Finding primes $p$ such that $p\nmid 2^n+3^m,\forall n,m\in\mathbb{N}$

Question

I want to determine prime numbers $p$ with this property $(P)$: $p\nmid 2^n+3^m,\forall n,m\in\mathbb{N}$. ($m,n$ are allowed to be zero)

I was able to find that primes $p$ with $p\equiv 23 (\text{mod}\: 24)$ have the desired property, as we have this equivalence regarding Legendre symbols: $$p\equiv 23 (\text{mod}\: 24)\iff -\left(\frac{-1}{p}\right)=\left(\frac{2}{p}\right)=\left(\frac{3}{p}\right)=1.$$ Indeed, if we assume that $p|2^n+3^m\Rightarrow 2^n\equiv-3^m (\text{mod} \: p)$. The left hand-side is a quadratic residue, whilst the right hand-side is not, hence the contradiction.

One can naturally ask about the converse:

Question: Must a prime $p$ satisfying $(P)$ have the form $p\equiv 23 (\text{mod}\: 24)$?

A prime $p$ which is not of the previous form, must satisfy $p\equiv 1,5,7,11,13,17,19 (\text{mod}\: 24)$. I checked the smallest primes, they don't satisfy $(P)$: $$2=2^0+3^0; 3=2^1+3^0; 5=2^1+3^1; 7=2^2+3^1; 11=2^3+3^1; 13=2^2+3^2;$$ $$17=2^4+3^0; 19=2^3+3^2; 29=2^1+3^3; 31=2^2+3^3; 37\cdot 7=2^8+3^1.$$ But still, I cannot find a pattern, and I don't know what to expect.

---

EDIT: Based on the code provided by @nocomment, I improved a bit the prime generator, up to $200000$, using the sieve of Eratosthenes:

N=200000
a=[0]*(N+1)#prime or not
prime=[2]#make the sieve first
for i in range(4,N+1,2):
    a[i]=1
for i in range(3,N+1,2):
    if a[i]==0:
        for j in range(i*i,N+1,i):
            a[j]=1
for i in range(3,N+1,2):
    if a[i]==0:
        prime.append(i)
#%%
def powers_modulo(base, mod):
    powers = set()
    power = 1
    while power not in powers:
        powers.add(power)
        power = power * base % mod
    return powers
def P(p):
    pow2 = powers_modulo(2, p)
    pow3 = powers_modulo(3, p)
    for a in pow2:
        if -a % p in pow3:
            return False
    return True
for p in prime:
    P_ = P(p)
    if (p % 24 == 23) != P_:
        print(p, p % 24, P_)

The result is the following (took me some $4$ minutes). So we generate only prime solutions that are not $23$ mod $24$. All of them, up to $200000$, are $1$ mod $24$.

601 1 True
4297 1 True
6553 1 True
11113 1 True
12073 1 True
12409 1 True
12457 1 True
15241 1 True
16633 1 True
17209 1 True
17881 1 True
20809 1 True
31033 1 True
33721 1 True
36529 1 True
40153 1 True
47017 1 True
47497 1 True
47881 1 True
49369 1 True
49417 1 True
50377 1 True
55609 1 True
56713 1 True
57649 1 True
58153 1 True
58921 1 True
59281 1 True
63241 1 True
63577 1 True
65929 1 True
66841 1 True
67033 1 True
70921 1 True
71761 1 True
72937 1 True
73417 1 True
77977 1 True
78697 1 True
83449 1 True
86857 1 True
88177 1 True
88681 1 True
91033 1 True
91129 1 True
97369 1 True
101737 1 True
104233 1 True
104953 1 True
105673 1 True
106681 1 True
107089 1 True
108649 1 True
108793 1 True
110569 1 True
114217 1 True
114889 1 True
121369 1 True
121993 1 True
130681 1 True
132409 1 True
132697 1 True
133321 1 True
139369 1 True
142057 1 True
148921 1 True
151057 1 True
152041 1 True
153529 1 True
155569 1 True
159097 1 True
161641 1 True
165721 1 True
167593 1 True
168409 1 True
172009 1 True
172681 1 True
172969 1 True
174121 1 True
178441 1 True
182041 1 True
182089 1 True
182929 1 True
184153 1 True
187513 1 True
189961 1 True
190633 1 True
191497 1 True
196681 1 True

Based on this numerical evidence, maybe one can theoretically prove what all the prime solutions are. Or that other residues (except $1$ and $23$) cannot appear.

