Current Pythagorean Leg

Number Theory Level 4

2018 is a leg of a Pythagorean Triple. For the corresponding primitive triple find it's level $l$ in the Tree of primitive Pythagorean triples. For level numbering use $1$ for the first triple $(3,4,5)$ .

For example triple $(9,40,41)$ is at level 4 on the tree.

In other words, let $A_1, A_2, A_3, \ldots , A_{n-1}$ be not necessarily distinct $3\times3$ matrices with integer elements.

Find the smallest value of $n$ such that $A_1 \times A_2 \times \cdots \times A_{n-1} \times \begin{bmatrix}{3} \\ {4} \\ {5}\end{bmatrix} = \begin{bmatrix}{p} \\ {q} \\ {r}\end{bmatrix} ,$ where $p,q,r$ are positive integers (one of which equals 2018) satisfying $p^2 + q^2 =r^2$ .

The answer is 504.

1 solution

Maria Kozlowska
Feb 11, 2018

2018 can not be a leg of a primitive triple. The corresponding triple will have a value of 1009 instead which is a prime number. Any primitive triple having a leg as a prime number $p$ will have to have the generating numbers $m=\dfrac{p+1}{2}, n=\dfrac{p-1}{2}$ . It will also have a property of $c=b+1$ which implies it will have to be generated in the tree by the following operation: $A^{n-1} * S$ where $S = \begin{bmatrix}{3} \\ {4} \\ {5}\end{bmatrix}$ and $A = \begin{bmatrix}{1} && {-2} && {2} \\ {2} && {-1} && {2} \\ {2} && {-2} && {3}\end{bmatrix}$

The node level is always equal to $n$ . In our case it is $\boxed{504}$

Current Pythagorean Leg

The answer is 504.

1 solution

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