference between two triangular numbers p=Δa−Δb occur only if a−b≤2.Proof.
UsingΔn=1/2nn+1 along with the identity nn+1−mm+1=n−mn+m+1 gives the following three equations:(12)Δ2n−Δ2m=n−m2n+2m+1,Δ2n−Δ2m+1=n+m+12n−2m−1,Δ2n+1−Δ2m+1=n−m2n+2m+3.
The first several primes which are the difference between two triangular numbers are2=Δ2−Δ1, 5=Δ3−Δ1, 7=Δ4−Δ2, etc.Corollary 1.
∑n=ban is prime only if a−b=1.Proof.
Using the well-known result∑m=0nm=1/2nn+1=Δn, it follows that ∑n=ban=Δa−Δb−1, so by the previous theorem, a prime of the form p=∑n=ban exists only if a−b−1≤2⇒a−b=1.Theorem 12.
Consider the triakis icosahedral numbersγn=1/2n19n2−27n+10. Then, γa−γb is prime only if a−b≤2.Proof.
Usingγn=1/2n19n2−27n+10 along with the factorization n19n2−27n+10−m19m2−27m+10=n−m19n2+19m2+19nm−27n−27m+10 gives the identities(13)γ2n−γ2m=2n−m38n2+38m2+38nm−27n−27m+5,γ2n−γ2m+1=2n−2m−138n2+38m2+38nm−8n+11m+1,γ2n+1−γ2m+1=n−m76n2+76m2+76nm+60n+60m+13.
The first several primes which are the difference between two triakis icosahedral numbers are31=γ2−γ1, 149=γ3−γ1, 463=γ5−γ4, etc.Theorem 13.
Consider the pentakis dodecahedral numbersρn=1/6n55n2−75n+26. Then, ρa−ρb is prime only if a−b≤6.Proof.
Usingρn=1/6n55n2−75n+26 along with the factorization n55n2−75n+26−m55m2−75m+26=n−m55n2+55m2+55nm−75n−75m+26 gives the identities(14)ρ6n−ρ6m=2n−m990n2+990m2+990nm−225n−225m+13,ρ6n−ρ6m+1=6n−6m−1330n2+330m2+330nm−20n+35m+1,ρ6n−ρ6m+2=23n−3m−1330n2+330m2+330nm+35n+145m+16,ρ6n−ρ6m+3=2n−2m−1990n2+990m2+990nm+270n+765m+148,ρ6n−ρ6m+4=23n−3m−2330n2+330m2+330nm+145n+365m+101,ρ6n−ρ6m+5=6n−6m−5330n2+330m2+330nm+200n+475m+171,ρ6n+1−ρ6m+1=n−m1980n2+1980m2+1980nm+540n+540m+41,ρ6n+1−ρ6m+2=6n−6m−1330n2+330m2+330nm+145n+200m+31,ρ6n+1−ρ6m+3=3n−3m−1660n2+660m2+660nm+400n+620m+147,ρ6n+1−ρ6m+4=2n−2m−1990n2+990m2+990nm+765n+1260m+403,ρ6n+1−ρ6m+5=3n−3m−2660n2+660m2+660nm+620n+1060m+427,ρ6n+2−ρ6m+2=2n−m990n2+990m2+990nm+765n+765m+193,ρ6n+2−ρ6m+3=6n−6m−1330n2+330m2+330nm+310n+365m+116,ρ6n+2−ρ6m+4=23n−3m−1330n2+330m2+330nm+365n+475m+186,ρ6n+2−ρ6m+5=2n−2m−1990n2+990m2+990nm+1260n+1755m+823,ρ6n+3−ρ6m+3=n−m1980n2+1980m2+1980nm+2520n+2520m+1061,ρ6n+3−ρ6m+4=6n−6m−1330n2+330m2+330nm+475n+530m+256,ρ6n+3−ρ6m+5=3n−3m−1660n2+660m2+660nm+1060n+1280m+707,ρ6n+4−ρ6m+4=2n−m990n2+990m2+990nm+1755n+1755m+1033,ρ6n+4−ρ6m+5=6n−6m−1330n2+330m2+330nm+640n+695m+451,ρ6n+5−ρ6m+5=n−m1980n2+1980m2+1980nm+4500n+4500m+3401.
The first several primes which are the difference between two pentakis dodecahedral numbers are31=ρ2−ρ1, 701=ρ6−ρ5, 823=ρ5−ρ2, etc.Theorem 14.
Consider the disdyakis triacontahedral numbersΛn=1/3n55n2−75n+23. Then, Λa−Λb is prime only if a−b≤3.Proof.
