Pythagorean Triangles with one side equal s and hypothenuse equal 2 s+1

[RamaWolf]

This thread is concerned with rectangular triangle with one side ( = s) and hypothenuse
(t = 2 s + 1) given and we are looking for the remaining side (all sides having integer values)

d$^{2}$ = t$^{2}$ - s$^{2}$ = (2 s + 1)$^{2}$ - s$^{2}$, with d, s, t $\in \mathbb{Z}$

Computational experiments show the first solutions to this problem as:

s = 0 | 8 | 120 | 1680 | 23408 | ...
t = 1 | 17 | 241 | 3361 | 46817 | ...
d = 1 | 15 | 209 | 2911 | 40545 | ...

Further mathematical treatment starts with:

Lemma: Let b$_{0}$ and b$_{1}$ be co-prime natural numbers and

b$_{k}$:=4 b$_{k-1}$ - b$_{k-2}$

then the pairs (b$_{k},b_{k+1}$) are all co-prime

Euclidean Rule for Pythagorean Numbers:
Let (m,n) be co-prime natural numbers (m<n), then

h := n$^{2}$ + m$^{2}$
e := 2 m n
d := n$^{2}$ - m$^{2}$

form the hypothenuse, the even and the odd leg
of a primitive Pythagorean triangle (PPT)


Now we have from b = {0, 1, 4, 15, 56, 209, 241, ...}

(m.n) -h- -e- -d-
---------------------------------
(0,1) -1- -0- -1-
(1,4) -17- -8- -15-
(4,15) -241- -120- -209-
(15,56) -3361- -1680- -2911-
(56,209) -46827- -23408- -40545-
(...,...)

and we identify the PPT's generated from the b-sequence as solutions to our problem.

 

[RamaWolf]

Further analysis in this topic will be much easier
with the following closed form equation for b$_{k}$


b$_{k}$ := $\frac{(2+\sqrt{3})^{k} - (2-\sqrt{3})^{k}}{2 \sqrt{3}}$

 

[RamaWolf]

With the equation for b$_{k}$ in closed form as above and the definitions
for then even resp odd leg and the hypothenuse of the Euclidean PTT rule,
we have:

d$_{k} := \frac{1}{6} ((3+2 \sqrt{3}) (2+\sqrt{3})^{2 k} + (3-2 \sqrt{3}) (2-\sqrt{3})^{2 k})$

e$_{k}:=\frac{1}{6} ((2+\sqrt{3})^{(1+2 k)}+(2-\sqrt{3})^{(1+2 k)} - 4)$

h$_{k} := \frac{1}{3} ((2+\sqrt{3})^{(1+2 k)}+(2-\sqrt{3})^{(1+2 k)} - 1)$

and from the last two equations, it is easy to see, that

h$_{k} = 2 e_{k}$ + 1

 

[RamaWolf]

Let b$_{0}$=0, b$_{1}$=1 and b$_{k}$:=4 b$_{k-1}$ - b$_{k-2}$ as above,
we have for the odd leg of aa PTT after the Euclidean rule for PTT's

d$_{k}$= b$_{k+1}$$^{2}$ - b$_{k}$$^{2}$

Numerical inspection of the so defined sequences show:

b = {0,1,4,15,56,209,780,2911,10864,40545,...} and

d = {15,209,2911,40545, ...}

the highly remarkable fact d$_{k}$= b$_{2k+1}$

 

[RamaWolf]

Related to our problem is:

Pythagorean Triangles with one side equal s and hypothenuse equal 2 s-1

Computational experiments of the first solutions (given are the even leg e, the odd leg d
and the hypothenuse h, with the corresponding (m,n) from the Êuclidean Rule for PPT):

i - (m.n) - d / e / h
-------------------
1 - (1,2) - 3 / 4 / 5
2 - (4,7) - 33 / 56 / 65
3 - (15,26) - 451 / 780 / 901
4 - (56,97) - 6273 / 10864 / 12545
5 - (209,362) - 87363 / 151316 / 174725

We see that the m$_{k}$ and the n$_{k}$ form the recurrence relation:

m$_{k}$ = 4 m$_{k-1}$ - m$_{k-2}$ and

n$_{k}$ = 4 n$_{k-1}$ - n$_{k-2}$

with different starting values.

