Current Pythagorean Hypotenuse

Number Theory Level 5

2018 is a hypotenuse of a Pythagorean Triple. For the corresponding primitive triple find it's level $l$ in the Tree of primitive Pythagorean triples and the parent node. For level numbering use $1$ for the first triple $(3,4,5)$ . For the parent node find the generating pair of integers $m>n$ using Euclid's formula. Enter the result as concatenated values of $lmn$ .

For example triple $(9,40,41)$ is at level 4 on the tree. Its parent node is $(7,24,25)$ with $m=4, n=3: m^2 + n^2=25, m^2-n^2=7, 2mn=24$ . The answer in this case would be $443$ .

The answer is 6152.

1 solution

Maria Kozlowska
Feb 12, 2018

Primitive triple always has an odd hypotenuse, therefore our corresponding primitive triple will have to have $c=1009$ . This is only possible for $28^2 + 15^2 = 1009 \rightarrow a=559, b=840$

We need matrices: $S = \begin{bmatrix}{3} \\ {4} \\ {5}\end{bmatrix}$ $E = \begin{bmatrix}{559} \\ {840} \\ {1009}\end{bmatrix}$ $A = \begin{bmatrix}{1} && {-2} && {2} \\ {2} && {-1} && {2} \\ {2} && {-2} && {3}\end{bmatrix}$ $B = \begin{bmatrix}{1} && {2} && {2} \\ {2} && {1} && {2} \\ {2} && {2} && {3}\end{bmatrix}$ $C = \begin{bmatrix}{-1} && {2} && {2} \\ {-2} && {1} && {2} \\ {-2} && {2} && {3}\end{bmatrix}$

We find that: ${A}^{-1} \times E = P$ where $P = \begin{bmatrix}{221} \\ {60} \\ {229}\end{bmatrix}$

After further trial and error, we find that: ${A}^{-1} \times {C}^{-1} \times {C}^{-1} \times {C}^{-1} \times {A}^{-1} \times E = S$

That means $l=6$

For the parent node triple $P$ we get $m=15,n=2$

This gives the answer $\boxed{6152}$ .

Note: More information about the Tree of primitive Pythagorean triples can be found in Wikipedia article .

Current Pythagorean Hypotenuse

The answer is 6152.

1 solution

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