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

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In summary, this conversation discusses a problem of finding the remaining side of a rectangular triangle with one side and the hypothenuse given, where all sides have integer values. Computational experiments show the first solutions to this problem and further mathematical treatment is provided, including a lemma and the Euclidean rule for Pythagorean numbers. It is also noted that the problem can be related to Pythagorean triangles with one side equal to s and hypothenuse equal to 2s + k. Different sequences can hold for the same value of k, and further analysis and investigation of this problem is suggested.
  • #1
RamaWolf
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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[itex]^{2}[/itex] = t[itex]^{2}[/itex] - s[itex]^{2}[/itex] = (2 s + 1)[itex]^{2}[/itex] - s[itex]^{2}[/itex], with d, s, t [itex]\in \mathbb{Z}[/itex]

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[itex]_{0}[/itex] and b[itex]_{1}[/itex] be co-prime natural numbers and

b[itex]_{k}[/itex]:=4 b[itex]_{k-1}[/itex] - b[itex]_{k-2}[/itex]

then the pairs (b[itex]_{k},b_{k+1}[/itex]) are all co-prime

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

h := n[itex]^{2}[/itex] + m[itex]^{2}[/itex]
e := 2 m n
d := n[itex]^{2}[/itex] - m[itex]^{2}[/itex]

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.
 
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  • #2
Further analysis in this topic will be much easier
with the following closed form equation for b[itex]_{k}[/itex]


b[itex]_{k}[/itex] := [itex]\frac{(2+\sqrt{3})^{k} - (2-\sqrt{3})^{k}}{2 \sqrt{3}}[/itex]
 
  • #3
With the equation for b[itex]_{k}[/itex] 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[itex]_{k} := \frac{1}{6} ((3+2 \sqrt{3}) (2+\sqrt{3})^{2 k} + (3-2 \sqrt{3}) (2-\sqrt{3})^{2 k})[/itex]

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

h[itex]_{k} := \frac{1}{3} ((2+\sqrt{3})^{(1+2 k)}+(2-\sqrt{3})^{(1+2 k)} - 1)[/itex]

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

h[itex]_{k} = 2 e_{k}[/itex] + 1
 
  • #4
Let b[itex]_{0}[/itex]=0, b[itex]_{1}[/itex]=1 and b[itex]_{k}[/itex]:=4 b[itex]_{k-1}[/itex] - b[itex]_{k-2}[/itex] as above,
we have for the odd leg of aa PTT after the Euclidean rule for PTT's

d[itex]_{k}[/itex]= b[itex]_{k+1}[/itex][itex]^{2}[/itex] - b[itex]_{k}[/itex][itex]^{2}[/itex]

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[itex]_{k}[/itex]= b[itex]_{2k+1}[/itex]
 
  • #5
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[itex]_{k}[/itex] and the n[itex]_{k}[/itex] form the recurrence relation:

m[itex]_{k}[/itex] = 4 m[itex]_{k-1}[/itex] - m[itex]_{k-2}[/itex] and

n[itex]_{k}[/itex] = 4 n[itex]_{k-1}[/itex] - n[itex]_{k-2}[/itex]

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[itex]\in \mathbb{Z}[/itex]
 
  • #6
related paper:

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

and some of my ideas:

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

http://dl.dropbox.com/u/13155084/Pythagorean%20lattice.pdf [Broken]
 
Last edited by a moderator:
  • #7
RamaWolf said:
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[itex]_{k}[/itex] and the n[itex]_{k}[/itex] form the recurrence relation:

m[itex]_{k}[/itex] = 4 m[itex]_{k-1}[/itex] - m[itex]_{k-2}[/itex] and

n[itex]_{k}[/itex] = 4 n[itex]_{k-1}[/itex] - n[itex]_{k-2}[/itex]

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[itex]\in \mathbb{Z}[/itex]
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.
 
  • #8
ramsey2879 said:
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) .
 

1. What is a Pythagorean Triangle?

A Pythagorean Triangle is a right triangle where all three sides are whole numbers. It follows the Pythagorean Theorem, which states that in a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the other two sides.

2. How can a triangle have one side equal to s and hypotenuse equal to 2s+1?

This type of triangle is called a Pythagorean Triangle with one side equal to s and hypotenuse equal to 2s+1. It is possible because the Pythagorean Theorem allows for the relationship between the sides of a right triangle to be expressed algebraically.

3. What values can s have in this type of triangle?

The value of s can vary depending on the constraints of the problem or situation. However, in general, s can be any positive integer as long as it satisfies the equation 2s+1 > s, which ensures that the triangle is a valid right triangle.

4. How can this type of triangle be useful in real-life situations?

Pythagorean Triangles with one side equal to s and hypotenuse equal to 2s+1 can be useful in many real-life situations, such as in construction, engineering, and architecture. They allow for precise measurements and calculations in right-angled structures, making them essential in these fields.

5. Is there a special name for this type of triangle?

Yes, this type of triangle is also known as a Pythagorean Triple, where the three sides are whole numbers and satisfy the Pythagorean Theorem. Specifically, this type of triangle can be referred to as a Pythagorean Triple with a ratio of 1:s:2s+1.

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