| Title: | Mathematics at DEC |
| Moderator: | RUSURE::EDP |
| Created: | Mon Feb 03 1986 |
| Last Modified: | Fri Jun 06 1997 |
| Last Successful Update: | Fri Jun 06 1997 |
| Number of topics: | 2083 |
| Total number of notes: | 14613 |
Hello,
This note will not create anything new just mearly a collection
of ideas. The question is:
"What right triangles have sides measured in integers less than
100? And can we get all of the primitive sets less than 100?"
Concider the following:
We all know and are familiar mostly with those right triangles that
come from bisecting a square or an equilateral triangle: their sides
are measured by 1, 1, sqrt(2) and by 1, 2, sqrt(3),
respectively. But the 3-4-5 right triangle is also well known.
From the 3-4-5 triangles it follows that 6,8,10 are sides of
a right triangle, also 9,12,15, and more generally 3k,4k,5k.
I any two of the sides, x,y,z, have a common factor, all three
will have it, since x^2 + y^2 = z^2.
When the greatest common factor has been divided out from the three
sides of a right triangle, the set of three quotients is called
a "primitive set".
Enjoy,
Kostas G.
<><><><><>
| T.R | Title | User | Personal Name | Date | Lines |
|---|---|---|---|---|---|
| 477.1 | CLT::GILBERT | Juggler of Noterdom | Thu May 01 1986 23:39 | 4 | |
| 477.2 | ENGINE::ROTH | Fri May 02 1986 01:22 | 7 | ||
Squaring any gaussian integer (x + i*y), x, y, integral, leads to the
formula in .1.
For much interesting info on Pythagorean triangles, see Beiler's book
'Recreations in the Theory of Numbers', published by Dover.
- Jim
| |||||
| 477.3 | Can we see all sets x,y,z each less than 100? | KEEPER::KOSTAS | Kostas G. Gavrielidis <o.o> | Fri May 02 1986 02:27 | 6 |
re. .1
Could you list all sets x,y,z each less than 100?
Kostas G.
| |||||
| 477.4 | BEING::POSTPISCHIL | Always mount a scratch monkey. | Tue May 06 1986 13:02 | 39 | |
Re .3: From the formula Peter mentioned, (x^2-y^2, 2xy, x^2+y^2), we have: x y a b c - - -- -- -- 2 1 3 4 5 3 1 8 6 10 3 2 5 12 13 4 1 15 8 17 4 2 12 16 20 4 3 7 24 25 5 1 24 10 26 5 2 21 20 29 5 3 16 30 34 5 4 9 40 41 6 1 35 12 37 6 2 32 24 40 6 3 27 36 45 6 4 20 48 52 6 5 11 60 61 7 1 48 14 50 7 2 45 28 53 7 3 40 42 58 7 4 33 56 65 7 5 24 70 74 7 6 13 84 85 8 1 63 16 65 8 2 60 32 68 8 3 55 48 73 8 4 48 64 80 8 5 39 80 89 9 1 80 18 82 9 2 77 36 85 9 3 72 54 90 9 4 65 72 97 -- edp | |||||
| 477.5 | CLT::GILBERT | Juggler of Noterdom | Tue May 06 1986 14:18 | 2 | |
How many primitive Pythagorean triples are there less than N?
(an asymptotic formula should suffice).
| |||||
| 477.6 | There are just 50 sets less than 100, 16 primitive | KEEPER::KOSTAS | Kostas G. Gavrielidis <o.o> | Wed May 07 1986 00:22 | 73 |
re. .4
Since I am not sure that you gave all the Pythagorean sets
x,y,z each less than 100, and you have not specified which
are the primitive sets I have duplicated the efford in this
reply.
There are just 50 sets, x,y,z, each less than 100,
for which x^2 + y^2 = z^2.
Only 16 of these sets are primitive.
We will use the notation: x - y - z for the primitive sets and
x, y, z for the none primitive ones.
5 - 4 - 3 primitive
10, 8, 6
13 - 12 - 5 primitive
15, 12, 9
17 - 15 - 8 primitive
20, 16, 12
25, 20, 15
25 - 24 - 7 primitive
26, 24, 10
29 - 21 - 20 primitive
30, 24, 18
34, 30, 16
35, 28, 21
37 - 35 - 12 primitive
39, 36, 15
40, 32, 24
41 - 40 - 9 primitive
45, 36, 27
50, 40, 30
50, 48, 14
51, 45, 24
52, 48, 20
53 - 45 - 28 primitive
55, 44, 33
58, 42, 40
60, 48, 36
61 - 60 - 11 primitive
65, 52, 39
65 - 56 - 33
65, 60, 25
65 - 63 - 16 primitive
68, 60, 32
70, 56, 42
73 - 55 - 48 primitive
74, 70, 24
75, 60, 45
75, 72, 21
78, 72, 30
80, 64, 48
82, 80, 18
85, 68, 51
85, 75, 40
85 - 77 - 36 primitive
85 - 84 - 13 primitive
87, 63, 60
89 - 80 - 39 primitive
90, 72, 54
91, 84, 35
95, 76, 57
97 - 72 - 65 primitive
Enjoy,
Kostas G.
<><><><><>
| |||||
| 477.7 | Fermat (1601 - 1665) ... | KEEPER::KOSTAS | Kostas G. Gavrielidis <o.o> | Wed May 07 1986 00:30 | 27 |
Some history ...
It has been proved that there are no integer values for x, y, z,
that satisfy x^3 + y^3 = z^3 or x^4 + y^4 = z^4, etc.
Fermat (1601-1665) wrote in the margin of one of his books that
he had found "a most wonderful proof" that there were no integers
that could satisfy the relation:
n n n
x + y = z
for values of n more than 2, but that "the margin was too small
to contain it."