UsingΛn=1/3n55n2−75n+23 along with the factorization n55n2−75n+23−m55m2−75m+23=n−m55n2+55m2+55nm−75n−75m+23 gives the identities(15)Λ3n−Λ3m=n−m495n2+495m2+495nm−225n−225m+23,Λ3n−Λ3m+1=3n−3m−1165n2+165m2+165nm−20n+35m+1,Λ3n−Λ3m+2=3n−3m−2165n2+165m2+165nm+35n+145m+31,Λ3n+1−Λ3m+1=n−m495n2+495m2+495nm+270n+270m+38,Λ3n+1−Λ3m+2=3n−3m−1165n2+165m2+165nm+145n+200m+61,Λ3n+2−Λ3m+2=n−m495n2+495m2+495nm+765n+765m+383.
The first several primes which are the difference between two disdyakis triacontahedral numbers are61=Λ2−Λ1, 2011=Λ7−Λ6, 6143=Λ8−Λ5, etc.Theorem 15.
Consider the pentagonal numberspn=1/2n3n−1. Then, pa−pb is prime only if a−b≤2.Proof.
(16)p2n−p2m=n−m6n+6m−1,p2n−p2m+1=2n−2m−13n+3m+1,p2n+1−p2m+1=n−m6n+6m+5.
The first several primes which are the difference between two pentagonal numbers are7=p3−p2, 11−p3−p1, 11−p3−p117=p4−p2, etc.
There is a universal generalization of some of these which is encompassed in the following several results.Lemma 2.
Leta0,a1,…,aR∈ℝ. Then, there exists the following algebraic identity:(17)n∑k=0Raknk−m∑k=0Rakmk=n−m∑j=0Rnj∑ℓ=jRaℓmℓ−j.
Now, the universal result is ready to be given.Theorem 16.
LetAnn=0∞ be a sequence of the form An=1/zg+n∑k=0Raknk satisfying the following:(1)
z,g∈ℤ and ai∈ℤ for all i=0:R(2)
z is the largest positive integer that divides n∑k=0Raknk for all positive integers n(3)
An is a positive integer for all n∈ℕ
Then,Aa−Ab is prime only if a−b≤z.Proof.
Letx,y∈ℕ such that 0≤x<z and 0≤y<z. Then, letting a=zn+x and b=zm+y in Lemma 2 gives Aa−Ab=Azn+x−Azm+y=1/zUV, where U=zn−zm+x−y and V=∑j=0Rzn+xj∑ℓ=jRaℓzm+yℓ−j. Now, there will be three scenarios.Scenario 1.
x=y and n>m. If this is true, then z divides U as U/z=n−m which when set equal to 1 gives n=m+1 which means a−b=z.Scenario 2.
x>y and n≥m. If this is true, there will be three cases:Case 1: some of the factors ofz divide U and some of the factors of z divide V but not all the factors of z divide U or V. This is impossible since it would imply that Aa−Ab is not an integer, but it clearly is an integer.Case 2:z divides U or z divides V. If z divides U, then we have Scenario 1. If z divides V, then the only values 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.Case 3: some of the factors ofz divide U and the rest of the factors of z divide V. If this is true, then U will reduce into z/gcdz,x−yn−m+x−y/gcdz,x−y which if set equal to 1 gives n=m+1/zgcdz,x−y+y−x. These are integers only if 1/zgcdz,x−y+y−x is an integer, and this can occur only if gcdx,x−y+y−x=0. This indicates that n=m, so a−b=x−y<z. Also, when V is reduced, then it can be 1 only if n=m=0, so a−b=x−y<z.Scenario 3.
x<y and n>m. If this is true, then there will be the same three cases as in Scenario 2.Case 1: this case was shown to be impossible.Case 2: this case is similar to case 2 in Scenario2.Case 3:U will reduce into z/gcdz,y−xn−m+x−y/gcdz,y−x which if set equal to 1 gives n=m+1/zgcdz,x−y+y−x. These are integers only if 1/zgcdz,y−x+y−x is an integer, and this can occur only if gcdx,y−x+y−x=1. This 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 n=m=0, but this will not happen since n>m.
All the polyhedra in this paper have associated polyhedral number sequences that are of the form referred to by the previous theorem. Many moreR−dimensional polyhedra not discussed here also have associated sequences that are of the type in Theorem 15.
