Primitive Pythagorean triangles where the hypotenuse is to a power.

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It is well-known that if $(a,b,c)$ is a solution to a Pythagorean triangle, where $c$ is the hypotenuse, then the Mersenne prime $3$ divides $ab$ , and the Fermat prime $5$ divides $abc$ . In this section, I show that if $\left(a,b,c^{2^j}\right)$ is a solution to a primitive Pythagorean triangle, where $j$ is a non-negative integer, then every Mersenne prime less than or equal to $\left(2^{j+2}-1\right)$ divides $ab$ , and every Fermat prime less than or equal to $\left(2^{j+2}+1\right)$ divides $abc$ .

Introduction

If $(a,b,c)$ is a solution to the Pythagorean triangle $a^2+b^2=c^2$ then $(a,b,c)$ is a Pythagorean triple. If, additionally, $a, b$ , and $c$ are pairwise relatively prime then $(a,b,c)$ is a primitive Pythagorean triple (PPT), and $a^2+b^2=c^2$ is a Primitive Pythagorean triangle.

All PPT's are given by the parametric equations

$$a=m^2-n^2, \quad b=2mn, \quad c=m^2+n^2$$ (15)

where $m$ and $n$ are relatively prime, positive integers of opposite parity, and $m > n$ .

When computing the PPT $\left(a,b,c^{2^j}\right)$ , it is convenient to express $a, b$ , and $c$ in terms of Gaussian integers $\mathbb{Z}[i]$ . To do so, let $m > n$ where $m$ and $n$ are relatively prime, positive integers having opposite parity. Let $z=m+ni$ and $\bar{z}=m-ni$ . And let $j$ be a non-negative integer. Then there exists positive integers $u$ and $v$ such that

$$u=\left\vert\frac{z^{2^j}+\bar{z}^{2^j}}{2}\right\vert=\left\vert\frac{(u+vi)+(u-vi)}{2}\right\vert,$$   and$$\quad v=\left\vert\frac{z^{2^j}-\bar{z}^{2^j}}{2i}\right\vert=\left\vert\frac{(u+vi)-(u-vi)}{2i}\right\vert.$$

• If $u<v$ then interchange the labels.
• Clearly, $\gcd(m,n)=1$ implies $gcd(u,v)=1$ .
• The parameters $m$ and $n$ have opposite parity. Therefore $2\nmid \left(m^2+n^2\right)^{2^j}=u^2+v^2$ . Hence $u$ and $v$ have opposite parity.

Thus, all primitive Pythagorean triples of the form $\left(a,b,c^{2^j}\right)$ are given by the parametric equations

$$a=u^2-v^2=\left\vert\frac{(m+ni)^{2^{j+1}}+(m-ni)^{2^{j+1}}}{2}\right\vert,$$ (16)

$$b=2uv=\left\vert\frac{(m+ni)^{2^{j+1}}-(m-ni)^{2^{j+1}}}{2i}\right\vert,$$ (17)

$$\quad c^{2^j}=u^2+v^2=\left(m^2+n^2\right)^{2^j} .$$ (18)

Then

$$2ab=\left\vert\frac{(m+ni)^{2^{j+2}}-(m-ni)^{2^{j+2}}}{2i}\right\vert.$$ (19)

Some prime divisors of primitive Pythagorean triangles.

If $a, b, c, k$ and $d$ are positive integers such that $a^2+b^2=\left(c^k\right)^2$ is a primitive Pythagorean triangle, and if $d$ divides $k,$ then we are going to show the following:

• If $4d-1$ is a prime $p,$ then $p$ divides exactly one of $a$ and $b.$
• If $4d+1$ is a prime $q,$ then $q$ divides exactly one of $a, b,$ and $c.$
• If $d=2^{\alpha},$ $\alpha$ a non-negative integer, then $2^{\alpha+2}$ divides exactly one of $a$ and $b.$

Previously, it was shown (here), that those three items are true for the case $d=1.$

I will, first, state and prove a theorem on the divisors of $ab$ , and of $abc$ , where $\left(a,b,c^k\right)$ is a PPT. Then the case where $k=2^j$ will be proven in the corollary. If $k=2^j$ in equation (19) then

$$2ab=\left\vert\frac{(m+ni)^{4k}-(m-ni)^{4k}}{2i}\right\vert.$$

Theorem 1   Let $a^2+b^2=\left(c^k\right)^2$ be a primitive Pythagorean triangle where $k\in \mathbb{Z}^+$ . Let $d$ be a positive integer divisor of $k$ . Let $p=4d-1$ and $q=4d+1$ .

(a)

If $p$ is a prime then $p\vert ab$ .

(b)

If $q$ is a prime then $q\vert abc$ .

