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I hope this question is clear. If not, please suggest a change to make
the
intention clearer (assuming you can work the intention out)...
Most of us are familiar with the (3, 4, 5) right-triangle. 5 is the
smallest integer hypotenuse which supports two other sides of a right- triangle with integer length. There are many other right-triangles with integer sides, such as: (5, 12, 13) and (8, 15, 17). These triples are considered primitive as the terms do not share a common factor.
On the other hand, although (6, 8, 10) is a right-triangle, it is NOT primitive as the elements share a common factor, 2.
Can you find the first four terms in the series where a(n) is the least hypotenuse of which 2^(n-1) Pythagorean triples are primitive? So, 5 is
the smallest and supports one triple. Can you find a hypotenuse that
supports two discrete primitive Pythagorean triples, four and eight?
Good luck.
I hope this question is clear. If not, please suggest a change to make the >intention clearer (assuming you can work the intention out)...
Most of us are familiar with the (3, 4, 5) right-triangle. 5 is the
smallest integer hypotenuse which supports two other sides of a right- >triangle with integer length. There are many other right-triangles with >integer sides, such as: (5, 12, 13) and (8, 15, 17). These triples are >considered primitive as the terms do not share a common factor.
On the other hand, although (6, 8, 10) is a right-triangle, it is NOT >primitive as the elements share a common factor, 2.
Can you find the first four terms in the series where a(n) is the least >hypotenuse of which 2^(n-1) Pythagorean triples are primitive? So, 5 is
the smallest and supports one triple. Can you find a hypotenuse that >supports two discrete primitive Pythagorean triples, four and eight?
Good luck.
On Fri, 20 Jun 2025 08:11:10 -0000 (UTC), David Entwistle ><qnivq.ragjvfgyr@ogvagrearg.pbz> wrote:
I hope this question is clear. If not, please suggest a change to make the >>intention clearer (assuming you can work the intention out)...
Most of us are familiar with the (3, 4, 5) right-triangle. 5 is the >>smallest integer hypotenuse which supports two other sides of a right- >>triangle with integer length. There are many other right-triangles with >>integer sides, such as: (5, 12, 13) and (8, 15, 17). These triples are >>considered primitive as the terms do not share a common factor.
On the other hand, although (6, 8, 10) is a right-triangle, it is NOT >>primitive as the elements share a common factor, 2.
Can you find the first four terms in the series where a(n) is the least >>hypotenuse of which 2^(n-1) Pythagorean triples are primitive? So, 5 is >>the smallest and supports one triple. Can you find a hypotenuse that >>supports two discrete primitive Pythagorean triples, four and eight?
Good luck.
Unfortunately, I do not have access to a decent progamming
language at the moment and so I messed around with spreadsheets.
Essentially, I used the well known Euclid expressions and searched.
a(2) = 145
with the primitive triples (17, 144, 145) and (24, 143, 145).
I could not make progress on the elegant front. For the above
a(2) problem, we need to find a pair of doublets (u1, v1)
and (u2, v2) such that
u1^2 + v1^2 = u2^2 + v2^2 (same hypotenuse)
u1 and v1 are coprime with only one being even and
u2 and v2 are coprime with only one being even.
Not sure if this is even the correct approach!
For n = 3, 4, .... the conditions of the triplets,
quadraplets, ..... grow and it seems a rather
unwieldly approach.
Waiting for others to make the breakthrough.
SPOILER
a(2) = 145
with the primitive triples (17, 144, 145) and (24, 143, 145).
Ooops! Just saw Ilan Mayer's post.
I could not make progress on the elegant front. For the above
a(2) problem, we need to find a pair of doublets (u1, v1)
and (u2, v2) such that
u1^2 + v1^2 = u2^2 + v2^2 (same hypotenuse)
u1 and v1 are coprime with only one being even and
u2 and v2 are coprime with only one being even.
Not sure if this is even the correct approach!
For n = 3, 4, .... the conditions of the triplets,
quadraplets, ..... grow and it seems a rather
unwieldly approach.
Waiting for others to make the breakthrough.
"a(n) is the product of the first n primes congruent to 1 (mod 4)".
Does anyone have any insight into why that would be so?
On Fri, 20 Jun 2025 13:30:12 -0400, Charlie Roberts
.....
"The number of "primitive" triples for any side of a Pythagorean
triple is 2^(n-1), where n is the number of unique prime factors of
that side length. There may be more imprimitives than this but not >primitives."
Well, the goose may have finally been cooked (for me, at least).
"The number of "primitive" triples for any side of a Pythagorean
triple is 2^(n-1), where n is the number of unique prime factors of
that side length. There may be more imprimitives than this but not primitives."
but no proof (or pointers to a proof) is given.
In article <1034dml$7dg8$1@dont-email.me>,
David Entwistle <qnivq.ragjvfgyr@ogvagrearg.pbz> wrote:
"a(n) is the product of the first n primes congruent to 1 (mod 4)".
Does anyone have any insight into why that would be so?
Warning, incoming maths! But I've tried to make it friendly and
legible.
It comes down to factorisations in the "Gaussian integers", which are
numbers of the form a+bi, with a, b integers and i = sqrt(-1).
For example, using these we can now factorise 5 = (2+i)(2-i), but we
can't factorise 7 any more than it already is.
