MHB Pythagorean Triples- Why is this the proof?

[evinda]

Hello! (Wave)

The solutions of $X^2+Y^2=Z^2$ are called Pythagorean Triples .

We suppose that $x$ odd and $y$ even.

All the solutions $(x,y,z)$ with $gcd(x,y,z)=1$ are given by:

$$(x,y,z)=(r^2-s^2,2rs,r^2+s^2)$$

$$r, s \in \mathbb{Z}, r>s, (r,s)=1$$

$$r \not\equiv s \pmod 2$$

Could you explain me why the following is the proof of the above sentence? (Sweating)

View attachment 3324

Line equation $AB$: $y-y_0=\lambda(x-x_0)$

so, $v=t(u+1)$.

Therefore, the point $B=(u,v)$ is a solution of the system:

$$\begin{Bmatrix}
u^2+v^2=1\\
v=t(u+1)
\end{Bmatrix}$$

Replacing $v$ at the first equation, we get:

$$(1+t^2)u^2+2t^2u+(t^2-1)=0$$

We solve the equation and find:

$$u=\frac{1-t^2}{1+t^2} , u=-1$$

$u=-1$ corresponds to the point $A=(-1,0)$.

We calculate $v=t(u+1)=\frac{2t}{t^2+1}$, so the coordinates of $B$ are:

$$B= \left ( \frac{1-t^2}{1+t^2}, \frac{2t}{1+t^2} \right )$$

Remark:

If the slope $t \in \mathbb{Q}$, then $B=(u,v) \in \mathbb{Q} \times \mathbb{Q}$.

Conversely, if the coordinates of $B=(u,v)$ are rational, then the line $AB$ has slope $t=\frac{v-0}{u+1} \in \mathbb{Q}$.

Therefore, the set of rational solutions of $x^2+y^2=1$ is:

$$\begin{Bmatrix}
(u,v)=\left ( \frac{1-t^2}{1+t^2}, \frac{2t}{1+t^2} \right )|t \in \mathbb{Q}
\end{Bmatrix} \cup \{ (-1,0)\}$$

Remark: If $t=\frac{r}{s}, r, s \in \mathbb{Z}, (r,s)=1$, we get the Pythagorean Triples.

But... how do we get them? (Sweating)

 

[RLBrown]

Hello! (Wave)
Therefore, the set of rational solutions of $x^2+y^2=1$ is:

$$\begin{Bmatrix}
(u,v)=\left ( \frac{1-t^2}{1+t^2}, \frac{2t}{1+t^2} \right )|t \in \mathbb{Q}
\end{Bmatrix} \cup \{ (-1,0)\}$$

Remark: If $t=\frac{r}{s}, r, s \in \mathbb{Z}, (r,s)=1$, we get the Pythagorean Triples.

But... how do we get them? (Sweating)

When h=1
$(u,v,h)=(\frac{1-t^2}{1+t^2}, \frac{2t}{1+t^2}, 1)$

For h>1
Multiply u,v and h by $1+t^2$
$(u',v',h')=(1-t^2, 2t, 1+t^2)$

Subst t=s/r
Multiply u',v' and h' by $r^2$

 

[mathbalarka]

A standard method to solve diophantine equations is to use fundamentals of analytic geometry (later which turns out to reveal a more general connection with arithmetic geometry and algebraic geometry) Here's an explanation : recall that $x^2 + y^2 = 1$ is the analytic definition of a circle with radius $1$ situated at the origin. So all rational points $(x, y)$ must lie on the circle in the Cartesian plane. Now draw a straightline from $(-1, 0)$ to some loci $(u, v)$ of the circle, i.e., it's a point which continuously "loops" around the origin.

View attachment 3328​

Recall the equation $y - y_0 = \lambda (x - x_0)$ of straightlines in Cartesian coordinates. $\lambda$ is the slope $\tan(\theta)$, which is $\overline{OA}$ here, $O = (0, 0)$ being the origin. $(x_0, y_0) = (-1, 0)$ so subbing in gives the equation $y = \lambda (x + 1)$. This must also hold at the moving point $(u, v)$ on the circle, for which $v = \lambda (u + 1)$. Thus, we have two relations between $u$ and $v$ while $(u, v)$ is moving on the path we set up for it : the standard $u^2 + v^2 = 1$ and $v = \lambda (u + 1)$. However, when $(u,v)$ stumbles onto a rational point on the circle (there are infinitely many, take $(3/5, 4/5)$), then from the second relation, $\lambda$ must be rational also (quotient of two rationals is a rational). Thus, ALL of rational points $(a, b)$ on the circle $x^2 + y^2 = 1$ satisfies

$a^2 + b^2 = 1$ AND
$a = t(b+1)$ for some *rational* $t$ (the slope).

Subbing the second into the first, $t^2 (b+1)^2 + b^2 = 1 \Longrightarrow (t^2+1)b^2 + 2bt^2 + t^2-1 = 0$ which factors as $((1+t^2)b-(1-t^2))(b+1)$. Solving for $b$ and finding $a$ from the second equation gives you the desired solution set (well, a *complete* solution set, i.e., all rational solutions are of that form).

 

[evinda]

A standard method to solve diophantine equations is to use fundamentals of analytic geometry (later which turns out to reveal a more general connection with arithmetic geometry and algebraic geometry) Here's an explanation : recall that $x^2 + y^2 = 1$ is the analytic definition of a circle with radius $1$ situated at the origin. So all rational points $(x, y)$ must lie on the circle in the Cartesian plane. Now draw a straightline from $(-1, 0)$ to some loci $(u, v)$ of the circle, i.e., it's a point which continuously "loops" around the origin.

