Pythagorean Sets of Consecutive Primes

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• There seems to be a limitless number of sets of four odd positive consecutive primes with distinct gaps, such that the sum of the squares of the 2 smallest
Message 1 of 7 , Aug 31, 2011
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There seems to be a limitless number of sets of four odd positive consecutive primes with distinct gaps, such that the sum of the squares of the 2 smallest gaps equals the square of the largest gap.

For example, the first occurrence of such a set is (677, 683, 691, 701). The distinct gaps in left to right order are 6, 8 and 10.

Call such sets Pythagorean Sets (PS).

Is there a limit on the length of a set S of odd positive consecutive primes, such that each subset of four consecutive primes is a PS?

Sets of 7 primes are reasonably easy to find, but sets of 8 quite hard (for me).

Here is the first occurrence of a set of 7 consisting of 4 PS's.
(316753, 316759, 316769, 316777, 316783, 316793, 316801).

Here is one for N=8 consisting of 5 PS's.
(2921173, 2921179, 2921189, 2921197, 2921203, 2921213, 2921221, 2921227).

Another question begs to be answered. Consider these 3 expressions namely, A= 4*n+2, B= 4*n*(n+1) and C= 4*n*(n+1)+2. They define even integer sides of right angle triangles. For any n, does a PS exist with gaps A, B and C?

I found a PS for n=1, 2 and by sheer random luck for n=4, but no PS so far for n=3 where A=14, B=48 and C=50. If there is one, the first prime of the PS must exceed 1000000021 according to my calculations.

Here is the PS for n=4: (12345678988555739, 12345678988555819, 12345678988555837, 12345678988555919), The distinct gaps in order are 80, 18 and 82.

Any one care to check me on this? Thanks folks.

Bill Sindelar

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• ... Here is the first set of 9: {v=[ 206710003, 206710013, 206710019, 206710027, 206710037, 206710043, 206710051, 206710061, 206710067];
Message 2 of 7 , Aug 31, 2011
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"w_sindelar@..." <w_sindelar@...> wrote:

> Sets of 7 primes are reasonably easy to find,
> but sets of 8 quite hard (for me).

Here is the first set of 9:

{v=[
206710003, 206710013, 206710019,
206710027, 206710037, 206710043,
206710051, 206710061, 206710067];
print(vector(#v-1,k,v[k+1]-v[k]));}

[10, 6, 8, 10, 6, 8, 10, 6]

David
• ... A gap cycle of 6, 8, 10 in some order can only get one longer than this before becoming inadmissible modulo 5. If we want more than 10 primes then we need
Message 3 of 7 , Aug 31, 2011
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David wrote:
> Here is the first set of 9:
>
> {v=[
> 206710003, 206710013, 206710019,
> 206710027, 206710037, 206710043,
> 206710051, 206710061, 206710067];
> print(vector(#v-1,k,v[k+1]-v[k]));}
>
> [10, 6, 8, 10, 6, 8, 10, 6]

A gap cycle of 6, 8, 10 in some order can only get one longer than
this before becoming inadmissible modulo 5.
If we want more than 10 primes then we need other gaps.

Puzzle:
If (a, b, c) and (b, c, d) are Pythagorean triples then is a=d?
I assume so in the following.
This means all solutions must have a gap cycle of 3 numbers.
If 5 divides the sum of the 3 numbers then the cycle never

(10, 24, 26) is the smallest even Pythagorean triple with sum
divisible by 5.
The first case of 12 consecutive primes for that cycle in some order:
490537270893409 + d,
for d = 0, 10, 34, 60, 70, 94, 120, 130, 154, 180, 190, 214.
The gaps are 10, 24, 26, 10, 24, 26, 10, 24, 26, 10, 24.
A few hundred cases of 12 non-consecutive primes were found
during the search.

I suspect the above is the first case of 12 consecutive primes
with any gaps.
A cycle with (10, 24, 26) cannot give more than 12 primes because
In order to get an admissible pattern with 13 primes I think we
would need gaps of a size which makes it very hard to find
consecutive primes.

If the k-tuple conjecture (*) is true then there are arbitrarily
long solutions.
For example, (3*n#, 4*n#, 5*n#) is a Pythagorean triple for all n,
and the gaps never become inadmissible modulo primes up to n.

