This thread is concerned with rectangular triangle with one side ( = s) and hypothenuse(adsbygoogle = window.adsbygoogle || []).push({});

(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 thehypothenuse, theeven 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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# Pythagorean Triangles with one side equal s and hypothenuse equal 2 s+1

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