Infinitely many Pythagorean triples question regarding $4xy$

Question

I understand that In any primitive Pythagorean triple, one leg is odd, one leg is even, and the hypotenuse is odd

Now it turns out that really there is a method for generating infinitely many pythagorean triples in an easy way. Lets use simple algebra here: \begin{align*} (x+y)^{2}&=x^{2}+y^{2}+2xy\tag{a.1}\\ (x-y)^{2}&=x^{2}+y^{2}-2xy\tag{a.2} \end{align*} The difference between (a.1) and (a.2) is $4xy$. So now we have a relationship that looks almost like a Pythagorean triple, namely:

One square i.e., $(x+y)^{2}$ equals another square i.e., $(x-y)^{2}$ plus lets say something that we wish were a square primarily $4xy$.

Now how do we ensure that $4xy$ is square? Well we can just choose $x$ and $y$ to be squares, so this is where I am absolutely baffled...

How do I relate this to the lemma:

Let $(a,b,c)$ be a primitive Pythagorean triple where $a$ is the even number, then $\tfrac{c+b}{2} $ and $\tfrac{c-b}{2}$ are perfect squares say $s^{2}$ and $t^{2}$ respectively and $s$ and $t$ are relatively prime.

I just want someone to guide me I can hopefully figure out the rest...

Answer 1

There are many ways to compute (or building an algorithem) Pythagorean triples, for example, here is a nice one.

Now, for your question, I didn't check if your characterization is right because all your question deals with $4xy$ as a square (I think this is what you wanted to say). Let $t\in\Bbb{Z}$, we can decompose $t$ into multiplication of primes (the fundamental theorem of arithmetic), say $t=p_1^{e_1}\cdot p_2^{e_2}\cdot ...\cdot p_n^{e_n}$, $p_i\neq p_j$ for $i\neq j$.

Such $t$ is a square if and only if for all $i\in\{1,2,..,n\}$, $e_i$ is even.

Using this characterization, we get that $4xy$ is a square if and only if $xy$ is a square. Write $x$ and $y$ as $$x=q_1^{f_1}\cdot q_2^{f_2}\cdot ...\cdot q_m^{f_m}$$ $$y=l_1^{g_1}\cdot l_2^{g_2}\cdot ...\cdot l_k^{g_k}$$ for $q_i, l_i$ primes. $x$ and $y$ might have common primes in the decomposition. If $x$ and $y$ don't have any common primes in the decomposition, then by the characterization above, we get that $xy$ is a square if and only if both $x$ and $y$ are squares. If $x$ and $y$ have common primes in the decomposition, say for example $l_1=p_1$ and $l_2=p_2$ and just them (without loss of generality), then by the characterization $xy$ is a square if and omly if $f_1+g_1$ and $f_2+g_2$ are even, and all the other powers ($f_i$ and $g_j$) are even numbers.

Answer 2

The usual formulas for primitive Pythagorean triples is $$ a = 2pq,\;\; b = p^2-q^2,\;\; c = p^2+q^2,\;\; a^2+b^2=c^2. $$ If we let $\,x=p^2,\;y=q^2,\,$ then the formulas become $$ a \!=\! \sqrt{4xy},\; b\!=\!x\!-\!y,\; c\!=\!x\!+\!y,\; 4xy\!+\!(x\!-\!y)^2\!=\!(x\!+\!y)^2. $$ Note that $\,x,y\,$ are relatively prime if and only if $\,p,q\,$ are relatively prime.

I think you can figure out the rest now.

Answer 3

The most common formula for generating Pythagorean triples is Euclid's formula, here shown as $\quad A=m^2-k^2\quad B=2mk\quad C=m^2+k^2.\qquad$ Side-B must be a multiple of $\space4\space$ so, to be a square, side-B must be the product of $\space4\space$ and another square, i.e. $\quad 4\times1=4,\space\space 4\times4=16,\space\space 4\times9=36, \space\cdots.\quad$ There are an infinite number of these and, to find the corresponding triples, we divide side-B by $\space 2\space$ and select the cofactors of the result that are mutually prime and where $\space m>k.\quad$ For example, if $\space B=36,\space$ we select the cofactor pairs of $\space 18:\space(9,2), (18,1),\space$ ignoring $\space (6,3)\space$ because $\space (6,3)\space$ are not mutually prime. Having done so, we find

$$f(9,2)=(77,36,85)\qquad f(18,1)=(323,36,325)$$