Thanks,
Zak
This is part of Problem D21 in Guy, Unsolved Problems in Number Theory.
The triple above was found by Charles L Shedd in 1945.
In 1986, Rathbun found three more, one of which has generators
(1610, 869), (2002, 1817), (2622, 143) [The triangle with generators
(a, b) is 2ab, a^2 - b^2, a^2 + b^2]. A 5th triple was found
independently by Hoey and Rathbun.
It appears to be unknown whether there is an infinity of such triples,
also whether there are any quadruples.
--
Gerry Myerson (ge...@maths.mq.edi.ai) (i -> u for email)
77, 38; 78, 55; 138, 5.
1610, 869; 2002, 1817; 2622, 143.
2035, 266; 3306, 61; 3422, 55.
2201, 1166; 2438, 2035; 3565, 198.
7238, 2465; 9077, 1122; 10434, 731.
(For anyone coming in late, each pair a, b expands to
a primitive Pythagorean triple, 2ab, a^2 - b^2, a^2 + b^2;
the three in each line above have the same value for 2ab(a^2 - b^2)
[hence, the same area when interpreted as right triangles];
these are the five triples of triples listed in D21 of Guy's
Unsolved Problems in Number Theory)
It is difficult to find the 6th one, as the area searched is out to 10^21.
Using Paul Whitlock's parametrization to simplify the search we find:
n,m = (5,6) Found 3 - (138,5) gcd:1 (78,55) gcd:1 (77,38) gcd:1
m,n = (8,3) Found 3 - (88,3) gcd:1 (55,13) gcd:1 (51,40) gcd:1
m,n = (16,7) Found 4 - (368,35) gcd:1 (299,259) gcd:1 (259,144) gcd:1 (259,155)
gcd:1
n,m = (22,31) Found 3 - (3565,198) gcd:1 (2438,2035) gcd:1 (2201,1166) gcd:1
m,n = (32,1) Found 3 - (1056,29) gcd:1 (992,35) gcd:1 (553,377) gcd:1
m,n = (46,11) Found 3 - (2622,143) gcd:1 (2002,1817) gcd:1 (1610,869) gcd:1
m,n = (58,1) Found 3 - (3422,55) gcd:1 (3306,61) gcd:1 (2035,266) gcd:1
m,n = (94,17) Found 3 - (10434,731) gcd:1 (9077,1122) gcd:1 (7238,2465) gcd:1
m,n = (146,97) Found 3 - (45114,41895) gcd:3 (42389,7154) gcd:1 (35478,14065)
gcd:1
n,m = (256,465) Found 3 - (767715,53504) gcd:1 (529635,184576) gcd:1
(506922,236282) gcd:2
n,m = (14,3125) Found 3 - (29340625,43554) gcd:1 (29253125,43946) gcd:1
(19664975,144670) gcd:5
m,n = (3632,3397) Found 4 - (46956731,853520) gcd:1 (25529328,22280923) gcd:1
(22280923,10002493) gcd:1 (22280923,15526835) gcd:1
The m,n search has been carried out to m,n > 20,000,19999 with no triples found
If a quadruple of primitive triangles exist, they must be extremely rare.
Randall
> > In article <1106082268....@c13g2000cwb.googlegroups.com>,
> > "zak" <seid...@yahoo.com> wrote:
> >
> >> These three primitive Pythagorean triangles
> >> have the same area
> >> {area, lesser leg, larger leg, hypotenuse}
> >> {13123110,1380,19019,19069},
> >> {13123110,3059,8580,9109},
> >> {13123110,4485,5852,7373}.
> >> Are any other such triples, 4,5-tuples?
> >
> > This is part of Problem D21 in Guy, Unsolved Problems in Number Theory.
> > The triple above was found by Charles L Shedd in 1945.
> > In 1986, Rathbun found three more, one of which has generators
> > (1610, 869), (2002, 1817), (2622, 143) [The triangle with generators
> > (a, b) is 2ab, a^2 - b^2, a^2 + b^2]. A 5th triple was found
> > independently by Hoey and Rathbun.
> >
> > It appears to be unknown whether there is an infinity of such triples,
> > also whether there are any quadruples.
> >
>
> It is difficult to find the 6th one, as the area searched is out to 10^21.
> Using Paul Whitlock's parametrization to simplify the search we find:
I don't know what Paul Whitlock's parametrization is. I do know
(you do too, of course) that if a and b are both odd then the resulting
triangle isn't primitive, as all three sides have even length. Some
of the entries on your list, namely,
> m,n = (8,3) Found 3 - (88,3) gcd:1 (55,13) gcd:1 (51,40) gcd:1
>
> m,n = (16,7) Found 4 - (368,35) gcd:1 (299,259) gcd:1 (259,144) gcd:1
> (259,155)
> gcd:1
>
> m,n = (32,1) Found 3 - (1056,29) gcd:1 (992,35) gcd:1 (553,377) gcd:1
>
> m,n = (3632,3397) Found 4 - (46956731,853520) gcd:1 (25529328,22280923) gcd:1
> (22280923,10002493) gcd:1 (22280923,15526835) gcd:1
suffer from this defect, so the gcd:1 fooled me for a while.
mea culpa. Yes, this I do know well, I should have put in a remark on the
posting, so no one would have been misled.
Thanks for the admonishment.
Randall