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

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

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

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

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

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

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

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

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	$3, 5, 2^3, 11$	$3, 5, 2^3, 11, 41$

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