Many statements follow as an immediate result from Theorem15, such as the statement nk−mk is prime only if n−m≤k!. Another one is that ∑k=0nkp−∑k=0mkp is prime only if n−m≤Q, where Q is the reduced denominator in the Faulhaber formula evaluated at a given positive integer p. Another statement that follows from Theorem 15 is that if gnk denotes the nth polygonal number from a polygon with k sides gnk=1/2nnk−2n−k+4, then gak−gbk is prime only if a−b=1 if k is even or a−b≤2 if k is odd. One more great statement that follows from Theorem 15 is that if Pnk denotes the nth number from a polygonal pyramid from a polygon with k sides Pnk=1/6nkn2−2n2+3n−k+5, then Pak−Pbk is prime only if a−b≤6/gcd3,k−2,5−k.
There is also another result for a sequence which is a case of the sequences referred to in Theorem15, but it has another very striking property, as shown in the following theorem.Theorem 17.
LetBnn=0∞ be a sequence of the form Bn=1/zn∑k=0Raknk satisfying(1)
z∈ℤ and ai∈ℤ for all i=0:R(2)
z divides n∑k=0Raknk for all positive integers n(3)
B−n=−Bn for all n∈ℤ(4)
Bn is a positive integer for all n∈ℕ
Then, for eachR∈ℕ, there are at most finitely many primes of the form p=Ba+Bb.Proof.
Replacem with −m in Lemma 2, and apply the assumption that every exponent in n∑k=0Raknk is odd.
This will give the identityn∑k=0Raknk+m∑k=0Rakmk=n+m∑j=0Rnj∑ℓ=jRaℓ−1ℓ−jmℓ−j.
So, now define positive integersx,y as 0≤x<z and 0≤y<z, and let a=zn+x and b=zm+y. Then, Ba+Bb=Bzn+x+Bzm+y=1/zμν, where μ=zn+zm+x+y and ν=∑j=0Rzn+xj∑ℓ=jRaℓ−1ℓ−jzm+yℓ−j. So, now there will be three cases:Case 1:z divides μ or z divides ν. If this is true, Ba+Bb will factor into two positive polynomials either of which will be 1 only if n=m=0, so there will be at most finitely many primes of the form p=Ba+Bb.Case 2: some of the factors ofz divide μ and the rest of the factors of z divide ν. If this is true, then like case 1, Ba+Bb will factor into two positive polynomials, either of them can be 1 only if n=m=0, so there will be at most finitely many primes of the form p=Ba+Bb.Case 3: some of the factors ofz divide μ and some of the factors of z divide ν but not all the factors of z divide μ or ν. This case is impossible because it would imply that Ba+Bb is not an integer, but it clearly is an integer.
Several more results include some other interesting properties of some polyhedral numbers.
The following identity can be verified with elementary algebra.Lemma 3.
(18)aa+ba+c+dc+d−bc+d=a+c+da2+d2+ab+cd−ad−bd.
Using some cases of this lemma leads to the following theorems.Theorem 18.
LetCn,r,k=1/knn+rn+2r for positive integers n,r, and k. Then, for all fixed r and k, there are at most finitely many primes of the form p=Cn,r,k+Cm,r,k.Proof.
First, note that from Lemma3, it follows that nn+rn+2r+mm+rm+2r=n+m+2rn2+m2+2rm+nr−nm−rm.
Now, for two positive integersa and b, let n=kn′+a and m=km′+b.
Then,Cn,r,k+Cm,r,k=Ckn′+a,r,k+Ckm′+b,r,k=1/kYZ, where Y=kn′+km′+a+b+2r and Z=a2+b2+ar+br−ab+k2n′2+k2m′2+kn′r+km′r+2kn′a+2km′b−k2n′m′−kn′b−km′a. Now, there will be two cases:Case 1:k divides a+b+2r or k divides a2+b2+ar+br−ab. Then, from this, k divides either Y or Z, so Cn,r,k+Cm,r,k runs through at most finitely many primes.Case 2:k does not divide a+b+2r and k does not divide a2+b2+ar+br−ab.
In this case, there are three scenarios.Scenario 4.
If there are no similar factors betweenk and a+b+2r and there are no similar factors between k and a2+b2+ar+br−ab, then k is coprime to both Y and Z which means An,r,k+Am,r,k is not an integer.Scenario 5.
If there are some similar factors betweenk and a+b+2r and the rest of the factors of k are similar to some factors of a2+b2+ar+br−ab, then 1/kYZ would simplify into the product of two polynomial factors, which means Cn,r,k+Cm,r,k runs through at most finitely many primes.Scenario 6.
If there are some similar factors betweenk and a+b+2r or some similar factors between k and a2+b2+ar+br−ab but not all the factors of k are similar to the factors of both a+b+2r and a2+b2+ar+br−ab, then 1/kYZ will reduce into 1/k′Y′Z′, but k′ is coprime to both Y′ and Z′ which means Cn,r,k+Cm,r,k is not an integer.