That gives hope for an interesting investigation in the problem:

Pythagorean Triangles with one side equal s and hypothenuse equal 2 s+k, k$\in \mathbb{Z}$

 

[JeremyEbert]

related paper:

http://conservancy.umn.edu/bitstream/4878/1/438.pdf

and some of my ideas:

http://dl.dropbox.com/u/13155084/2D/Fourier.html

http://dl.dropbox.com/u/13155084/Pythagorean%20lattice.pdf

 

[ramsey2879]

Related to our problem is:

Pythagorean Triangles with one side equal s and hypothenuse equal 2 s-1

Computational experiments of the first solutions (given are the even leg e, the odd leg d
and the hypothenuse h, with the corresponding (m,n) from the Êuclidean Rule for PPT):

i - (m.n) - d / e / h
-------------------
1 - (1,2) - 3 / 4 / 5
2 - (4,7) - 33 / 56 / 65
3 - (15,26) - 451 / 780 / 901
4 - (56,97) - 6273 / 10864 / 12545
5 - (209,362) - 87363 / 151316 / 174725

We see that the m$_{k}$ and the n$_{k}$ form the recurrence relation:

m$_{k}$ = 4 m$_{k-1}$ - m$_{k-2}$ and

n$_{k}$ = 4 n$_{k-1}$ - n$_{k-2}$

with different starting values.

That gives hope for an interesting investigation in the problem:

Pythagorean Triangles with one side equal s and hypothenuse equal 2 s+k, k$\in \mathbb{Z}$

Let k = 3a^2-b^2 then the series for m begins {a,2a+b,7a+4b,...} and the series for n begins {b,3a+2b,12a+7b,...}. This does not preclude other values of m and n from also solving the relationship one side = s and the hypothenuse equals 2s + k though. Example, m^2+n^2 - 2(n^2-m^2) = 3m^2-n^2 = 3*(2a+b)^2 - (3a+2b)^2 = 3*(7a+4b)^2 - (12a-7b)^2. As one can see, the relation m_n = 4 m_(n-1) - m_(n-2) holds for this pattern. Since -13 = 3*1^2-4^2 = 3*2^2 -5^2 one can see how different sequences may hold for the same value of k. Still working on other relationships e.g. for the twice the even side + k = the hypothenuse.

 

[ramsey2879]

Let k = 3a^2-b^2 then the series for m begins {a,2a+b,7a+4b,...} and the series for n begins {b,3a+2b,12a+7b,...}. This does not preclude other values of m and n from also solving the relationship one side = s and the hypothenuse equals 2s + k though. Example, m^2+n^2 - 2(n^2-m^2) = 3m^2-n^2 = 3*(2a+b)^2 - (3a+2b)^2 = 3*(7a+4b)^2 - (12a-7b)^2. As one can see, the relation m_n = 4 m_(n-1) - m_(n-2) holds for this pattern. Since -13 = 3*1^2-4^2 = 3*2^2 -5^2 one can see how different sequences may hold for the same value of k. Still working on other relationships e.g. for the twice the even side + k = the hypothenuse.

For the hypothenuse - twice the even side we have a^2 + b^2 -4ab = b^2 + (4b-a)^2 - 4b*(4b-a). In other words m_0 = a, m_1 = b m_2 = 4b-a and n_0 = b, n_1 = 4b-a, n_2 = 15b-4a. Both have the recurrence relation S(n) = 4*S(n-1) - S(n-2)

For the even side -the odd side, 4ab - b^2 + a^a = 4*(4b+a)*(17b+4a)-(17b+4a)^2 + (4b+a)^2. In other words, m_0 = a, m_1 = a+4b, m_2 = 17a + 72b and n_0 = b, n_1 = 17b + 4a, n_2 = 305b + 72a. Both have the recurrence relation S_n = 18*S(n-1)-S(n-2) .