It's All Triples 2

This problem was created using a general recursive sequence that I derived using primitive pythagorean triples.

Below is the recursive sequence.

Let

$a_{1} = m_{1}^2 + n_{1}^2$

$b_{1} = 2m_{1}n_{1}$

$c_{1} = m_{1}^2 + n_{1}^2$ ,

where $(m_{1},n_{1}) = 1, m_{1} > n_{1}$ and $m_{1}$ is even and $n_{1}$ is odd.

for $(1 \leq j \leq n - 1)$ let

$a_{j + 1} = c_{j}$

$b_{j + 1} = \dfrac{(c{j} + 1)(c_{j} - 1)}{2}$

$c_{j + 1} = \dfrac{c_{j}^2 + 1}{2}$

then for each $(1 \leq k \leq n) \:\ (a_{k},b_{k},c_{k})$ is a primitive pythagorean triple.

Let $p$ be a positive even integer with $m_{1} = p$ and $n_{1} = 1$ .

Using the above the first ppt triple is $(p^2 - 1,2p,p^2 + 1) \implies A_{1} = \dfrac{1}{2}(2p)(p^2 - 1) = 60 \implies p^3 - p - 60 = 0 \implies (p - 4)(p^2 + 4p + 15) = 0 \implies p = 4$

$\implies (p^2 - 1,2p,p^2 + 1) = (15,8,17)$ .

The second ppt triple is $(p^2 + 1,\dfrac{(p^2 + 2)*p^2}{2},\dfrac{p^4 + 2p^2 + 2}{2}) = (17,144,145)$

The third ppt triple is $(\dfrac{p^4 + 2p^2 + 2}{2}, \dfrac{p^2(p^2 + 2)(p^4 + 2p^2 + 4)}{8}, \dfrac{p^8 + 4p^6 + 8p^4 + 8p^2 + 8}{8}) = (145,10512,10513)$ .

$\cos(\theta_{1}) = \dfrac{8}{17}, \sin(\theta_{1}) = \dfrac{15}{17}$

$\cos(\theta_{2}) = \dfrac{17}{145}, \sin(\theta_{2}) = \dfrac{144}{145}$

$\cos(\theta_{3}) = \dfrac{145}{10513}, \sin(\theta_{3}) = \dfrac{10512}{10513}$

$\implies \cos(\theta_{1} + \theta_{2} + \theta_{3}) =$

$\cos(\theta_{1})\cos(\theta_{2})\cos(\theta_{3}) - \sin(\theta_{1})\sin(\theta_{2})\cos(\theta_{3}) - \sin(\theta_{1})\sin(\theta_{3})\cos(\theta_{2}) - \sin(\theta_{2})\sin(\theta_{3})\cos(\theta_{1}) =$

$\dfrac{8}{10513} - \dfrac{(15)(144)}{(17)(10513)} - \dfrac{(15)(10512)}{(10513)(145)} - \dfrac{(144)(10512)(8)}{(145)(10513)(17)} = -\dfrac{15083864}{25914545}$

$\implies$

$d_{2} = AE = \sqrt{8^2 + 10513^2 + \dfrac{(2)(8)(10513)(15083864)}{25914545}} = \sqrt{\dfrac{272681111169}{2465}} \approx \boxed{10517.65850528}$ .

$$

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Note: For the general recursive sequence above the sum of the squares $a_{1}^2 + \sum_{j = 1}^{n} b_{j}^2 = c_{n}^2$ .

For this problem with $p = 4$ we have: $15^2 + 8^2 + 144^2 + 10512^2 = 10513^2.$

Also note given $(a_{1},b_{1},c_{1})$ and setting $m_{j + 1} - n_{j + 1} = 1$ and $m_{j + 1} + n_{j + 1} = c_{j} \implies m_{j + 1} = \dfrac{c_{j} + 1}{2}$ and $n_{j + 1} = \dfrac{c_{j} - 1}{2}$ and replacing $m_{j + 1},n_{j + 1}$ into $a_{j + 1} = (m_{j + 1} - n_{j + 1})(m_{j + 1} + n_{j + 1}), \:\ b_{j + 1} = 2m_{j + 1}n_{j + 1}, \:\ c_{j + 1} = (m_{j + 1})^2 +(n_{j + 1})^2$

$\implies$

$a_{j + 1} = c_{j}$

$b_{j + 1} = \dfrac{(c_{j} + 1)(c_{j} - 1)}{2}$

$c_{j + 1} = \dfrac{c_{j}^2 + 1}{2}$

and for each $j$ such that $(1 \leq j \leq n) \:\ c_{j}$ is odd $\implies a_{j},b_{j}$ and $c_{j}$ are positive integers and $(c_{j}, \dfrac{c_{j}^2 - 1}{2}, \dfrac{c_{j}^2 - 1}{2} + 1) = 1$ and $a_{j}^2 + b_{j}^2 = c_{j}^2$ .