---

EDIT $2$: Thanks to @Einar Rødland, we were able to find a faster test. Indeed, a prime $p$ has property $(P)$ iff $2^q\equiv 3^q\equiv 1( \text{mod} \:p)$, where $p−1=2^rq,2\nmid q$. The improved Python code is the following, it generates (in less than a minute) $2946$ prime numbers up to $10^7$, exactly as Einar did, all of them being $1$ mod $24$:

N=10000000
a=[0]*(N+1)#prime or not
prime=[2]#make the sieve first
for i in range(4,N+1,2):
    a[i]=1
for i in range(3,N+1,2):
    if a[i]==0:
        for j in range(i*i,N+1,i):
            a[j]=1
for i in range(3,N+1,2):
    if a[i]==0:
        prime.append(i)
#%%
#or the pow(,,) function, already implemented
def expo(base,exp,mod):
    if exp==0:
        return 1
    if exp%2==0:
        return (expo(base,exp//2,mod)2)%mod
    else:
        return (base*expo(base,(exp-1)//2,mod)2)%mod
def P(p):
    q=p-1
    while q%2==0:
        q=q//2
pow2 = expo(2,q,p)
pow3 = expo(3,q,p)
return (pow2==pow3 and pow2==1)

Nr=0
for p in prime:
    P_ = P(p)
    if (p % 24 == 23) != P_:
        print(p, p % 24)
        Nr=Nr+1
print("{0} solutions have been generated.".format(Nr))

---

EDIT $3$: I posted my own answer, to summarize and clarify all of our findings. Indeed, a prime satisfying $(P)$ must be $1$ or $23$ modulo $24$. The ony "mistery" left is to give a complete description of the primes which are $1$ modulo $24$ satisfying $(P)$, but probably this is too difficult.

Answer 1

Question: Must a prime $p$ satisfying $(P)$ have the form $p\equiv 23 (\text{mod}\: 24)$?

No. For example, $p=601$ satisfies $(P)$, and $p\equiv 1 (\text{mod}\: 24)$.

Test script:

def powers_modulo(base, mod):
    powers = set()
    power = 1
    while power not in powers:
        powers.add(power)
        power = power * base % mod
    return powers
def P(p):
    pow2 = powers_modulo(2, 601)
    pow3 = powers_modulo(3, 601)
    return all(
        (a+b) % p != 0
        for a in pow2
        for b in pow3
    )
p = 601
print(P(p), p % 24)

Output (Attempt This Online!):

True 1

All exceptions up to 50000 (showing $p$, $p$ modulo 24, and the test result):

601 1 True
4297 1 True
6553 1 True
11113 1 True
12073 1 True
12409 1 True
12457 1 True
15241 1 True
16633 1 True
17209 1 True
17881 1 True
20809 1 True
31033 1 True
33721 1 True
36529 1 True
40153 1 True
47017 1 True
47497 1 True
47881 1 True
49369 1 True
49417 1 True

Found with this script in ~44 seconds:

import math
def powers_modulo(base, mod):
    powers = set()
    power = 1
    while power not in powers:
        powers.add(power)
        power = power * base % mod
    return powers
def P(p):
    pow2 = powers_modulo(2, p)
    pow3 = powers_modulo(3, p)
    for a in pow2:
        if -a % p in pow3:
            return False
    return True
def primes_under(N):
    for n in range(2, N):
        for d in range(2, math.isqrt(n)+1):
            if not n % d:
                break
        else:
            yield n
for p in primes_under(50000):
    P_ = P(p)
    if (p % 24 == 23) != P_:
        print(p, p % 24, P_)

Attempt This Online!

Answer 2

Let me summarize all the answers received here, and many thanks to @Oscar, @Einar, @nocomment for their contribution. Remember that we want primes $p$ with property $(P)$: $p\nmid 2^n+3^m,\forall n,m\in\mathbb{N}$.

