Integer right triangles
Rich Grise
whatever it's called. One day, they were talking about right triangles,
and all their foibles.
Well, that got me to thinking - everybody knows about the 3,4,5 and the
5,12,13 right triangles; I'm wondering if, other than their multiples
(6,8,10; 10,24,26 etc.) are there any other integer right triangles?
If so, is(are?) there an infinite number of them? (I think it should be
"is" - "an infinite number" is a singular phrase, after all. But that's
not the question, just checking my own English. ;-) )
Thanks,
Rich
Mark Brader
> Well, that got me to thinking - everybody knows about the 3,4,5 and the
> 5,12,13 right triangles; I'm wondering if, other than their multiples
> (6,8,10; 10,24,26 etc.) are there any other integer right triangles?
Yes.
> If so, is(are?) there an infinite number of them?
Yes. See e.g. http://mathworld.wolfram.com/PythagoreanTriple.html
> (I think it should be
> "is" - "an infinite number" is a singular phrase, after all. But that's
> not the question, just checking my own English. ;-) )
"Are". "A number of Xes" is construed to be about the Xes, not the
number, so it's plural. "The number of Xes" is about the number, so
*it's* singular.
--
Mark Brader | "...the average homeowner should expect...
Toronto | meteor damage every hundred million years."
m...@vex.net | --Robert Nemiroff & Jerry Bonnell
My text in this article is in the public domain.
Eric Sosman
> Rich Grise:
>> Well, that got me to thinking - everybody knows about the 3,4,5 and the
>> 5,12,13 right triangles; I'm wondering if, other than their multiples
>> (6,8,10; 10,24,26 etc.) are there any other integer right triangles?
>
> Yes.
>
>> If so, is(are?) there an infinite number of them?
>
> Yes. See e.g. http://mathworld.wolfram.com/PythagoreanTriple.html
>
>> (I think it should be
>> "is" - "an infinite number" is a singular phrase, after all. But that's
>> not the question, just checking my own English. ;-) )
>
> "Are". "A number of Xes" is construed to be about the Xes, not the
> number, so it's plural. "The number of Xes" is about the number, so
> *it's* singular.
Hence, "A Rare Entries contest has a number of questions,
which are ten."
(Hey -- Wait a minutes, ...)
--
Eric Sosman
eso...@ieee-dot-org.invalid
bartc
"Mark Brader" <m...@vex.net> wrote in message
news:pOKdnfEv29vAWLvW...@vex.net...
> Rich Grise:
>> Well, that got me to thinking - everybody knows about the 3,4,5 and the
>> 5,12,13 right triangles; I'm wondering if, other than their multiples
>> (6,8,10; 10,24,26 etc.) are there any other integer right triangles?
>
> Yes.
>
>> If so, is(are?) there an infinite number of them?
>
> Yes. See e.g. http://mathworld.wolfram.com/PythagoreanTriple.html
OK, that's a little bit more in-depth than my attempt:
n:=100
for a:=1 to n do
for b:=a to n do
for c:=b to n*2 do
if (a%3<>b%4) and (a%5<>b%12) then # %=integer divide
if a*a+b*b=c*c then
println a,b,c
fi
fi
od
od
od
--
Bartc
The Qurqirish Dragon
Since the difference between any two adjacent squares is an odd
number, and every odd number is a difference, it should be easy for
you to show that any odd number greater than 3 will get you a triple
that is not in the family of any other found in this manner.
We note that (n+1)^2 - n^2 = 2n - 1 to get these examples:
5^2 - 4^2 = 9, so 3,4,5 works
13^2 - 12^2 = 25, so 5,12,13 works
25^2 - 24^2 = 49, so 7,24,25 works
41^2 - 40^2 = 81, so 9,40,41 works
and so on.
Each one creates a family of multiples. (9,40,41; 18,80,82;
27,120,123; &c.)
Further, this is not the only collection of families:
we can see that (n+2)^2 - n^2 = 4n + 4, which gets all multiples of 4.
