Puzzle 1014. Observation about three consecutive primes

Problems & Puzzles: Puzzles

Puzzle 1014. Observation about three consecutive primes

Sebastián Martín Ruiz noted that:

For any three consecutive primes p, q, r such that p<q<r

No one of the following conditions are satisfied if p>13

a) p divides (q^2 +r^2)

b) q divides (p^2 +r^2)

c) r divides (p^2 +q^2)

Q. Find an explanation or a counterexample, for each condition.

On the week 23-"9 Aug, 2020, contributions came from Oscar Volpatti, Fausto Morales and Emmanuel Vantieghem

***

Oscar wrote:

Sebastián's observation can be explained (but not proved) as follows.

Consider the prime gaps g=q-p and h=r-q.

a) if p divides q^2+r^2 = (p+g)^2+(p+g+h)^2, then p also divides g^2+(g+h)^2 = x;

b) if q divides p^2+r^2 = (q-g)^2+(q+h)^2, then q also divides g^2+h^2 = y;

c) if r divides p^2+q^2 = (r-g-h)^2+(r-h)^2, then r also divides (g+h)^2+h^2 = z.

According to Cramer-Granville conjecture, g and h grow about as fast as (ln p)^2, (ln q)^2, (ln r)^2, up to some constant factor.
If so, the positive quotients x/p, y/q, z/r must converge to 0 as p,q,r grow to infinity.

In particular, for p,q,r large enough, the three quotients must be smaller than 1. Then we have:

a1) 0 < x < p, so p can't divide x;

b1) 0 < y < q, so q can't divide y;

c1) 0 < z < r, so r can't divide z.

Numerical computations suggest precise lower bounds:
condition a1 possibly holds for every prime p > 2477;
condition b1 possibly holds for every prime q > 293;

condition c1 possibly holds for every prime r > 2203.

The finite range 2<=p<=2477 can be easily checked:

condition a is satisfied for p=2, p=5, and p=13;

conditions b and c are never satisfied.

***

Fausto wrote:

For any re-naming (x, y, z) of the triplet {p, q, r}, we have that, for some pair of (positive or negative) integers i and j:

x + i = y + j = z,

so that

x^2 + y^2 = 2z^2 + i^2 + j^2 - 2z*(i+j).

Now, if z is to divide the LHS of this expression, then it must divide i^2 + j^2 aswell. In particular, a requirement implied is:

i^2 + j^2 >= z,

something that has not occurred in the range

2521 < z < 10^8.

Accordingly, my guess is that Sebastian's observations must hold for all z, since primes in consecutive triplets appear to have become too close together for any counterexamples to exist.

***

Emmanuel wrote:

Let a = q - p, b = r - q, c = r - p.
Then, modulo p we have : q^2 + r^2 == a^2 + c^2
So, if p divides q^2 + r^2 we have that p divides a^2 + c^2.
As a, b, c are even we then have : p divides (a^2 + c^2)/4.
An immediate intuitive approach tells me that this will be impossible because (a^2 + c^2)/4 is too small.
But I find no proof of this in the literature. So, that is a conjecture.
Similarly for the other cases.

To give an idea of the size of (a^2 + c^2)/4 in comparizon with the size of p, q, r I computed a record table :

[p, q, r} (a^2 + c^2)/4
{17, 19, 23} 10
{19, 23, 29} 29
{31, 37, 41} 34
{47, 53, 59} 45
{83, 89, 97} 58
{109, 113, 127} 85
{113, 127, 131} 130
{199, 211, 223} 180
{523, 541, 547} 225
{887, 907, 911} 244
{1129, 1151, 1153} 265
{1321, 1327, 1361} 409
{1327, 1361, 1367} 689
{5591, 5623, 5639} 832
{9551, 9587, 9601} 949
{15683, 15727, 15731} 1060
{18803, 18839, 18859} 1108
{19609, 19661, 19681} 1972
{31397, 31469, 31477} 2896
{58789, 58831, 58889} 2941
{155893, 155921, 156007} 3445
{155921, 156007, 156011} 3874
{188029, 188107, 188137} 4437
{338033, 338119, 338137} 4553
{360653, 360749, 360769} 5668
{370261, 370373, 370387} 7105
{396733, 396833, 396871} 7261
{492113, 492227, 492251} 8010
{1349533, 1349651, 1349669} 8105
{1357201, 1357333, 1357337} 8980
{1671781, 1671907, 1671941} 10369
{2010733, 2010881, 2010887} 11405
{4652353, 4652507, 4652513} 12329
{10937921, 10938023, 10938119} 12402
{15203977, 15204131, 15204169} 15145
{17051707, 17051887, 17051899} 17316
{20831323, 20831533, 20831557} 24714
{47326693, 47326913, 47326919} 24869
{90438133, 90438343, 90438379} 26154

***

Records | Conjectures | Problems | Puzzles