Problems & Puzzles: Puzzles

Puzzle 1207 Primes of the form 648*k^6+1 From the always interesting site Prime Curios! from G. L. Honaker, we take this Curio: If a number can be written both as 2*a^2 + 1 and as 3*b^3 + 1, it must be of the form 648*k^6 + 1. The smallest prime of this form is not only 472393, where k = 3, but curiously k = 4 and k = 5 also (are primes). [Hartley] Questions: Q1. Demonstrate that "If a number can be written both as 2*a^2 + 1 and as 3*b^3 + 1, it must be of the form 648*k^6 + 1" Q2. Are there other three consecutive k values such that 648*k^6 + 1 are primes?

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From Feb 8-14, 2025, contributions came from Michael Branicky, Giorgos Kalogeropoulos, Gennady Gusev, Adam Stinchcombe, Oscar Volpatti, Simon Cavegn, Emmanuel Vantieghem

 

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Michael wrote:

Q2. He sent the first 1000 triplets:

3, 4, 5)
(358, 359, 360)
(1698, 1699, 1700)
(5129, 5130, 5131)
(6340, 6341, 6342)

...

8772525, 8772526, 8772527)
(8781142, 8781143, 8781144)
(8786618, 8786619, 8786620)
(8795054, 8795055, 8795056)
(8809539, 8809540, 8809541)

He also sent the first 100 quadruplets:

(31299, 31300, 31301, 31302)
(64649, 64650, 64651, 64652)
(364225, 364226, 364227, 364228)
(429147, 429148, 429149, 429150)
(514993, 514994, 514995, 514996)

...

(21405683, 21405684, 21405685, 21405686)
(21410611, 21410612, 21410613, 21410614)
(21580618, 21580619, 21580620, 21580621)
(21683842, 21683843, 21683844, 21683845)
(21803181, 21803182, 21803183, 21803184)

And also:

The first quintuplet is (6440416, 6440417, 6440418, 6440419, 6440420).

The first sextuplet is (52583973, 52583974, 52583975, 52583976, 52583977, 52583978).

the first septuple: (186451168, 186451169, 186451170, 186451171, 186451172, 186451173, 186451174)

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Giorgos wrote:

Q1. N = 2*a^2 + 1  = 3*b^3 + 1

             2*a^2 = 3*b^3

Since 2 and 3 are coprime, 3 must divide a and 2 must divide b.

Let a=3m and b=2n. Substituting these into the equation:

2(3m)^2 = 3(2n)^3

18m^2 = 24n^3

3m^2=4n^3

Again, since 3 and 4 are coprime, 3 must divide n and 4 must divide m. 

Let n=3p and m=2q. 

Substituting these into the equation:
3(2q)^2 = 4(3p)^3

12q^2 = 108p^3

q^2 = 9p^3

Let q=3r

9r^2 = 9p^3 and r^2=p^3.

So, p must be a perfect square.

Let p=s^2 then r=s^3.

Substituting back, we get q=3s^3

Tracking back through all substitutions:

m=2q=6s^3
a=3m=18s^3
n=3p=3s^2
b=2n=6s^2

Substituting a=18s^3 to N=2a^2+1 we get

N=648s^6+1

If we do the same for b=6s^2 we also get

N=648s^6+1

So, N is of this form.

Q2. We can easily find many triplets of 3 consecutive k values that give 3 primes.

Here are the first few

{3,4,5}, {358,359,360}, {1698,1699,1700}, {5129,5130,5131}, {6340,6341,6342}, {13064,13065,13066}, {17955,17956,17957}, {18120,18121,18122}, {21519,21520,21521}...

I also searched for more consecutive values that return primes.

Here are the results for 4,5,6 and 7 consecutive values of k:

4->{31299,31300,31301,31302}
5->{6440416,6440417,6440418,6440419,6440420}
6->{52583973,52583974,52583975,52583976,52583977,52583978}
7->{186451168,186451169,186451170,186451171,186451172,186451173,186451174}

I am now trying to find 8 consecutive k and this would be the end.

We cannot have 9 consecutive k because one of the values of 648k^6+1 will be divisible by 11. Why?

We can show that when k = 1 or 10 mod 11 then 648k^6+1=0 mod11.

So, the best case for k is to be 2,3,4,5,6,7,8,9mod11 which is 8 consecutive numbers.

proof

First, reduce 648 modulo 11:
648 ÷ 11 = 58 with a remainder of 10, so 648 ≡ 10 mod 11.
Thus, 648k^6 + 1 ≡ 10k^6 + 1 mod 11.