Thus, any even number will give a family in the same manner as above,
although some of them may be a multiple of the first collection:
5^2 - 3^2 = 16, so 4,3,5 works (already have this one)
10^2 - 8^2 = 36, so 6,8,10 works (a multiple of 3,4,5)
17^2 - 15^2 = 64, so 8,15,17 works (and this is a new one!)
37^2 - 35^2 =144, so 12,35,37 works.
See if you can find other collections. (just note that as I showed
above, they may not be disjoint)
Mensanator
> "Mark Brader" <m...@vex.net> wrote in message
>
> news:pOKdnfEv29vAWLvW...@vex.net...
>
> > Rich Grise:
> >> Well, that got me to thinking - everybody knows about the 3,4,5 and the
> >> 5,12,13 right triangles; I'm wondering if, other than their multiples
> >> (6,8,10; 10,24,26 etc.) are there any other integer right triangles?
>
> > Yes.
>
> >> If so, is(are?) there an infinite number of them?
>
>
> OK, that's a little bit more in-depth than my attempt:
>
> n:=100
>
> for a:=1 to n do
> for b:=a to n do
> for c:=b to n*2 do
> if (a%3<>b%4) and (a%5<>b%12) then # %=integer divide
> if a*a+b*b=c*c then
> println a,b,c
> fi
> fi
> od
> od
> od
Wher's 3,4,5?
>
> --
> Bartc
Dan Hoey
> Rich Grise:
>> Well, that got me to thinking - everybody knows about the 3,4,5 and the
>> 5,12,13 right triangles; I'm wondering if, other than their multiples
>> (6,8,10; 10,24,26 etc.) are there any other integer right triangles?
>
> Yes.
>
>> If so, is(are?) there an infinite number of them?
>
> Yes. See e.g. http://mathworld.wolfram.com/PythagoreanTriple.html
>
If u > v are positive integers, relatively prime and not both
odd, then the numbers
2uv, u^2-v^2, u^2+v^2
form a Pythagorean triple, and every Pythagorean triple is generated
by exactly one such pair (u,v).
Dan
bartc
The OP said he wasn't interested in 3,4,5 or 5,12,13 triangles or their
multiples, so the line full of %s ignores those (multiples of other
triangles will still appear though).
This is the output for short sides up to 100:
7 24 25
9 40 41
11 60 61
13 84 85
14 48 50
16 30 34
16 63 65
18 80 82
20 21 29
20 99 101
21 72 75
24 45 51
24 70 74
28 45 53
28 96 100
32 60 68
33 56 65
36 77 85
39 80 89
40 42 58
40 75 85
48 55 73
48 90 102
56 90 106
60 63 87
60 91 109
65 72 97
80 84 116
--
Bartc
Matthew Russotto
>That page has a lot of stuff, making it easy to miss equation 11:
>
>If u > v are positive integers, relatively prime and not both
>odd, then the numbers
>
> 2uv, u^2-v^2, u^2+v^2
>
>form a Pythagorean triple, and every Pythagorean triple is generated
>by exactly one such pair (u,v).
>
Every primitive pythagorean truple is generated by such a pair, and
some of the non-primitive ones, but not all of them.
To get all of them exactly once you need
k(2uv), k(u^2 - v^2), k(u^2 + v^2),
k > 0, u > v > 0, gcd(u,v) = 1, exactly one of u,v, odd.
--
The problem with socialism is there's always
someone with less ability and more need.
Philippe 92
> In article <hg8prk$1ll$1...@ra.nrl.navy.mil>, Dan Hoey <hao...@aol.com> wrote:
>> That page has a lot of stuff, making it easy to miss equation 11:
>>
>> If u > v are positive integers, relatively prime and not both
>> odd, then the numbers
>>
>> 2uv, u^2-v^2, u^2+v^2
>>
>> form a Pythagorean triple, and every Pythagorean triple is generated
>> by exactly one such pair (u,v).