As noted in my post, $p=2$ and $p=3$ must be discarded. Hence, any other prime is of the form: $p\equiv 1,5,7,11,13,17,19,23 (\text{mod} \: 24)$. I proved that any prime with $p\equiv 23 (\text{mod} \: 24)$ has property $(P)$. @Einar Rødland found a nice characterisation, and let me give my own proof (which also shows how to choose $n,m$ if they exist):

Lemma. Let $p\geq 3$ be a prime number. Then $p$ has $(P)$ if and only if $2^q\equiv 3^q\equiv 1 (\text{mod} \: p)$, where $p-1=2^rq,r\in\mathbb{N}^\ast, 2\nmid q$.

Proof: $(\Leftarrow)$ Suppose that there are $n,m\in\mathbb{N}, p|2^n+3^m\Rightarrow 2^n\equiv -3^m (\text{mod} \: p)$. But then: $(2^n)^q\equiv (-3^m)^q\Rightarrow 1\equiv (2^q)^n\equiv -(3^q)^m\equiv -1 (\text{mod} \: p)$, a contradiction.

$(\Rightarrow)$ Suppose that $2^q\not\equiv 1 (\text{mod} \: p)$. If we denote $k:=\gamma_p(2)$ the Gaussian/multiplicative order, then we have: $k\nmid q$. Since $k|p-1=2^rq$, it implies that $k=2^is$ for some $i\in\overline{1,r}$ and $s|q$. But this means: $2^{k/2}\equiv -1\Rightarrow p|2^{k/2}+3^0$, contradicting $(P)$. Similarly, if we assume that $3^q\not\equiv 1 (\text{mod} \: p)$, then for $l:=\gamma_p(3)$ we have: $p|2^0+3^{l/2}$, contradicting $(P)$. $\Box$

As @Oscar noted, we do have a restriction on the possible residues modulo $24$, which is also consistent with our previous computer simulations:

Proposition. Let $p$ be a prime number, satisfying $(P)$. Then $p\equiv 1,23 (\text{mod} \: 24)$.

Proof: Using the notations from the previous Lemma, we have: $2^q\equiv 3^q\equiv 1 (\text{mod} \: p)$. Since $q|(p-1)/2$, this implies: $2^{(p-1)/2}\equiv 3^{(p-1)/2}\equiv 1 (\text{mod} \: p)$. Using Euler's criterion, this is equivalent to: $$\left(\frac{2}{p}\right)=\left(\frac{3}{p}\right)=1.$$ The first Legendre symbol is translated as: $p\equiv 1,7 (\text{mod} \: 8)$, and the other symbol as: $p\equiv 1,11 (\text{mod} \: 12)$. Out of the four total combinations of congruences, only two are possible:
$(i)$ $p\equiv 1 (\text{mod} \: 8)$ and $p\equiv 1 (\text{mod} \: 12)$, this leads to $p\equiv 1 (\text{mod} \: 24)$;

$(ii)$ $p\equiv 7 (\text{mod} \: 8)$ and $p\equiv 11 (\text{mod} \: 12)$, this leads to $p\equiv 23 (\text{mod} \: 24)$. $\Box$

Answer 3

We prove nondivisibility when $p=601$.

The proof is similar to that given here for $p=23$, except it involves higher power residues.

Modulo $601$, which is understood in the equivalences that follow, we have $729\equiv128\implies 3^6\equiv2^7$. Now suppose that $2^n+3^m\equiv0$. Since $(a+b)|(a^7+b^7)$, we have $2^{7n}+3^{7m}\equiv3^{6n}+3^{7m}\equiv0$. Then $3^k\equiv-1$ where $k=|6n-7m|$ is a whole number, but no such $k$ actually exists because $3$ turns out to have odd (multiplicative) order (order = $75$).

---

The above method of proof works in two situations:

• An odd power of $3$ is equivalent to some power of $2\bmod p$ and $2$ is then found to have odd order, or

• An odd power of $2$ is equivalent to some power of $3\bmod p$ and $3$ is then found to have odd order.

These options actually imply that both $2$ and $3$ have odd multiplicative order $\bmod p$, which (for $p>3$) requires both $2$ and $3$ to be quadratic residues $\bmod p$. This implies that nondivisibility can he proved by this method only if $p\equiv\pm1\bmod24$.

For $p\equiv-1\bmod24$ a quadratic residue will always have odd order, as the Euler totient of $p$ is twice an odd number, so nondivisibility of $2^n+3^m$ always holds. For $p\equiv+1\bmod24$ we need a higher power residue to guarantee odd order, so only some of these primes will give this nondivisibility.