View attachment 3328​

Recall the equation $y - y_0 = \lambda (x - x_0)$ of straightlines in Cartesian coordinates. $\lambda$ is the slope $\tan(\theta)$, which is $\overline{OA}$ here, $O = (0, 0)$ being the origin. $(x_0, y_0) = (-1, 0)$ so subbing in gives the equation $y = \lambda (x + 1)$. This must also hold at the moving point $(u, v)$ on the circle, for which $v = \lambda (u + 1)$. Thus, we have two relations between $u$ and $v$ while $(u, v)$ is moving on the path we set up for it : the standard $u^2 + v^2 = 1$ and $v = \lambda (u + 1)$. However, when $(u,v)$ stumbles onto a rational point on the circle (there are infinitely many, take $(3/5, 4/5)$), then from the second relation, $\lambda$ must be rational also (quotient of two rationals is a rational). Thus, ALL of rational points $(a, b)$ on the circle $x^2 + y^2 = 1$ satisfies

$a^2 + b^2 = 1$ AND
$a = t(b+1)$ for some *rational* $t$ (the slope).

Subbing the second into the first, $t^2 (b+1)^2 + b^2 = 1 \Longrightarrow (t^2+1)b^2 + 2bt^2 + t^2-1 = 0$ which factors as $((1+t^2)b-(1-t^2))(b+1)$. Solving for $b$ and finding $a$ from the second equation gives you the desired solution set (well, a *complete* solution set, i.e., all rational solutions are of that form).

Why do we prove in that way that all the solutions with $gcd(x,y,z)=1$ are given by:

$$(x,y,z)=(r^2-s^2,2rs,r^2+s^2)$$

$$r,s \in \mathbb{Z}, r>s , (r,s)=1, r \not \equiv 2(2)$$

? (Thinking) (Sweating)

 

[mathbalarka]

Note that $a^2 + b^2 = c^2$ can be transformed into $x^2 + y^2 = 1$ where $x = a/c$ and $y = b/c$. Thus, from our previous calculations,

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

For all $(a, b, c) \in \Bbb Z^3$ for some rational $t$. As $t$ is rational, we can sub in $t = s/r$ for integers $s, r$. That gives us

$\displaystyle \frac{1-t^2}{1+t^2} = \frac{1-s^2/r^2}{1+s^2/r^2} = \frac{1-s^2/r^2}{1+s^2/r^2} \cdot \frac{r^2}{r^2} = \frac{r^2 - s^2}{r^2 + s^2}$ and $\displaystyle \frac{2t}{1+t^2} = \frac{2s/r}{1+s^2/r^2} = \frac{2s/r}{1+s^2/r^2} \cdot \frac{r^2}{r^2} = \frac{2sr}{r^2 + s^2}$

Hence, $\displaystyle (a, b) = \left ( \frac{r^2 - s^2}{r^2 + s^2} \cdot c, \frac{2rs}{r^2 + s^2} \right )$. But $a, b$ are integers, thus $r^2 + s^2$ divides $c$, i.e., $c = d\cdot (r^2 + s^2)$ for some integer $d$. Subbing that in, we get $(a, b) = (d \cdot (r^2-s^2), d \cdot 2rs)$. Hence,

$$(a, b, c) = (d \cdot (r^2 - s^2), d \cdot 2rs, d \cdot (r^2 + s^2))$$

For all integer $a, b, c$. Apparently, if $\text{gcd}(a, b, c) = 1$ then $d = 1$ and $\text{gcd}(r, s) = 1$. Other conditions also follows likewise (left as an exercise for you).

 

[RLBrown]

Hello! (Wave)

The solutions of $X^2+Y^2=Z^2$ are called Pythagorean Triples .

We suppose that $x$ odd and $y$ even.

All the solutions $(x,y,z)$ with $gcd(x,y,z)=1$ are given by:
$$(x,y,z)=(r^2-s^2,2rs,r^2+s^2)$$
$$r, s \in \mathbb{Z}, r>s, (r,s)=1$$
$$r \not\equiv s \pmod 2$$

By the way:
The method that you provide (Sometimes called Euclid's Formula) produces a subset of Pythagorean Triples called Primitive Pythagorean Triples. Although it is attributed to Euclid, artifacts like Plimpton 322 hint that this formula was known to the Babylonians.

A simple way to remember Euclids formula is this...

Using complex numbers
Let z = r + si
Then x + yi = z²
(with the same restrictions on r,s as before)

SUMMARY: The square root of any Primitive Pythagorean Triple represented as x+yi (with x odd) is a complex number with integer components.

 

[RLBrown]

Primitive Pythagorean Triple (x odd) generation not using Euclid's Formula.Let a, b and c be three integers. Then,
x = a+c
y = b+c
z = a+b+c

The prime factorization of a, b, and c is critical.

a = 1 * p_1a² * p_2a² * p_3a² * ... * p_na²
b = 2 * p_1b² * p_2b² * p_3b² * ... * p_mb²
c = 2 * (p_1a * p_2a * p_3a * ... * p_na) * (p_1b * p_2b * p_3b * ... * p_mb)

Repeated use of the same prime is allowed, as long as a and b have no prime in common.
n=m=0 corresponds to the (3,4,5) triangle.