>>
>
> Every primitive pythagorean truple is generated by such a pair, and
> some of the non-primitive ones, but not all of them.
Hi,
no non primitive triangle can be generated by that.
(re-read the 1st sentence)
Dan Hoey just forgot "primitive" Pythagorean triple in the conclusion.
And of course all Pythagorean triples are generated by multiples of
the primitive ones : k(2uv), k(u^2 - v^2), k(u^2 + v^2))
As you say, ommitting the factor k and **relaxing** the conditions on
u and v (otherwise it would generate only primitive triangles) does not
suffice to generate all (non primitive) triangles.
For instance a triangle (3*3, 3*4, 3*5) cant't be generated in that way
as 3*5 = u^2 + v^2 has no solution, you need then the factor k = 3.
Triangle (2*3, 2*4, 2*5), may be generated in two different ways :
u = 3, v = 1, k = 1 : 2*3*1 = 6, 3^2 - 1^2 = 8, 3^2 + 1^2 = 10
and u = 2, v = 1, k = 2 :
2*(2*2*1) = 8, 2*(2^2 - 1^1) = 6, 2*(2^2 + 1^1) = 10
Hence the condition on u and v, with the factor k, to generate all
triangles *only once*.
And also, as the Wolfram site is very rich, and others gave lots of
valuable answers, I won't add a lot about this problem.
Just to mention two specific families of solutions,
The "Pythagora"(*) family :
a^2 + b^2 = (b+1)^2 (generated by v = u+1)
The "Plato" family :
a^2 + b^2 = (a+2)^2 (generated by v = 1)
And there are triangles wich don't belong to these families, the
smallest (primitive) being 29^2 = 20^2 + 21^2
I mention this because these are the keys for disecting a square into
2 smaller squares in just 4 pieces (general case requires 5 pieces)
See Frederikson's book "Dissection: Plane & Fancy", or my web site at :
<http://mathafou.free.fr/pbg_en/sol110d.html>
(*) the name of these families are from Frederickson's book, Wolfram
mentions the Plato's as being from Pythagora, hence confusing...
(who is right ???) See :
<http://en.wikipedia.org/wiki/Pythagorean_triple>
"A special case: the Platonic sequence"
<http://www.math.nmsu.edu/~history/book/euclidpt.pdf>
Seems to be a well documented historical reference.
Regards.
--
Philippe C., mail : chephip, with domain free.fr
site : http://mathafou.free.fr/ (mathematical recreations)
Bill
Here' s one infinite set. A = any odd number; (B+C) = A^2; C - B =
`1.
Here's another A = an even number; (B+C) = (A^2)/2 ; C
- B = `2.
regards, Bill J
Edward McArdle
<8e0fbc7c-b793-4e7f...@m38g2000yqd.googlegroups.com>, Bill
<b92...@yahoo.com> wrote:
>On Dec 14, 2:56=A0pm, Rich Grise <richgr...@example.net> wrote:
>> Lately, on insomniac TV, I've been watching "Geometry for teachers" or
>> whatever it's called. One day, they were talking about right triangles,
>> and all their foibles.
>>
>> Well, that got me to thinking - everybody knows about the 3,4,5 and the
>> 5,12,13 right triangles; I'm wondering if, other than their multiples
>> (6,8,10; 10,24,26 etc.) are there any other integer right triangles?
>>
>> If so, is(are?) there an infinite number of them? (I think it should be
>> "is" - "an infinite number" is a singular phrase, after all. But that's
>> not the question, just checking my own English. ;-) )
>>
>> Thanks,
>> Rich
>
>Here' s one infinite set. A =3D any odd number; (B+C) =3D A^2; C - B =3D
>`1.
>Here's another A =3D an even number; (B+C) =3D (A^2)/2 ; C
>- B =3D `2.