Answer 4

Adding some more info about the solutions $\equiv 1 \pmod {24}$: Because $2$ and $3$ both have odd order, they are 8th power residues modulo $p$. In particular, $4$th (biquadratic) and theorems of Gauss give that $p$ is of the form $a^2 + (24 b)^2$. https://en.wikipedia.org/wiki/Quartic_reciprocity.

In the case of $601$: $5^2 + 24^2$, showing that you cannot get more factors beyond $24$.

You can also give a characterization in terms of splitting properties in a certain field extension, as in this question: About the parity of $ord_p(7)$ or this MO question: https://mathoverflow.net/questions/372450/parity-of-the-multiplicative-order-of-2-modulo-p and this allows you to compute the density of such primes.

Let $j\geq 3$, then the following are equivalent with at most finitely many exceptions:

• $p\equiv 1 \pmod {24}$ and $2^j\Vert p-1$ and $2$ and $3$ are $2^j$th powers modulo $p$.
• $p$ is split in the field $K_j = \mathbb Q(1^{1/2^j},2^{1/2^j},3^{1/2^j})$ but not in $L_j=K_j(1^{1/2^{j+1}})$.

(The finitely many exceptions come from my inability to write down an integral basis or to prove that the elements I wrote down generate the ring of integers.)

The density of the primes $\equiv 1 \pmod {24}$ that we are looking for is then $$\sum_{j\geq 3} \left(\frac1{[K_j:\mathbb Q]}-\frac1{[L_j:\mathbb Q]}\right) \,.$$

For $j\geq 3$ the following should be true:

• $[L_j:K_j]=2$
• $[K_j:\mathbb Q]= 2^{3j-2}$

Assuming this, the density is $$\sum_{j\geq3} \frac1{2^{3j-1} } = \frac1{2^{5}\cdot7}=\frac1{224} \,.$$

This predicts about $$\frac{50000}{224\log(50000)}\approx 20.63$$ such primes below 50000, compared to the actual 21 found in another answer here.

In other words, among all the primes that are $\equiv 1 \pmod {24}$, of them $\frac1{28}$ are solutions to the problem.

Answer 5

Just to add some comments to the existing solution.

For a given prime $p$, there are $n,m$ so that $p\,|\,2^n+3^m$ if and only if there is either an $n$ for which $2^n\equiv-1\pmod{p}$ or $3^m\equiv-1\pmod{p}$.

Let $p-1=2^rq$ with $q$ odd. If $2^q\equiv1\pmod{p}$, no such $n$ exists; similarly, if $3^q\equiv1\pmod{p}$, no such $m$ exists. In fact, for any $x$ not divisible by $p$, the powers $x^{(p-1)/2^k}$ will either all be 1 modulo $p$, or the smallest $k$ for which it is not 1 it will be $-1$ modulo $p$.

This gives us a faster test: just compute $2^q$ and $3^q\pmod{p}$, and if they are both equal to $1$ modulo $p$, there are no $n,m$ for which $p\,|\,2^n+3^m$. Otherwise, such $n,m$ exist.

Using SAGE, this method used 58 seconds to find 2946 primes below 10,000,000 and not equal to $23\pmod{24}$ with this property; up to 100,000,000 took 8 min 34 sec and found 25707 such primes. These primes are all equal to $1\pmod{24}$, and the majority (appr 7/8) are $25\pmod{48}$.

As for predicting such primes without having to test them one by one, my best idea would be to do something similar to Legendre symbols, but with $x^q\pmod{p}$ for $x=2,3$ instead of $x^{p-1}\equiv\left(\frac{x}{p}\right)\pmod{p}$.

---

Here is my SAGE code:

def Ptest(p, S=[2,3], log=True):
    R = Zmod(p)
    Sol = []
    q = p-1
    while q%2==0:
        q /= 2
    Sol = [s for s in S if R(s)^q!=1]
    if log: print('%s: q=%s, Sol=%s'%(p,q,Sol))
    return Sol
SOL = []
for p in prime_range(5,10000000):
    sol = Ptest(p, log=False)
    if not sol and p%24!=23:
        SOL.append(p)
print('Primes found: %s'%len(SOL))
```