It can be shown (proof omitted!) that all integer primes which are 1 mod
4 factorise as (a+bi)(a-bi) for some a, b, and moreover that there is
only one such factorisation. (Well, only one up to moving irrelevant
factors of -1 or +-i around, e.g. writing 5 = (1+2i)(1-2i) instead).
The prime 2 also factorises, as (1+i)(1-i), but that's not going to
matter to us.
And integer primes which are 3 mod 4 don't factorise. This is easier to show. Suppose that p = (a+bi)(c+di) for prime p which is 3 mod 4. Multiplying both sides by their complex conjugates gives p^2 = (a^2+b^2)(c^2+d^2), a factorisation in the integers. So we must have
a^2+b^2 = c^2+d^2 = p. However, you can't write any number which is 3
mod 4 as a sum of two squares: working mod 4, even numbers square to 0
and odd numbers square to 1, so a^2+b^2 can only ever be 0 or 1 or 2 mod
4, and so can never equal p since that's 3 mod 4.
Suppose that we have a primitive Pythagorean triple: a^2 + b^2 = c^2.
First, c must be odd. If c is even then the right-hand side is 0 mod
4, and then a and b must both be even. (As above: if exactly one is
odd then the left-hand side is 1 mod 4, and if both are odd then it's
2 mod 4.) But if all three are even then the triple isn't primitive.
Can c have any prime factor p which is 3 mod 4? No. Suppose it did,
and factorise the left-hand side as (a+bi)(a-bi). Since p itself
doesn't factorise in the Gaussians, we get that each of a+bi and a-bi
is a multiple of p, so that a and b themselves are multiples of p, and
so again the triple isn't primitive.
So c's prime factors are all of the form 1 mod 4. Now we factorise each
of those primes in the Gaussians. To make a^2+b^2, we aim for
(a+bi)(a-bi), by gathering together one of each pair of complex factors making up each prime to form a+bi, and then gathering their conjugates
to form a-bi.
For example, c = 1105 = 5 x 13 x 17.
In the Gaussians, c = (2+i)(2-i) x (2+3i)(2-3i) x (4+i)(4-i)
So the factorisation of c^2 has each of these twice.
Grouping gives the following choices for a+bi.
[(2+i)(2+3i)(4+i)]^2 = [-4 + 33i]^2 = -1073 + 264i
[(2+i)(2+3i)(4-i)]^2 = [12 + 31i]^2 = -817 + 744i
[(2+i)(2-3i)(4+i)]^2 = [32 - 9i]^2 = 943 - 576i
[(2+i)(2-3i)(4-i)]^2 = [24 - 23i]^2 = 47 - 1104i
(While there are eight choices of the plus or minus signs for the three factors, there are only four products listed because in each case our
product gives one of a+bi and a-bi and then the other is forced.
Equivlently, we could assume that we'll choose plus for the first factor
and then choose the other two freely. This is where 2^{n-1} comes from
the final answer.)
So we get 1105^2 equals:
1073^2 + 264^2
817^2 + 744^2
943^2 + 576^2
47^2 + 1104^2
Note that on the way we also found all ways to write 1105 itself as a
sum of two squares:
4^2 + 33^2
12^2 + 31^2
32^2 + 9^2
24^2 + 23^2
If the prime factors aren't distinct then we can still do this but we
get less choice. For example, c = 325 = 5 x 5 x 13.
In the Gaussian, c = (2+i)^2 x (2-i)^2 x (2+3i) x (2-3i)
We can now only group them as:
[(2+i)^2 (2+3i)]^2 = [-6+17i]^2 = -253 - 204i
[(2+i)^2 (2-3i)]^2 = [18-i]^2 = 323 - 36i
We can't choose (2+i)(2-i)(2+3i) because then both it and its complex conjugate are multiple of 5, and the triple isn't primitive.
So we get 325^2 has two non-primitive ways:
253^2 + 204^2
323^2 + 36^2
Hence, to get the smallest number with a given number of distinct
ways, we need c to be a product of the first N primes which are 1 mod
4, and then there are 2^{N-1} ways.
Gareth
Warning, incoming maths! But I've tried to make it friendly and
legible.
It comes down to factorisations in the "Gaussian integers", which are
numbers of the form a+bi, with a, b integers and i = sqrt(-1).
For example, using these we can now factorise 5 = (2+i)(2-i), but we
can't factorise 7 any more than it already is.
It can be shown (proof omitted!) that all integer primes which are 1 mod
4 factorise as (a+bi)(a-bi) for some a, b, and moreover that there is
only one such factorisation. (Well, only one up to moving irrelevant
factors of -1 or +-i around, e.g. writing 5 = (1+2i)(1-2i) instead). The prime 2 also factorises, as (1+i)(1-i), but that's not going to matter
to us.
In article <sm7r5k1qn6jir93tn8jl4nlo75j5uqiufq@4ax.com>,
Charlie Roberts <croberts@gmail.com> wrote:
Hello. Yesterday, I posted some maths waffle in a reply elsewhere in
this thread. I mention it partly in case you missed it, but partly in
case it hasn't shown up at all. (I haven't posted to a newsgroup for
ages and might have got it wrong!)
Gareth