>
>regards, Bill J
I think you get them all from
A^2 + B^2
A^2 - B^2
2AB
There is also a formula for 60 and 120 degree triangles, which I found
once, but I can't remember at the moment. It was in and old issue of the
Journal of Recreational Mathematics. It's around somewhere.
--
Edward McArdle
Bill
wrote:
If K is a divisor of A^2 and K < A,; then B = (A^2 - K^2) / (2*K) &;
C = B + K.
Note; K = 1 is a valid K for all A.
.
This generates the complete family of a every A for 3 <= A <
infinity.
Philippe 92
> On Dec 15 2009, 4:12ï¿œam, The Qurqirish Dragon <qurqiri...@aol.com>
> wrote:
>> On Dec 14, 5:56ï¿œpm, Rich Grise <richgr...@example.net> wrote:
>>
>>> Lately, on insomniac TV, I've been watching "Geometry for teachers" or
>>> whatever it's called. One day, they were talking about right triangles,
>>> and all their foibles.
>>
>>> If so, is(are?) there an infinite number of them?
>>
>> number, and every odd number is a difference, it should be easy for
>> you to show that any odd number greater than 3 will get you a triple
>> that is not in the family of any other found in this manner.
>> We note that (n+1)^2 - n^2 = 2n - 1 to get these examples:
>> Further, this is not the only collection of families:
>> we can see that (n+2)^2 - n^2 = 4n + 4, which gets all multiples of 4.
>>
>> See if you can find other collections. (just note that as I showed
>> above, they may not be disjoint)
>
> If K is a divisor of A^2 and K < A,; then B = (A^2 - K^2) / (2*K) &;
> C = B + K.
> Note; K = 1 is a valid K for all A.
A = 2, K = 1, B = 3/2 ???
Correction :
If K is a divisor of A^2 and K and A are of same parity etc...
. ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
This is a valuable method to get triples (by scanning all divisors
of A^2) from a given side A, and seems better than my own method at
<http://mathafou.free.fr/exe_en/exeside.html>, as it handles the two
cases in only one and doesn't require eliminating duplicates.
However this method doesn't generate all *primitive* triples *only
once*
as does the "standard method" already mentionned in this thread :
All primitive triples (GCD(a,b,c)=1) are generated once each by :
a = m^2 - n^2
b = 2*m*n
c = m^2 + n^2
m, n any integer with :
GCD(m,n) = 1 and m+n != 0 mod 2 (m and n of opposite parity)
Multiplying by k then generates all triples once each.
Philippe 92
> Bill <b92...@yahoo.com> wrote:
>> Rich Grise <richgr...@example.net> wrote:
>>>
>>> Well, that got me to thinking - everybody knows about the 3,4,5 and the
>>> 5,12,13 right triangles; I'm wondering if, other than their multiples
>>> (6,8,10; 10,24,26 etc.) are there any other integer right triangles?
>>>
>>> If so, is(are?) there an infinite number of them?
>>
>> Here's another
>
> A^2 + B^2
> A^2 - B^2
> 2AB
>
Hi,
Not really, to get really all of them you need to include a factor K.
For instance the (9, 12, 15) triple can't be obtained by your formula
as 15 can't be a sum of two squares A^2 + B^2
But it is 3*(1^2 + 2^2), so that this triple is obtained by
K*(A^2 + B^2)
K*(A^2 - B^2)
2*K*A*B
with K = 3, A = 2, B = 1
But your formula is good (without the K) if you restrict to
*primitive* triples (that is GCD = 1). But if you want *only*
primitives, then you must also restrict A and B :
GCD(A,B) = 1 and A+B != 0 mod 2
As this question has been debated widely along this thread, my answer
is mainly because :
> There is also a formula for 60 and 120 degree triangles, which I found
> once, but I can't remember at the moment. It was in and old issue of the
> Journal of Recreational Mathematics. It's around somewhere.
Yes, it is also discussed sometimes in newsgroups.
