Integer Triangles and Diophantine Equations?

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I did it a different way.
a/sin(3x) = b/sin(x) = c/sin(4x)

Taking a/sin(3x) = b/sin(x) gives :
a/b = sin(3x)/sin(x)
= [3sin(x) - 4sin^3(x)] / (sin(x)
= 3 - 4sin^2(x)
from which we get : sin(x) = (1/2)sqrt[(3b - a)/b]
so, cos(x) = (1/2)sqrt[(a + b)/b] using sin^2(A) + cos^2(A) = 1

Taking b/sin(x) = c/sin(4x) gives :
c/b = sin(4x)/sin(x)
= [4sin(x)cos(x) - 8sin^3(x)cos(x)] / sin(x)
= 4cos(x) - 8sin^2(x)cos(x)
= 4cos(x) - 8[1 - cos^2(x)]cos(x)
= 8cos^3(x) - 4cos(x)

Substituting cos(x) = (1/2)sqrt[(a + b)/b] gives :
c/b = [(a + b)/b]sqrt[(a + b)/b] - 2sqrt[(a + b)/b]

Multiplying by b and simplifying gives :
c = (a - b)sqrt[(a + b)/b]

From here, I cheated a little by running a program to find
relevant a's and b's that would yield integer c's and got
this table of values for a, b, c :

10, 8, 3
48, 27, 35
132, 64, 119
195, 125, 112
280, 125, 279
270, 216, 81
357, 343, 20

and stopped there, but I'm sure there are more.
The set 270, 216, 81 is a multiple of 10, 8, 3 so it's a bit
superfluous, and of course, there are many multiples I
didn't include, but the interesting thing is that b seems
to be always a cube. I haven't resolved that yet, meaning
I haven't come up with a generating formula and I'm not
sure if I will, but I'll keep plugging away for a while.

EDIT: I mean, b is a cube, except for the multiples.

EDIT 2: Will look into it in a day or two. At the moment, must
find some sleep time. My program didn't miss those values.
I did, in my rush to publish. I didn't eliminate multiples in the
program, so had to search and eliminate using rather
tired eyes.

EDIT 3: Just a small point - shouldn't my first equation at
least be in the correct format, by the way you've worded the
question? And if I replace my x with B, then the correct
equation really should be : a/sin(3B) = b/sin(B) = c/sin(4B).
My middle name is 'Pedantic', by the way.

Anyway, the last couple of hours haven't proved very
fruitful in completely solving the Diophantine equation,
but I did make a little progress.

Starting with bc^2 = (a + b)(a - b)^2, I expanded RHS to get
bc^2 = a^3 - a^2b - ab^2 + b^3 and then dividing by b gives
c^2 = a^3/b - a^2 - ab + b^2.

Now I assume that b is a cube, i.e. b = p^3 (no proof, just
observation). Thus, c^2 = a^3/p^3 - a^2 - ap^3 + p^6. Now,
a^3/p^3 = (a/p)^3, so a/p must be an integer, so let a/p = J.
So, a = Jp.

Substituting a = Jp gives c^2 = J^3 - J^2p^2 - Jp^4 + p^6.
Factorising : c^2 = J^2(J - p^2) - p^4(J - p^2)
so, c^2 = (J - p^2)(J^2 - p^4) = (J - p^2)(J - p^2)(J + p^2)
Therefore, finally, c = (J - p^2)*sqrt(J + p^2), which is not a
lot different from the original, but it got rid of that pesky b.

This is where I felt I wouldn't be doing a perfect job, so I
settled on a partial one and looked for patterns in 'J'.
Some of the first few values of J nicely fitted the equation
J = 3(p - 1)^2 + 2(p - 1), simplified as J = 3p^2 - 4p + 1.

Substituting back gives :
a = 3p^3 - 4p^2 + p
b = p^3
c = (2p^2 - 4p + 1)*(2p - 1)

Letting p = 2 to 10 gives this table for a, b, c :

p = 2 : 10, 8, 3
p = 3 : 48, 27, 35
p = 4 : 132, 64, 119
p = 5 : 280, 125, 279
p = 6 : 510, 216, 539
p = 7 : 840, 343, 923
p = 8 : 1288, 512, 1455
p = 9 : 1872, 729, 2159
p = 10: 2610, 1000, 3059

Looks like an infinite number of triangles can be made,
but not quite as big as the infinity of all the possibilities.
Oh! well, now I'm ready to see the answer.
Thanks for your comments, Duke, and for an interesting
question, and hopefully, for a "how to solve this problem".

EDIT 4: The crux of the matter is finding a general formula
that makes sqrt(J + p^2) an integer. Not even Dickson's
Theory of Numbers Vol II helped me, although the answer
may be in there somewhere. Still going around in circles!
Have just seen your hint - will get back.

EDIT 5:
bc^2 = (a + b)(a - b)^2 = a^3 - a^2b - ab^2 + b^3

Dividing by b^3 gives c^2/b^2 = a^3/b^3 - a^2/b^2 - a/b + 1

Let y = c/b and z = a/b, then

y^2 = z^3 - z^2 - z + 1
y^2 = z^2(z - 1) - (z - 1)
y^2 = (z - 1)(z^2 - 1)
y^2 = (z + 1)(z - 1)^2

so, y = (z - 1)*sqrt(z + 1)

Now it looks easy if we let z = n^2 - 1

Plugging back gives :
y = n(n^2 - 2)

So now, given b, we have :
a = b(n^2 - 1)
c = bn(n^2 - 2)

n is obviously > sqrt(2) and it may be fractional,
so let n = u/v.
Then, a = b(u^2 - v^2) / v^2
and, c = bu(u^2 - 2v^2) / v^3

The denominators disappear if we let b = v^3. We get :
a = v(u^2 - v^2)
b = v^3
c = u(u^2 - 2v^2)

Now, b ≠ 1, because there are no integral side triangles
with all sides unequal, such that one side has length 1.
Thus, v ≠ 1, so v ≥ 2. Which means u ≥ 3.

Those equations are pretty general, but I don't know how
to fiddle with them, so that only primitive solutions turn up.
We also have to make sure that in any triangle, a + b > c
AND a + c > b AND b + c > a. I think however, that those
equations do contain all the solutions, unfortunately,
together with a lot of superfluous multiples, as well as
a lot of impossible triangles. I'm not sure how to fix that yet.
Maybe I'll come back to it later.

EDIT 6: Ah! ... co-primes - the penny's dropped - if it can
be proved, of course, but that definitely won't be by me....Show more