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Primitive Pythagorean triangles where the odd leg is to a power.


Generalizations on the well known Pythagorean triple divisors, $ 3, 4,$ and $ 5.$


Let the ordered triple $ (a^k,b,c)$ be a primitive Pythagorean triple where $ k$ is a positive integer and $ a$ is the odd leg. Then there exists relatively prime odd positive integers $ m$ and $ n$ where $ m > n$ such that

$\displaystyle a^k=mn,\quad b=\frac{m^2-n^2}{2},$   and$\displaystyle \quad
c=\frac{m^2+n^2}{2}.
$ (20)

Since $ m$ and $ n$ are relatively prime, $ m=u^k$ and $ n=v^k$ for some odd positive integers $ u$ and $ k$ . Then equation (20) becomes

$\displaystyle a^k=u^k v^k,\quad b=\frac{u^{2k}-v^{2k}}{2},$   and$\displaystyle \quad c=\frac{u^{2k}+v^{2k}}{2}.$ (21)


Theorem 3   If $ a$ is odd and $ (a^k,b,c)$ is a primitive Pythagorean triple, $ k$ and $ d$ are positive integers such that $ d$ divides $ k$ then:
1.
If $ p=2d+1$ is a prime, $ p$ divides one of $ a$ or $ b$ .
2.
If $ q=4d+1$ is a prime, $ q$ divides one of $ a, b,$ or $ c$ .

Proof. From equation (21),

$\displaystyle 2ab=uv\left(u^{2k}-v^{2k}\right)$   and$\displaystyle \quad 4abc=uv\left(u^{4k}-v^{4k}\right).$ (22)

If $ d$ is a divisor of $ k$ then $ k=k_1d$ where $ k_1$ is the integer $ k/d$ . Thus, we can rewrite equations (22) as

$\displaystyle 2a^{k_1}b=u^{k_1}v^{k_1}\Bigl(\left(u^{k_1}\right)^{2d}-\left(v^{k_1}\right)^{2d}\Bigr)$   and$\displaystyle \quad 4a^{k_1}bc=u^{k_1}v^{k_1}\Bigl(\left(u^{k_1}\right)^{4d}-\left(v^{k_1}\right)^{4d}\Bigr).$ (23)

Note: If a prime $ P$ divides $ a^{k_1}$ then $ P$ divides $ a$ .

Therefore, from equation (23) and Fermat's little theorem, if prime $ p=2d+1$ then $ p$ divides $ ab$ . And if prime $ q=4d+1$ then $ q$ divides $ abc$ . That is, since $ a, b,$ and $ c$ are pairwise relatively prime, $ p=2d+1$ divides exactly one of $ a$ or $ b$ and $ q=4d+1$ divides exactly one of $ a, b,$ or $ c.$ $ \qedsymbol$


Theorem 4   If $ \left(a^{2^j},b,c\right)$ is a primitive Pythagorean triple where $ j$ is a nonnegative integer and $ a$ is odd then $ 2^{j+2}$ divides $ b$ .

Proof. (By induction on $ j$ ) The theorem is true for $ j=0$ . Assume true for $ j$ . That is, assume if

$\displaystyle \left(a^{2^j}\right)^2=c^2-b^2
$

then $ 2^{j+2}$ divides the even side $ b$ .

$\displaystyle \left(a^{2^{j+1}}\right)^2=\left(a^{2^j}\right)^2\left(a^{2^j}\right)^2
=\left(c^2-b^2\right)^2=\left(c^2+b^2\right)^2-(2cb)^2.
$

Therefore $ 2\left(2^{j+2}\right) = 2^{(j+1)+2}$ divides $ 2cb$ . $ \qedsymbol$


Examples



Table 5: Divisors of primitive Pythagorean triples
Triangle $ k$ $ d$ Divisors of $ a$ or $ b$ Divisors of $ a, b,$ or $ c$
$ a^2+b^2=c^2$ 1 1 $ 3, 2^2$ $ 3, 2^2, 5$
$ \left(a^2\right)^2+b^2=c^2$ 2 1,2 $ 3, 5, 2^3$ $ 3, 5, 2^3$
$ \left(a^3\right)^2+b^2=c^2$ 3 1,3 $ 3, 2^2, 7$ $ 3, 2^2, 5, 7, 13$
$ \left(a^4\right)^2+b^2=c^2$ 4 1,2,4 $ 3, 5, 2^4$ $ 3, 5, 2^4, 17$
$ \left(a^5\right)^2+b^2=c^2$ 5 1,5 $ 3, 2^2, 11$ $ 3, 2^2, 5, 11$
$ \left(a^6\right)^2+b^2=c^2$ 6 1,2,3,6 $ 3, 5, 7, 2^3, 13$ $ 3, 5, 7, 2^3, 13$
$ \left(a^7\right)^2+b^2=c^2$ 7 1,7 $ 3, 2^2$ $ 3, 2^2, 5, 29$
$ \left(a^8\right)^2+b^2=c^2$ 8 1,2,4,8 $ 3, 5, 17, 2^5$ $ 3, 5, 17, 2^5$
$ \left(a^9\right)^2+b^2=c^2$ 9 1,3,9 $ 3, 2^2, 7, 19$ $ 3, 2^2, 5, 7, 13, 19, 37$
$ \left(a^{10}\right)^2+b^2=c^2$ 10 1,2,5,10 $ 3, 5, 2^3, 11$ $ 3, 5, 2^3, 11, 41$



next up previous contents
Next: Multiplying Pythagorean Triples Up: Pythagorean Triples, etc. Previous: Primitive Pythagorean triangles where   Contents
f. barnes 2008-04-29
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