Here is my own derivation about this 60ￜ triangles :
<http://mathafou.free.fr/pba_en/sol021b.html>
The resulting formulas generating all primitive triples (GCD = 1)
only once each is in two cases :
for any r,s with r < s or r > 3s , GCD(r,s) = 1 and r != 0 mod 3
r and s odd
u = |3s^2 - r^2 - 2rs|/4
v = |3s^2 - r^2 + 2rs|/4
x = (r^2 + 3s^2)/4
else
u = |3s^2 - r^2 - 2rs|
v = |3s^2 - r^2 + 2rs|
x = r^2 + 3s^2
There are other derivations, based upon the "Eisenstein integers".
These are the key for representing numbers as u^2 + 3v^2, in the same
manner than Gauss integers are the key for u^2 + v^2,
Spamless
> On Dec 15 2009, 4:12=A0am, The Qurqirish Dragon <qurqiri...@aol.com>
> wrote:
>> On Dec 14, 5:56=A0pm, Rich Grise <richgr...@example.net> wrote:
>>
>> > Lately, on insomniac TV, I've been watching "Geometry for teachers" or
>> > whatever it's called. One day, they were talking about right triangles,
>> > and all their foibles.
>>
>> > Well, that got me to thinking - everybody knows about the 3,4,5 and the
>> > 5,12,13 right triangles; I'm wondering if, other than their multiples
>> > (6,8,10; 10,24,26 etc.) are there any other integer right triangles?
>>
>> > If so, is(are?) there an infinite number of them? (I think it should be
>> > "is" - "an infinite number" is a singular phrase, after all. But that's
>> > not the question, just checking my own English. ;-) )
Pythagnorean Triples! You can search for that term.
Primitive Pythagorean Triples:
Take two positive integers, m and n, m>n, one of them even
and one odd, relatively prime (no common divisor).
Set c=m^2 + n^2, a=2mn, b=m^2 - n^2.
(a,b,c) is a primitive pythagorean triple. Others are multiples
of the primitive ones.
E.g. m=2,n=1 : c=5, a=4, b=3: (3,4,5) is a primitive Pythagorean
Triple. 2*(3,4,5) = 6,8,10 is a non-primitive one.
m=7,n=4 gives c=65,a=56,b=33. 65^2=33^2+56^2 - A Pythagorean Triple/
Now consider almost equilateral Pythagorean Triples,
ones where b-a = +/-1 (the two sides, as integers, are
as close as possible - differing just by one).
These are clearly primititve and one has m^2-n^2-2mn=+/-1,
(n+m)^2 - 2m^2 = +/-1, Pell's equation!
(A^2-D*B^2=1) with infinitely many solutions.
Try x+y*SQRT(2)=(1+SQRT(2))^k and m=y, n=x-y, and a=abs(m^2-n^2)
(to make it positive) [x, y integers], b=2mn, c=m^2+n^2
E.g.
k = 4: (119,120,169): 119^2 + 120^2 = 169^2 - A Pythagorean Triple
k = 5: (696,697,985): 696^2 + 697^2 = 985^2 - A Pythagorean Triple
k = 6: (4059,4060,5741): 4059^2 + 4060^2 = 5741^2 - A Pythagorean Triple
Spamless
>
> Pythagnorean Triples! You can search for that term.
>
> Primitive Pythagorean Triples:
> Take two positive integers, m and n, m>n, one of them even
> and one odd, relatively prime (no common divisor).
> Set c=m^2 + n^2, a=2mn, b=m^2 - n^2.
> (a,b,c) is a primitive pythagorean triple. Others are multiples
> of the primitive ones.
>
> E.g. m=2,n=1 : c=5, a=4, b=3: (3,4,5) is a primitive Pythagorean
> Triple. 2*(3,4,5) = 6,8,10 is a non-primitive one.
>
> m=7,n=4 gives c=65,a=56,b=33. 65^2=33^2+56^2 - A Pythagorean Triple/
>
> Now consider almost equilateral Pythagorean Triples,
ISOCELES da.n it.