Man, that Hirschhorn is dangerous! billgosper$ python rattri.py 5 3 | head 2098817 2016840 220597 In[524]:= ratang[823543, 1024303, 220597] Out[524]= {(\[Pi] - ArcCos[1461083/1647086])/(\[Pi] - ArcCos[-(11472481/11529602)]), (\[Pi] - ArcCos[-(11472481/11529602)])/(\[Pi] - ArcCos[-(13/14)]), (\[Pi] - ArcCos[-(13/14)])/(\[Pi] - ArcCos[1461083/1647086])} In[525]:= N@% Out[525]= {26.7290445719883, 0.261887400469822, 0.142857142857143} In[526]:= FI /@ {823543, 1024303, 220597} Out[526]= {7^7,7 41 43 83,13 71 239} Anecdote: The only time I spoke with Erdös was at the Ramanujan Centenary at UIUC. I told him my conjecture that any sum of four distinct odd squares was the sum of four distinct even squares. After about 30 sec, he said "This is not my problem." Later I repeated the conjecture to Hirschhorn, who handed me a handwritten algebraic proof the next day. --rwg On 2018-07-16 10:03, Tomas Rokicki wrote:
Here's a python script that generates integer triangles with angles in any rational ratio (pass the numerator and denominator on the command line).
I'm sure someone skilled in Python can write this much more simply.
Given this, the only thing we need to show is that these are the only such integer triangles.
Sample usage:
abacus:inttri rokicki$ python rattri.py 1 6 | head
1771561 2271060 566839
46656 72930 30421
4826809 8761896 4632263
117649 240240 146329
11390625 25591020 17151679
262144 637560 458081
# # Given two integers h and k, find integer triangles with angles in the # ratio h:k. Based on the generic parameterization given here: # #
https://en.wikipedia.org/wiki/Integer_triangle#Integer_triangles_with_one_an...
# import sys from fractions import gcd import math # h = int(sys.argv[1]) ; k = int(sys.argv[2]) ; q = 1 coscrit = math.cos(math.pi/(h+k)) def choose(a, b): if b < 0 or b > a: return 0 if b + b > a: b = a - b r = 1 for i in range(1, b+1): r = r * (a - i + 1) / i return r ; while True: for p in range(1, q): if q*coscrit >= p: continue if gcd(p, q) > 1: continue a = 0 b = 0 c = 0 for i in range(0, (h+1)/2): a += ((-1)**i)*choose(h, 2*i+1)*(p**(h-2*i-1))*((q*q-p*p)**i) for i in range(0, (k+1)/2): b += ((-1)**i)*choose(k, 2*i+1)*(p**(k-2*i-1))*((q*q-p*p)**i) for i in range(0, (h+k+1)/2): c += ((-1)**i)*choose(h+k, 2*i+1)*(p**(h+k-2*i-1))*((q*q-p*p)**i) a *= q**k b *= q**h g = gcd(gcd(a, b), c) a /= g b /= g c /= g print a, b, c q = q + 1
On Sun, Jul 15, 2018 at 2:55 PM Bill Gosper <billgosper@gmail.com> wrote:
Wow. I haven't even digested WDS's thing. --Bill Maybe it's time for MathOverflow, e.g.?
On Sun, Jul 15, 2018 at 1:36 PM Tomas Rokicki <rokicki@gmail.com> wrote:
Any more progress on this?
I have a large number of additional examples if you are interested, such as this triple for 6:
117649 146329 240240
and these for 5:
231 1024 1220
3024 3125 5555
7776 12155 17214
16807 34320 42833
36400 59049 89001
32768 79695 92168
59049 162656 178695
100000 302499 320210
161051 206460 323851
I have not yet tried to find recurrences for these . . .
It's abundantly clear empirically that your hypothesis works both ways (with suitable caveats). There should be a way to prove this.
-tom
On Sat, Jul 14, 2018 at 4:14 PM Bill Gosper <billgosper@gmail.com> wrote:
OMG! Nick Baxter somehow found 1024 | 1220 | 231 -> {23.09066960680706, 0.2165376788608164, 1/5}, 46656 | 72930 | 30421 -> {9.357241092137794, 0.6412146423203051, 1/6}
23.1--a toothpick! Even more toothpicular is 357 | 343 | 20 -> {18.99091913441869, 0.01755224857596323, 3}, because it's 1/19 of a smaller angle.
[Massive chop!]