(*) There are variations of the k-tuple conjecture.
We only need one saying that all admissible patterns have at least
one occurrence.
Then it follows that they also have occurrences with consecutive primes.

--
Jens Kruse Andersen
• ... Yes. Proof: Suppose the converse. Then we have 4 possibilities: 1) a^2 = b^2 + c^2, d^2 = b^2 - c^2 2) a^2 = b^2 + c^2, d^2 = c^2 - b^2 3) a^2 = b^2 - c^2,
Message 4 of 7 , Sep 1, 2011
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"Jens Kruse Andersen" <jens.k.a@...> wrote:

> Puzzle:
> If (a, b, c) and (b, c, d) are Pythagorean triples then is a=d?

Yes.

Proof: Suppose the converse. Then we have 4 possibilities:
1) a^2 = b^2 + c^2, d^2 = b^2 - c^2
2) a^2 = b^2 + c^2, d^2 = c^2 - b^2
3) a^2 = b^2 - c^2, d^2 = b^2 + c^2
4) a^2 = c^2 - b^2, d^2 = b^2 + c^2
Hence there exists a pair of coprime integers (x,y)
such that x^2+y^2 and x^2-y^2 are both odd squares.
But each odd square is congruent to 1 modulo 8.
Hence 2*x^2 = 2 mod 8 and x is even.
Hence both y^2 and -y^2 are congruent to 1 modulo 4,
which is absurd.

• ... [Then I screwed up, sorry: I *should* have said:] and x is odd. I am pondering the unfortunate consequences.... David (feeling more odd then even :-)
Message 5 of 7 , Sep 1, 2011
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> > Puzzle:
> > If (a, b, c) and (b, c, d) are Pythagorean triples then is a=d?
>
> Suppose the converse. Then we have 4 possibilities:
> 1) a^2 = b^2 + c^2, d^2 = b^2 - c^2
> 2) a^2 = b^2 + c^2, d^2 = c^2 - b^2
> 3) a^2 = b^2 - c^2, d^2 = b^2 + c^2
> 4) a^2 = c^2 - b^2, d^2 = b^2 + c^2
> Hence there exists a pair of coprime integers (x,y)
> such that x^2+y^2 and x^2-y^2 are both odd squares.
> But each odd square is congruent to 1 modulo 8.
> Hence 2*x^2 = 2 mod 8

[Then I screwed up, sorry: I *should* have said:]

and x is odd.

I am pondering the unfortunate consequences....

David (feeling more odd then even :-)
• ... See http://www.mathmeth.com/tom/files/pyth.pdf which claims a proof by infinite descent that x^2+y^2 and x^2-y^2 cannot both be odd squares. David (who
Message 6 of 7 , Sep 1, 2011
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> > Puzzle:
> > If (a, b, c) and (b, c, d) are Pythagorean triples then is a=d?
> >
> > Suppose the converse. Then we have 4 possibilities:
> > 1) a^2 = b^2 + c^2, d^2 = b^2 - c^2
> > 2) a^2 = b^2 + c^2, d^2 = c^2 - b^2
> > 3) a^2 = b^2 - c^2, d^2 = b^2 + c^2
> > 4) a^2 = c^2 - b^2, d^2 = b^2 + c^2
> > Hence there exists a pair of coprime integers (x,y)
> > such that x^2+y^2 and x^2-y^2 are both odd squares.
> > But each odd square is congruent to 1 modulo 8.
> > Hence 2*x^2 = 2 mod 8
>
> [Then I screwed up, sorry

See
http://www.mathmeth.com/tom/files/pyth.pdf
which claims a proof by infinite descent that
x^2+y^2 and x^2-y^2 cannot both be odd squares.

David (who miserably failed to find a simpler proof)
• ... It was apparently harder than I thought. I didn t have a proof but guessed it would be relatively easy when there is a known parametrization of all
Message 7 of 7 , Sep 1, 2011
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David wrote:
> See
> http://www.mathmeth.com/tom/files/pyth.pdf
> which claims a proof by infinite descent that
> x^2+y^2 and x^2-y^2 cannot both be odd squares.
>
> David (who miserably failed to find a simpler proof)

It was apparently harder than I thought.
I didn't have a proof but guessed it would be relatively easy when
there is a known parametrization of all Pythagorean triples.
I only did a brute force search to see there were no small solutions.

--
Jens Kruse Andersen
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