Case 1: k ≡ 1 mod 11

If k ≡ 1 mod 11, then:
k^6 ≡ 1^6 ≡ 1 mod 11.
Substitute into 10k^6 + 1 mod 11:
10k^6 + 1 ≡ 10 * 1 + 1 ≡ 11 ≡ 0 mod 11.
Thus, 648k^6 + 1 ≡ 0 mod 11, meaning it is divisible by 11.

Case 2: k ≡ 10 mod 11
If k ≡ 10 mod 11, note that 10 ≡ -1 mod 11. Thus:
k^6 ≡ (-1)^6 ≡ 1 mod 11.
Substitute into 10k^6 + 1 mod 11:
10k^6 + 1 ≡ 10 * 1 + 1 ≡ 11 ≡ 0 mod 11.
Again, 648k^6 + 1 ≡ 0 mod 11, meaning it is divisible by 11.

Also, here are the moduli for k=2,3,4,5,6,7,8,9 mod 11 then 648k^6+1 ≡ 3, 9, 8, 7, 7, 8, 9, 3 mod 11

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Gennady wrote:

Q1.

If n=2a^2+1 and n=3b^3+1, then 2a^2=3b^3.

Then 'a' must be a multiple of 3 and 'b' a multiple of 2.

Let a=3u and b=2v, we have 3u^2=4v^3.

Similarly, then let u=2x and v=3y.

We get x^2=9y^3.

Let x=3s, then s^2=y^3.

This is possible if s=k^3.

Moving back, we get x=3k^3, u=6k^3, v=3k^2, and finally

a=18k^3, b=6k^2.

n=2a^2+1=3b^3+1=2*(18k^3)^2+1=3*(6k^2)^3+1=648k^6+1.

Q2.

1100 triples of numbers were found for k<=10^7 (not counting: 3, 4, 5). See the attached file

358	359	360
1698	1699	1700
5129	5130	5131
6340	6341	6342
13064	13065	13066

...

9965872	9965873	9965874
9972184	9972185	9972186
9972185	9972186	9972187
9975210	9975211	9975212
9993634	9993635	9993636

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Adam wrote:

Suppose 2a^2+1=3*b^3+1, so 2a^2=3b^3.  Consequently, 3 divides a, so 3^2 divides 2a^2=3b^3 so 3 divides b^3 and 3 divides b.  Also 2 divides 2a^2=3b^3 so 2 divides b.  Write b=2*3*B, then 3b^3+1 = 3*2^3*3^3*B^3+1 =648b^3+1.  We can then write the original 2a^2+1=3b*3+1 as 

2*(18A)^2+1=3*(6B)^3+1.  For a prime p (including 2 or 3 with attentive analysis), p divides A means p divides B so p^3 divides A^2 so p^2 divides A and p^4 divides A^2 and also B^3, so p^2 divides B.  Consequently, you can march up any prime up to p^6 divides both A^2 and B^3, in other words these are 6th powers.  The prime is of the form 648b^6+1.

I find many such examples.  Up to b=114804523, I found 8007 instances of 3 in a row, 343 instances of 4 in a row, and 22 instances longer than 4.

The sequences of length 3 start with first b value of 3, or 358, or 1698,or 5129 and ended with a list of the last 3 starting b values 114765829, or 114792452, or 114795014.

The length 4 sequences started at 31299,64649,364225 and ended at 114006898,114146587,114158719.

The longer list, with data [length, start value] are:

[5, 6440416], [5, 10074354], [5, 10583214], [5, 28090098], [5, 29352127], [5, 33304648], [5, 34114577], [5, 50272962], [6, 52583973], [5, 53254247], [5, 63513837], [5, 73140456], [6, 73989588], [5, 74055668], [5, 81753423], [5, 86490407], [5, 91810592], [5, 94574144], [5, 97327971], [5, 105998059], [5, 106349995], [5, 108466932]

A couple of length 6 sequences.  If you investigate the prime p=11, you see there is a max length of 8, in between the residues b=2 mod 11 and b=9 mod 11.

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Oscar wrote:

Q1.

Assume that number n can be written both as 2*a^2+1 and as 3*b^3+1.

Consider the factorization of number c = n-1 = 2*a^2 = 3*b^3.

For each prime p_i dividing c, let p_i^x_i, p_i^y_i, and p_i^z_i be the largest powers of p_i dividing a, b, and c respectively.

