Pythagorean Triples- Why is this the proof?

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In summary, the solutions of $X^2+Y^2=Z^2$ are called Pythagorean Triples. All solutions with $gcd(x,y,z)=1$ can be expressed as $(x,y,z)=(r^2-s^2,2rs,r^2+s^2)$, where $r$ and $s$ are integers, $r>s$, and $(r,s)=1$. This can be proven by transforming the equation into $x^2+y^2=1$ and using analytic geometry to find all rational solutions. Then, rational $t$ values can be substituted with integers $r$ and $s$ to find the solutions for $x$, $
  • #1
evinda
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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)
 

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  • #2
evinda said:
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 $r^2$
 
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  • #3
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.


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).
 

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  • #4
mathbalarka said:
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.


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) :confused:
 
  • #5
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).
 
  • #6
evinda said:
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.
 
  • #7
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 * p1a2 * p2a2 * p3a2 * ... * pna2
b = 2 * p1b2 * p2b2 * p3b2 * ... * pmb2
c = 2 * (p1a * p2a * p3a * ... * pna) * (p1b * p2b * p3b * ... * pmb)

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.
 
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FAQ: Pythagorean Triples- Why is this the proof?

1. What is a Pythagorean Triple?

A Pythagorean Triple is a set of three positive integers (a, b, c) that satisfy the Pythagorean Theorem, which states that in a right triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.

2. Why is the Pythagorean Theorem important?

The Pythagorean Theorem is important because it is a fundamental concept in mathematics and has numerous applications in geometry, trigonometry, and physics. It is also the basis for the Pythagorean Triples, which have been studied for thousands of years and have many interesting properties.

3. What is the proof behind Pythagorean Triples?

The proof behind Pythagorean Triples is based on the Pythagorean Theorem, which can be proven using various methods including algebra, geometry, and trigonometry. The most commonly used proof is the geometric proof, which involves drawing squares on each side of a right triangle and showing that the area of the square on the hypotenuse is equal to the sum of the areas of the other two squares.

4. How can Pythagorean Triples be generated?

There are many methods for generating Pythagorean Triples, including the Euclid's Formula, which states that any Pythagorean Triple can be generated by multiplying two integers m and n, where m > n, and plugging them into the formula a = m^2 - n^2, b = 2mn, c = m^2 + n^2. Other methods include the use of primitive Pythagorean Triples, Pythagorean Triple trees, and the Chakravala method.

5. What are some real-world applications of Pythagorean Triples?

Pythagorean Triples have many real-world applications, including in architecture, engineering, surveying, and navigation. They are also used in solving problems involving distance, such as calculating the diagonal distance of a rectangle or the distance between two points on a coordinate plane. Additionally, Pythagorean Triples have been used in cryptography and coding theory.

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