From this, it follows that ifCn,r,k+Cm,r,k is an integer for some fixed r and k, it must factor into two positive polynomials, so there can only be finitely many primes of the form p=Cn,r,k+Cm,r,k.Corollary 2.
Consider the numbersFn=1/4nn+1n+2 for n∈ℕ. Then, the only prime of the form p=Fn+Fm is 3=F1+F1.Proof.
Lettingr=1 and k=4 in Theorem 17 gives the identities(19)F4n+F4m=22n+2m+14n2+4m2−4nm+n+m,F4n+F4m+1=124n+4m+38n2+8m2−8nm+6m+1,F4n+F4m+2=n+m+116n2+16m2−16nm−4n+20m+6,F4n+F4m+3=4n+4m+54n2+4m2−4nm−2n+7m+3,F4n+1+F4m+1=n+m+116n2+16m2−16nm+8n+8m+3,F4n+1+F4m+2=124n+4m+58n2+8m2−8nm+2n+8m+3,F4n+1+F4m+3=122n+2m+316n2+16m2−16nm+24m+11,F4n+2+F4m+2=22n+2m+34n2+4m2−4nm+3n+3m+2,F4n+2+F4m+3=4n+4m+74n2+4m2−4nm+2n+5m+3,F4n+3+F4m+3=n+m+216n2+16m2−16nm+16n+16m+15.Theorem 19.
Consider the triangular numbersΔn=nn+1/2, and let n,m,andk be nonnegative integers. Then, the only primes that are of the form p=n+k+1Δn+mΔm+k are 2=1+0+1Δ1+0Δ0+0 and 3=1+1+1Δ1+0Δ0+1=1+0+1Δ1+1Δ1+0=0+1+1Δ0+1Δ2.Proof.
LetGn,m,k=n+k+1Δn+mΔm+k. Using Δn=nn+1/2 and the fact that nn+1n+k+1+mm+km+k+1=n+m+k+1n2+m2−nm+n+km, it is easy to verify the following identities:(20)G2n,2m,2k=2n+2m+2k+12n2+2m2−2nm+2mk+n,G2n,2m,2k+1=2n+m+k+12n2+2m2−2nm+2mk+n+m,G2n,2m+1,2k=n+m+k+14n2+4m2−4nm+4mk+4m+2k+1,G2n,2m+1,2k+1=2n+2m+2k+32n2+2m2−2nm+2mk+3m+k+1,G2n+1,2m,2k=2n+m+k+12n2+2m2−2nm+2mk+3n−m+1,G2n+1,2m,2k+1=2n+2m+2k+32n2+2m2−2nm+2mk+3n+1,G2n+1,2m+1,2k=2n+2m+2k+32n2+2m2−2nm+2mk+2n+m+k+1,G2n+1,2m+1,2k+1=n+m+k+24n2+4m2−4nm+4mk+4n+4m+2k+3.Lemma 4.
Consider the tetrahedral numbersTn=1/6nn+1n+2. Then, for all n∈ℕ, the following identity can be shown with elementary algebra:(21)Tnn+3+Tn+1n+4=2Tn+1n+3.Theorem 20.
Still usingTn=1/6nn+1n+2, there exists the identity(22)∑m=1n−1n+mTmm+2=12Tnn+3.Proof.
FromTnn+3+Tn−1n+2=2Tnn+2, we can assemble the list of equations:(23)Tnn+3+Tn−1n+2=2Tnn+2,−Tn−1n+2−Tn−2n+1=−2Tn−1n+1,Tn−2n+1+Tn−3n=2Tn−2n,−Tn−3n−Tn−4n−1=−2Tn−3n−1,−1n+1T14=2−1n+1T13.
Adding up all these gives2∑m=1n−1n+mTmm+2=Tnn+3.


## 2. Conclusion

We now have a result to determine very easily whether the difference between two elements from a polyhedral number sequence may be eligible to be a prime. We also see that the 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.There exist some additional following identities relating some of these sequences to each other. LetΔn=1/2nn+1 denote the nth triangular number, Tn=1/6nn+1n+2 denote the nth tetrahedral number, On=1/3n2n2+1 denote the nth octahedral number, and Pn=1/6nn+12n+1 denote the nth pyramidal number.Lemma 5.
(24)2n+14+n+14=2n+44+n4.Lemma 6.
(25)Tn+2+Pn+1=Tn+Pn+2.Lemma 7.
(26)Tn2−2+n4=Tn2.Lemma 8.
(27)Tn+22+OΔn=Tn+12+OΔn+2.



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*Source: 9956343-2021-08-09.xml*