proof of Pythagorean triplet

Consider the $4AB={(A+B)}^2-{(A-B)}^2$. Assume that $A,B$ are coprime. (This is no : If $d$ is the greatest common divisor of $A,B$, then one can write $A=d{A}^{{}^{\prime }},B=d{B}^{{}^{\prime }}$ to get $4d^2{(A^{{}^{\prime }}{B}^{{}^{\prime }})}^2=d^2\left({(A^{{}^{\prime }}+B^{{}^{\prime }})}^2-{(A^{{}^{\prime }}-B^{{}^{\prime }})}^2\right)$ and cancel $d^2$.)

For $4AB,A+B,A-B$ to form a Pythagorean triple, each of $A,B$ must be squares. So $A=m^2,B=n^2$ where $m,n$ are coprime, . So we have

$$a=2mn,b=m^2-n^2,c=m^2+n^2$$ (1)

and $\{a,b,c\}$ are a Pythagorean triple. But this needn’t be primitive: If $m,n$ are odd, then $2\mid m^2\pm n^2$, so not all of $a,b,c$ are relatively prime.

Suppose $2mn,m^2-n^2,m^2+n^2$ are pairwise coprime. Then $\gcd (2mn,m^2+n^2)=\gcd (2mn,{(m+n)}^2)=1=\gcd (2mn,{(m-n)}^2)$, and it follows that $\gcd (2mn,m+n)=1,\gcd (2mn,m-n)=1$. Thus $\gcd (2,m+n)=1$, i.e. $m\pm n$ is odd. Furthermore, $\gcd (2mn,m+n)=1$ implies $\gcd (m,n)=1$. And since the sum/difference of two integers is odd iff one is even, and the other is odd, only one of $m,n$ is odd. Thus, $\gcd (m^2-n^2,m^2+n^2)=\gcd (m^2+n^2,2n^2)=1$. Conversely, if $m,n$ are coprime, and exactly one of $m,n$ is odd, then $\gcd (2,m\pm n)=1$; thus, $2mn$,$m^2-n^2$ are coprime. From the fact that $\gcd (a+b,a-b)=\gcd (a,b)$ if $a,b$ have opposite parity and $\gcd (m^2,n^2)=1$ it follows that $m^2-n^2$,$m^2+n^2$ are also coprime. And since $\gcd (2mn,m^2+n^2)=\gcd (2mn,{(m+n)}^2)$ and $\gcd (2mn,m+n)=1$ it follows that $2mn,m^2+n^2$ are coprime. So the conditions the Pythagorean triple $\{2mn,m^2-n^2,m^2+n^2\}$ is primitive, $\gcd (2mn,m+n)=1$ and $m,n$ are coprime and exactly one of them is odd are equivalent.

So if $a,b,c$ satisfy $a^2+b^2=c^2$ and $a,b,c$ are pairwise coprime, then $c$ is odd, and exactly one of $a,b$ are odd and the other is even.

Let $n,m$ be coprime positive integers of opposite parity, . Set

$$n^{{}^{\prime }}=m+n,m^{{}^{\prime }}=n-m$$ (2)

in equation 1 gives

$$b=m^{{}^{\prime }}{n}^{{}^{\prime }},c=\frac{n^2+m^2}{2}=\frac{m^{{}^{\prime }}{}^2+n^{{}^{\prime }}{}^2}{2},a=\frac{n^{{}^{\prime }}{}^2-m^{{}^{\prime }}{}^2}{2}$$ (3)

since $n=\frac{m^{{}^{\prime }}+n^{{}^{\prime }}}{2}$, $m=\frac{n^{{}^{\prime }}-m^{{}^{\prime }}}{2}$. Clearly, $\gcd (m^{{}^{\prime }},n^{{}^{\prime }})=1$.

Now we prove that any primitive Pythagorean triple can be generated choosing odd coprime integers.

Remark 1.

