- David wrote:
> Here is the first set of 9:

A gap cycle of 6, 8, 10 in some order can only get one longer than

>

> {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]

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

becomes inadmissible modulo 5.

(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

it becomes inadmissible modulo 7.

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 - --- In primenumbers@yahoogroups.com,

"Jens Kruse Andersen" <jens.k.a@...> wrote:

> Puzzle:

Yes.

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

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.

David Broadhurst - --- In primenumbers@yahoogroups.com, "djbroadhurst" <d.broadhurst@...> wrote:

> > Puzzle:

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

> > 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

and x is odd.

I am pondering the unfortunate consequences....

David (feeling more odd then even :-) - --- In primenumbers@yahoogroups.com,

"djbroadhurst" <d.broadhurst@...> wrote:

> > Puzzle:

See

> > 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

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) - David wrote:
> See

It was apparently harder than I thought.

> 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)

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