A great number of mathematicians have done a vast amount of work
in trying to prove Fermat's theorem, and it has been proved for
a very extensive range of values of n, but not for all values.
Does any one remember the proof for n = 4?
Enjoy,
Kostas G.
| |||||
| 477.8 | Here's a pointer | METOO::YARBROUGH | Wed May 21 1986 15:31 | 5 | |
There is a reasonably concise proof in Beiler's "Recreations in
the Theory of Numbers". In summary, it assumes that such a solution
exists and then shows how that implies that a smaller solution also
must exist. Since the solutions are positive integers, no solution
can exist.
| |||||
| 477.9 | John Pell's (1610 - 1685) equation . . . | 7481::KOSTAS | Kostas G. Gavrielidis <o.o> | Sat May 24 1986 15:45 | 50 |
Pells equation:
2 2
x - qy = 1 (q <> 0)
for x, y > 0
This equation could be called Fermat's equation. However due to
Euler's mistake in attributing the equation to the English
mathematician John Pell (1610 - 1685) this equation is called Pell's
equation.
Solutions to Pell's equations for nonsquare integers q satisfying
1 < q < 30, and x,y > 0 follow.
q x y
------------------
2 3 2
3 2 1
5 9 4
6 5 2
7 8 3
8 3 1
10 19 6
11 10 3
12 7 2
13 649 180
14 15 4
15 4 1
17 33 8
18 17 4
19 170 39
20 9 2
21 55 12
22 197 42
23 24 5
24 5 1
26 51 10
27 26 5
28 127 24
29 9801 1820
Enjoy,
Kostas G.
| |||||
| 477.10 | positive solutions of Pell's eq. x^2-42y^2 =1? | 7480::KOSTAS | Kostas G. Gavrielidis <o.o> | Sat May 24 1986 16:03 | 15 |
Before I forget,
What are the positive solutions of the Pell's equation:
2 2
x - 42y = 1
Enjoy,
Kostas G.
| |||||
| 477.11 | And now for a hard Pellian problem! | METOO::YARBROUGH | Tue May 27 1986 16:45 | 1 | |
re .-1: x=13, y=2. Care to try for q=61? That's a real toughie. | |||||
| 477.12 | also | 7481::KOSTAS | Kostas G. Gavrielidis <o.o> | Wed May 28 1986 01:58 | 9 |
re. .-1
for x^2 - 42*y^2 = 1
another solution in addition to x = 13, y = 2
x = 337, y = 52.
| |||||
| 477.13 | one solution to Pell's equation when q=61 | 7481::KOSTAS | Kostas G. Gavrielidis <o.o> | Wed May 28 1986 02:01 | 15 |
re. .-2
2 2
for Pell's equation (i.e. x - 61y = 1 )
x = 176631049
and
y = 226153980
Enjoy,
Kostas G.
| |||||
| 477.14 | And now for the hardest Pellian problem! | 7481::KOSTAS | Kostas G. Gavrielidis <o.o> | Wed May 28 1986 02:05 | 28 |
re. .11
This has to be the hardest Pellian problem
when q = 1621, which means find solutions to the equation:
2 2
x - qy = 1
when q = 1621.
Note:
When q = 1620 then x = 161 and y = 4
But when q = 1621 x has 76 digits and
y has 77 digits
Enjoy,
Kostas G.
| |||||
| 477.15 | Nah, that's not the worst | METOO::YARBROUGH | Wed May 28 1986 12:52 | 5 | |
Beiler's "Recreations in the Theory of Numbers" gives the 75-digit
solution to this problem and also the 150+ - digit solution to the
case Q= 9781. I won't try to transcribe them here. Obviously there
is no 'worst case' here; solutions just keep getting (randomly) larger,
with lots of easy cases in between.
| |||||
| 477.16 | Another solution to X^2 - 42y^2 = 1 | 7481::KOSTAS | Kostas G. Gavrielidis <o.o> | Wed May 28 1986 13:26 | 12 |
re. .11
another solution to x^2 - 42y^2 = 1
in addition to : x = 13, y = 2
x = 337, y = 52
is also x = 8749, y = 1350
KGG
| |||||
| 477.17 | pythag. triples - source of endless fascination | AUSSIE::GARSON | Sun Feb 21 1993 01:49 | 28 | |
| 477.18 | Restatement of primitive form | MIMS::GULICK_L | When the impossible is eliminated... | Thu Feb 25 1993 04:08 | 14 |
RE: -1
Most of the proofs follow directly from the fact that:
a = m**2 - n**2
b = 4mn
c = m**2 + n**2
The primitive solution which is necessary and sufficient for
all solutions with gcd(a,b,c)=1.
Lew
| |||||
| 477.19 | AUSSIE::GARSON | Thu Feb 25 1993 08:11 | 6 | ||
re .18
>Most of the proofs follow directly from the fact that:
Indeed true but proving those relationships becomes bigger than the
original statement.
| |||||
| 477.20 | b = 2mn | RANGER::BRADLEY | Chuck Bradley | Thu Feb 25 1993 12:35 | 2 |
re .18 i think you mean b=2mn, not 4mn. | |||||
| 477.21 | Yes. 2mn | MIMS::GULICK_L | When the impossible is eliminated... | Fri Feb 26 1993 06:38 | 11 |
> i think you mean b=2mn, not 4mn. Yes. Thanks. Re: .19 Of course, but this is in most elementary number theory books, and I didn't want to simply reproduce here. Lew | |||||
| 477.22 | for completeness | AUSSIE::GARSON | Fri Feb 26 1993 22:06 | 32 | |