Compare the overall powers of p_i on each side of the equation.

i = 1, p_1 = 2.

z_1 = 1+2*x_1 = 3*y_1; z_1 odd, y_1 odd.

y_1 = 1+2*w_1, x_1 = 1+3*w_1, z_1 = 3+6*w_1. 

i = 2, p_2 = 3.

z_2 = 2*x_2 = 1+3*y_2; z_2 even, y_2 odd.

y_2 = 1+2*w_2, x_2 = 2+3*w_2, z_2 = 4+6*w_2.

i > 2, p_i > 3.

z_i = 2*x_i = 3*y_i; z_i even, y_i even.

y_i = 2*w_i, x_i = 3*w_i, z_i = 6*w_i.

Let k be the product of all terms p_i^w_i. Then:

c = (2^3)*(3^4)*(k^6) = 648*k^6;

n = c+1 = 648*k^6+1 = f(k).

QED

Q2.

Next few sequences of three consecutive primes have the following starting indexes k:  

358, 1698, 5129, 6340, 13064, 17955, 18120, 21519, 30663.

Next big sequence contains four primes, but longer sequences can be found too.

The first sequence with length m>3 has starting index k and involves primes with d digits, according to the following table:

m  k  (d)

4  31299  (30)

5  6440416  (44)

6  52583973  (50)

7  186451168  (53)

8  828738089016  (75)

The given table is complete, as there can be no sequence of nine consecutive primes or more.

Proof.

Looking at the first few values f(k), prime 11 is a proper factor of f(1) and of f(10):

f(1) = 649 = 11*59

f(10) = 648000001 = 11*58909091

More generally, prime 11 divides f(k) whenever k == 1 mod 11 or k == 10 mod 11, always as a proper factor, because f(k) is strictly increasing and is already greater than 11 for k = 1. 

Hence there can only be:

isolated primes in position k=11*q (like k = 33,77,88,...);

or sequences with length m<=8, within intervals of the form [11*q+2,11*q+9].

QED.

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Simon wrote:

Q2.
7 consecutive k:
186451168
9101420848

6 consecutive k:
52583973
73989588
186451168
474704123
547351631
1694729522
2288371848
3078063584
3352923972
4221399570
4377507412
5758339074
6916947096
7894701960
9101420848
9251081976
10077737409
11067865725
11714328002
11842263084
12826559242
17869898126
19651246881
19969594438
21768918790
22490377846
22984704239
24017685386
24220269153
24277987450
24385424494
25347029184
27323850062
27413382077
29314713431
33865093100
38047935059
38301311282
39230962730
45378909897
49004999012
53943047854
54533279077
54715518552
56404866861
56463780245
57127147684
58618591463
58853988967
59100549346
59374424651
62383076120
65499812359
66921993054
67807014619
67823698285
68957133216
69427842167
70203294143
71378857453
71729282143
72093977650
72257460365
72519810772
73020464376
73346032410
78347895067
80281894389
80799357884
82787166015
82978137246
83673593425
83976341155
84553722917
84716682076
85267469840
89590916385
90061965897
92761046226
93478716148
94572385461
97329583486
99073986772
99350226978
99568861174

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Emmanuel wrote:


Q1
If a number  m  is at the same time of the form  2 a^2 + 1  and of the form  3 b^3 +1, then
m - 1  is at the same time of the form  2 a^2  and of the form  3 b^3.
Thus, in the  prime factorization of  m - 1,
   the exponent of  2  is an odd number divisible by 3 and hence of the form 3 + 6v;
   the exponent of  3  is an even number of the from 3m + 1 and hence of the form  4 + 6w
   the exponent of any other prime is divisible by 6.
Thus, m - 1 = (2^3)*(3^4)*h^6= 648 k^6, Q.E.D.

Q2
Put  F(k) = 648 k^6.+ 1.
Here are some values of  k  such that  F(k), F(k + 1) and  F(k + 2)  are prime :
  358, 1698, 5129, 6340, 13064, 17955, 18120, 21519, 30663,  ...  many more.  
From here on I found also values of  k such that  F(k), F(k + 1), F(k + 2)  and  F(k + 3)  are prime : :    
 31299, 64649, 364225, 429147, 514993, 701366, 812143, 821319, 878254, 986526, ...
Further I found :
  6440416  is the start of five consecutive numbers with prime F-value.
  52583973  is the start of six consecutive numbers with prime F-value.
  186451168 is the start of seven consecutive numbers with prime F-value.

No doubt this might go on ...

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