Proof. [Proof of part (a)] Since $m+ni\in \mathbb{Z}[i]$ , then $\left((m+ni)^{\frac{k}{d}}\right)^{4d}=(M+Ni)^{4d}$ for some $M ,N\in \mathbb{Z}$ . If $p=4d-1$ is a prime, then since $-i=i^{4d-1}=i^p$ ,

$$p\vert(M+Ni)^p-(M-Ni)=\bigl((M+Ni)^p-M^p-(Ni)^p\bigr)+(M^p-M)-i(N^p-N).$$

Which implies

$$p\vert(M+Ni)^{p+1}-\left(M^2+N^2\right).$$

Similarly,

$$p\vert(M-Ni)^{p+1}-\left(M^2+N^2\right).$$

Which implies

$$p$$    divides $$(M+Ni)^{p+1}-(M-Ni)^{p+1}=(m+ni)^{4k}-(m-ni)^{4k}=4abi.$$

Therefore $p\vert ab$ . $\qedsymbol$

Proof. [Proof of part (b)] If $q=4d+1$ is a prime, then since $i=i^{4d+1}=i^q$ ,

$q\vert(M+Ni)^q-(M+Ni)$ where $M+Ni=(m+ni)^{\frac{k}{d}}$ and $M-Ni=(m-ni)^{\frac{k}{d}}$ . Which implies

$$q\vert(M+Ni)\bigl((M+Ni)^{q-1}-1\bigr).$$

Similarly

$$q\vert(M-Ni)\bigl((M-Ni)^{q-1}-1\bigr).$$

Then, if

$$q\nmid (M+Ni)(M-Ni)=M^2+N^2=c^{\frac{k}{d}},$$

$$q$$ divides $$(M+Ni)^{q-1}-(M-Ni)^{q-1}=(m+ni)^{4k}-(m-ni)^{4k}=4abi.$$

Therefore $q\vert abc$ . $\qedsymbol$

Corollary 1  

Let $a^2+b^2=\left(c^{2^j}\right)^2$ be a primitive Pythagorean triangle where $j$ is a nonnegative integer. Let $M$ be any Mersenne prime less than or equal to $2^{j+2}-1$ . And, let $F$ be any Fermat prime less than or equal to $2^{j+2}+1$ . Then

$$\boxed{M \mbox{divides} ab, \mbox{ and }F \mbox{ divides }abc.}$$

Proof. Let $k=2^j$ in theorem (1). Then

$$d =\left\{ 2^0, 2^1, 2^2,\ldots ,2^{j-1}, 2^j \right\},$$

$$4d-1 =\left\{ 2^2-1, 2^3-1, 2^4-1,\ldots ,2^{j+1}-1, 2^{j+2}-1 \right\},$$

$$4d+1 = \left\{ 2^2+1, 2^3+1, 2^4+1,\ldots ,2^{j+1}+1, 2^{j+2}+1 \right\}.$$

From Theorem (1), part (a), every prime in the set $4d-1$ divides $ab$ . From Theorem (1), part (b), every prime in the set $4d+1$ divides $abc$ . $\qedsymbol$

The divisor $\mathbf{4=2^{0+2}}$

Theorem 2   If

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

is a primitive Pythagorean triangle where $j$ is a nonnegative integer then $2^{j+2}$ divides $ab$ .

Proof. (by induction on $j$ ) If $j=0$ then

$$a^2+b^2=\left(c^{2^0}\right)^2=c^2 .$$

So there exists integers $m$ and $n$ , one odd the other even, such that $a=m^2-n^2$ and $b=2mn$ . Hence $4=2^{0+2}$ divides $ab$ . Assume true for $j$ , then

$$\left(c^{2^{j+1}}\right)^2=\left[\left(c^{2^j}\right)^2\right]^2 =\left(a^2+b^2\right)^2 =\left(a^2-b^2\right)^2+(2ab)^2$$

. Let $a_1=a^2-b^2$ and $b_1=2ab$ . Then if $2^{j+2}$ divides $ab$ , $2^{(j+1)+2}$ divides $a_1b_1$ . $\qedsymbol$

Examples

Table 4: 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	$3, 2^2$	$3, 2^2, 5$
$a^2+b^2=\left(c^2\right)^2$	2	1,2	$3, 7, 2^3$	$3, 5, 7, 2^3$
$a^2+b^2=\left(c^3\right)^2$	3	1,3	$3, 2^2, 11$	$3, 2^2, 5, 11, 13$
$a^2+b^2=\left(c^4\right)^2$	4	1,2,4	$3, 7, 2^4$	$3, 5, 7, 2^4, 17$
$a^2+b^2=\left(c^5\right)^2$	5	1,5	$3, 2^2, 19$	$3, 2^2, 5, 19$
$a^2+b^2=\left(c^6\right)^2$	6	1,2,3,6	$3, 7, 2^3, 11, 23$	$3, 5, 7, 2^3, 11, 13, 23$
$a^2+b^2=\left(c^7\right)^2$	7	1,7	$3, 2^2$	$3, 2^2, 5, 29$
$a^2+b^2=\left(c^8\right)^2$	8	1,2,4,8	$3, 7, 31, 2^5$	$3, 5, 7, 17, 31, 2^5$
$a^2+b^2=\left(c^9\right)^2$	9	1,3,9	$3, 2^2, 11$	$3, 2^2, 5, 11, 13, 37$
$a^2+b^2=\left(c^{10}\right)^2$	10	1,2,5,10	$3, 7, 2^3, 19$	$3, 5, 7, 2^3, 19, 41$

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