Let $m\mathrm{,}n$ be odd coprime integers, . Let $f_{\mathrm{1}}\mathrm{=}\frac{n^{\mathrm{2}}\mathrm{-}{m}^{\mathrm{2}}}{\mathrm{2}}\mathrm{,}{f}_{\mathrm{2}}\mathrm{=}\frac{n^{\mathrm{2}}\mathrm{+}{m}^{\mathrm{2}}}{\mathrm{2}}$. Then $\gcd \mathit{}\mathrm{(}{f}_{\mathrm{1}}\mathrm{,}{f}_{\mathrm{2}}\mathrm{)}\mathrm{=}\mathrm{1}$.

Proof.

Since $\gcd (n^2,m^2)=1=\gcd (f_1+f_2,f_1-f_2)$, the statement follows from the fact that $f_1,f_2$ have opposite parity since in this case $\gcd (f_1,f_2)=\gcd (f_1+f_2,f_1-f_2)$. Since $4\mid n^2-m^2$, $f_2$ is odd, and since $f_1=f_2+n^2$ and $n$ is odd, $f_1,f_2$ have opposite parity. ∎

Substituting $C=n^2$, $B=m^2$ in $BC={\left(\frac{C+B}{2}\right)}^2-{\left(\frac{C-B}{2}\right)}^2$ yields that $mn,\frac{n^2-m^2}{2},\frac{n^2+m^2}{2}$ is a primitive Pythagorean triple.

To see that any primitive Pythagorean triple is of this form:

Theorem 1.

Let $a\mathrm{,}b$ be positive coprime integers satisfying $a^{\mathrm{2}}\mathrm{+}{b}^{\mathrm{2}}\mathrm{=}{c}^{\mathrm{2}}$. Then $a\mathrm{,}b$ have opposite parity, and $c$ is odd. Furthermore $\mathrm{(}a\mathrm{,}c\mathrm{)}\mathrm{=}\mathrm{(}b\mathrm{,}c\mathrm{)}\mathrm{=}\mathrm{1}$.

Proof.

$a,b$ cannot both be even since $\gcd (a,b)=1$. If both $a,b$ were odd we had $c^2\equiv 2(mod4)$ which is impossible since the square of any number is either congruent (http://planetmath.org/Congruences) 0 or 1 modulo 4. Thus, $c$ must be odd. Now for any integers $a,b$ the congruence $a^2+b^2\equiv {(a+b)}^2(mod2)$ holds. Together with $c^2\equiv 1(mod2)$ this gives $a+b\equiv 1(mod2)$, so $a,b$ have opposite parity. ∎

Suppose $a$ is odd. Since $a^2=(c+b)(c-b)$ is a square, and $(c+b,c-b)=(c+b,2b)$ and $(2,c+b)=1$ it follows that $c\pm b$ are coprime and consequently each of them is square. This gives $c-b=n^2,c+b=m^2$ where $m,n$ are odd coprime integers, and we get

$a^2=m^2{n}^2\iff$ (4)

$a$ $=$ $mn,$ (5)

$b$ $=(c+b-(c-b))/2$ $={\displaystyle \frac{m^2-n^2}{2}},$ (6)

$c$ $={\displaystyle \frac{n^2+m^2}{2}}.$ (7)

Now let $A=\frac{ab}{2}$ be a square. Without loss of generality we can set $a=mn$, $b=\frac{n^2-m^2}{2}$ where $m,n$ are odd coprime integers. So we have $A=mn\frac{n^2-m^2}{4}$, and since $mn$ and $\frac{n^2-m^2}{4}$ are coprime, each of them must itself be a square. So we have

$$c^2={\left(\frac{n^2+m^2}{2}\right)}^2={(mn)}^2+{\left(\frac{n^2-m^2}{2}\right)}^2$$ (8)

where the right-hand side numbers are biquadratic integers. So the question if the area of a right triangle with integer sides is square is equivalent to asking if $x^4+y^4=z^2$ has a solution in positive integers.

Title	proof of Pythagorean triplet
Canonical name	ProofOfPythagoreanTriplet
Date of creation	2013-03-22 14:06:52
Last modified on	2013-03-22 14:06:52
Owner	Thomas Heye (1234)
Last modified by	Thomas Heye (1234)
Numerical id	12
Author	Thomas Heye (1234)
Entry type	Proof
Classification	msc 11D09
Related topic	ContraharmonicMeansAndPythagoreanHypotenuses