Does a relation by addition and subtraction always hold for four adjacent prime numbers?

0. References

[1] https://ja.wikipedia.org/wiki/ダランベールの収束判定法

D'Alembert's Convergence Judgment Method last accessed on November 30, 2021, at 13:25.

1. Conjecture

For many primes p, there will be primes q, r, and s that are neighbors of each other, and in particular, if a, b, c, and d are in decreasing order, then there will be a case where a+d=b+c.

This expression is a little difficult to understand. For example, if p=5, we have (q,r,s)=(2,3,7), (3,7,11), (7,11,13). In this case, by choosing (7,11,13), 5+13=7+11 is valid.

2. Proving that some equality relations do not hold for four neighboring primes

In this conjecture, we say that a+d=b+c will hold for primes a,b,c,d, (a<b<c<d). On the other hand, does a+b=c+d, a+c=b+d, d=a+b+c not hold?

2.1 Proof that a+b=c+d does not hold

From a<b<c<d

→a+b<a+c

→a+b+d<a+c+d

-d<-a, and since both sides can be added, then

→a+b<c+d

This means that a+b=c+d is not true.

2.2 Proof that a+c=b+d does not hold

From a<b<c<d, we have

→a+c<b+c

→a+c+d<b+c+d

-d<-c, then adding to both sides of the equation, we get

→a+c<b+d

This means that a+c=b+d is not true.

2.3 Proof that d=a+b+c does not hold

2.3.1 Ingham's Theorem

According to Ingham, for a sufficiently large natural number n

n^3<p<(n+1)^3

it is known that there exists at least one prime number satisfying Another form of the famous conjecture is

For any natural number n

n^2<p<(n+1)^2

is that for any natural number n, there will be at least one prime that satisfies the Le-Jandre conjecture.

We will assume that this sufficiently large natural number n is "valid for any number n" for the following discussion.

2.3.2 Proof 1 (Assuming that the conjecture about Ingham's theorem that it holds for any number n is correct)

If the existence of just one prime number in the interval (n^3, (n+1)^3) occurs three times in a row (let's call it the Ingham limit), then the following relation is satisfied

If d=a+b+c, then

The only natural number n that satisfies this is n<=5.

When n=1

Not suitable from 1<2<3<8

When n=2

Not suitable from 8<11<13<27

When n=3

Not suitable from 27<29<31<64

For n=4

Not suitable from 64<67<71<125

For n=5

Unsuitable from 125<129<131<216

From the above, we know that d=a+b+c does not hold for a, b, c, and d, which make the Ingham limit hold.

2.3.3 Proof 2 (Proof using mod)

For four consecutive primes p,q,r,s, let p≡-1(mod6), q≡1(mod6), r≡1(mod6), s≡1(mod6), and assume that p+q+r=s, p<q<r<s.

Here we have

a=(p+1)/2, b=(q+1)/2, c=(r+1)/2, d=(s+1)/2

α=(p-1)/2, β=(q-1)/2, γ=(r-1)/2, δ=(s-1)/2

and γ=(r-1)/2, δ=(s-1)/2.

Then

(a-α)(a+α)+(b-β)(b+β)+(c-γ)(c+γ)=(d-δ)(d+δ)

and p+q+r=s

is an equivalence.

Here we have

and then evaluate k_1 and k_2.

2.3.3(1) k_1=k_2=k

Let us assume that 8k ≡ h(mod 6).

If we substitute this result into k_1, the left side ≡ 1 and the right side ≡ 0 or 3, resulting in a contradiction.

2.3.3(2) If k_1+k_2=k, then

Now, let 4k ≡ h(mod 6).

Then, in the same way as before, we have h ≡ 0 (mod 6), and we know that k is a multiple of 12.

Now

This means that we only need h_1+h_2=0 or 12.

Now, let's rewrite p, q, r, and s in the notation with 12 as the law.

The possible combinations are as follows.

Then, in the last column, there is no such combination as h_1+h_2=0 (mod 12), which is a contradiction.

From the above, we know that there is no prime number p,q,r,s, (p<q<r<s), p+q+r=s. 3.

3. numerical experiments to prove that expectation A is true in many cases (43.7% of all cases)

Numerical experiments show that the conjecture is easily valid for primes up to 199. The percentage of primes up to 30011 for which the prediction A is true is p(30011)=(the number of primes for which the prediction is true)/(the number of primes up to 30011)*100=43.7%. When we examined the rate of change of this percentage, we found that it was small enough for each of the 29989 and 30011 cases.

ln|p(29989)/p(30011)|=0.00039687

Therefore, from D'Alembert's Convergence Judgment Method, we concluded that the numerical result of 43.7% would be a significant value.