Primitive Pythagorean triangles where a - b is a constant.

This section in pdf form. triples1.pdf

If $a$ , $b$ , and $c$ are positive integers and $a^2+b^2=c^2$ then the solution $(a,b,c)$ is a Pythagorean triple. If $\gcd(a,b,c)=1$ then $(a,b,c)$ is a primitive (or reduced) Pythagorean triple, and the equation $a^2+b^2=c^2$ is a primitive Pythagorean triangle. Finding primitive Pythagorean triangles with a given difference between the lengths of the hypotenuse and a leg is a simple matter; from the parametrization

$$a=m^2-n^2 ,\quad b=2mn,$$   and$$\quad c=m^2+n^2 ,$$

$c-a=2n^2$ and $c-b=(m-n)^2$ . So, just choose the proper $m$ and/or $n$ .

However, finding Pythagorean triangles with constant difference between the lengths of the two smaller legs leads to a Pell equation,

$$a-b=m^2-n^2-2mn=(m-n)^2-2n^2=d.$$

Hence, if one solution is found, other solutions can be found recursively. From which a closed form can be obtained.

The following scheme does not find solutions for every integer $d$ , only those where $d=\pm (2t^2-1)$ , and not all of those. For example, in the Pythagorean triangle $35^2+12^2=37^2$ , there exists no integer $t$ such that $2t^2-1=23$ , and the scheme does not generate the triangle $15^2+8^2=17^2$ where $2 (2)^2-1=7$ .

I'll use the alternative parametrization: $(a,b,c)$ is a primitive Pythagorean triple if and only if there exists relatively prime, odd positive integers $m$ and $n$ ,
$m > n$ , such that

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

I'll give a closed form for $a$ and $b$ where $a-b = \pm (2t^2-1)$ . Let

$$\colorbox{Goldenrod}{\boxed{f(s)=\frac{\left(1+\sqrt{2} \right)^{s}+\left(1-\sqrt{2} \right)^{s}}{2} , }}$$ (12)

and

$$\colorbox{Goldenrod}{\boxed{g(s)=t f(s)-(t-1) f(s-1).}}$$ (13)

where $s$ and $t$ are positive integers. Set

$$\colorbox{Goldenrod}{\boxed{a=g(s+1) g(s) ,\quad b=\frac{g(s+1)^2-g(s)^2}{2} ,\quad \mbox{and} \quad c=\frac{g(s+1)^2+g(s)^2}{2} .}}$$ (14)

Claim 4   For positive integers $s$ and $t$ where $(a,b,c)$ is as defined in (14),

(i)

     $2 f(s)+f(s-1)=f(s+1)$ .

(ii)

     $2 g(s)+g(s-1)=g(s+1)$ .

(iii)

     $\gcd(g(s),g(s+1))=1$ .

(iv)

    $(a,b,c)$ is a primitive Pythagorean triple.
And

(v)

     $a-b=(-1)^s(2t^2-1)$ .

Proof. [ of (i)]

$$2 f(s)+f(s-1) =\frac{\left(1+\sqrt{2} \right)^{s-1}\left[2\left(1+\sqrt{2} \r... ...ht]+ \left(1-\sqrt{2}\right)^{s-1}\left[2\left(1-\sqrt{2} \right)+1\right]}{2}$$

$$=\frac{\left(1+\sqrt{2} \right)^{s-1}\left(1+\sqrt{2} \right)^2+ \left(1-\sqrt{2} \right)^{s-1}\left(1-\sqrt{2} \right)^2}{2}$$

$$=\frac{\left(1+\sqrt{2} \right)^{s+1}+\left(1-\sqrt{2} \right)^{s+1}}{2}$$

$$=f(s+1).$$

$\qedsymbol$

Proof. [of (ii)]

$$2 g(s)+g(s-1) =2[t f(s)-(t-1) f(s-1)]+[t f(s-1)-(t-1) g(f-2)]$$

$$=t[2 f(s)+f(s-1)]-(t-1)[2 f(s-1)+f(s-2)]$$

$$=t f(s+1)-(t-1) f(s) )$$

$$=g(s+1).$$

$\qedsymbol$

Proof. [ of (iii)]     $d\vert g(s+1)$ and $d\vert g(s)$ implies $d\vert g(s+1)-2 g(s)=g(s-1)$ , from (ii). Then $d\vert g(s)$ and $d\vert g(s-1)$ implies $d\vert g(s-2)$ . And so on, until ultimately, $d\vert g(0)=-1$ . $\qedsymbol$

Proof. [of (iv)]    $g(s+1)$ and $g(s)$ are relative prime, odd positive integers, $g(s+1)>g(s)$ where $a^2+b^2=c^2$ . $\qedsymbol$

Proof. [of (v)], by induction on s]     If $s=1$ , then

$$f(s+1) =f(2)=3.$$

$$f(s) =f(1)=1.$$

$$f(s-1) =f(0)=1.$$

Thus

$$a =g(s+1)=t f(2)-(t-1) f(1)=3t-(t-1)=2t+1.$$

$$b =g(s)=t f(1)-(t-1) f(0)=t-(t-1)=1.$$

Then, from (14),

$$a-b=(2t+1)(1)-\left(\frac{(2t+1)^2-1^2}{2}\right)=(-1)^1\left(2t^2-1\right).$$

So the claim is true for $s=1$ .

Assume it's true for $s=r$ . That is, assume

$$a-b=\frac{2 g(r+1) g(r)-g(r+1)^2+g(r)^2}{2}=(-1)^r(2t^2-1).$$

Then, for $s=r+1$ ,

$$a-b =\frac{2 g(r+2) g(r+1)-g(r+2)^2+g(r+1)^2}{2}$$

$$=\frac{g(r+2)[(2 g(r+1)-g(r+2)]+g(r+1)^2}{2}$$

$$=\frac{[2 g(r+1)+g(r)](-g(r))+g(r+1)^2}{2}$$

$$=\frac{-2 g(r+1) g(r)-g(r)^2+g(r+1)^2}{2}$$

$$=-\left(\frac{2 g(r+1) g(r)-g(r+1)^2+g(r)^2}{2}\right)$$

$$=-(-1)^r(2t^2-1)=(-1)^{r+1}(2t^2-1).$$

$\qedsymbol$

Examples

Table 1: $a-b=\pm 1$

$t$	$s$	$g(s+1)$	$g(s)$	$a$	$b$	$c$
1	1	3	1	3	4	5
1	2	7	3	21	20	29
1	3	17	7	119	120	169
1	4	41	17	697	696	985
1	5	99	41	4059	4060	5741

Table 2: $a-b=\pm 7$

$t$	$s$	$g(s+1)$	$g(s)$	$a$	$b$	$c$
2	1	5	1	5	12	13
2	2	11	5	55	48	73
2	3	27	11	297	304	425
2	4	65	27	1755	1748	2477
2	5	157	65	10205	10212	14437

Table 3: $a-b=\pm 97$

$t$	$s$	$g(s+1)$	$g(s)$	$a$	$b$	$c$
7	1	15	1	15	112	113
7	2	31	15	465	368	593
7	3	77	31	2387	2484	3445
7	4	185	77	14245	14148	20077
7	5	447	185	82695	82792	117017

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