4
$\begingroup$

I was wondering how to prove the following fact about primitive Pythagorean triples:

Let $(z,u,w)$ be a primitive Pythagorean triple. Then there exist relatively prime positive integers $a,b$ of different parity such that $$z = a^2-b^2, \quad u = 2ab, \quad \text{and} \quad w = a^2+b^2.$$

I see how $z,u,w$ must form the sides of a triangle, where $c$ is the hypotenuse, but how do we show that $\gcd(z,u,w) = 1$ and that these generate all the triples?

$\endgroup$
5
$\begingroup$

It is easy to see that if $(z,u,w)$ is a primitive pythagorean triple, then $w$ is odd and either $z$ is odd and $u$ is even or $z$ is even and $u$ is odd. WLOG suppose that $u=2x$ is even, so $z$ is odd. Then

$$u^2=4x^2=w^2-z^2=(w+z)(w-z).$$

It is plain that $\gcd(w+z,w-z)=2$, so $y_1:=(w+z)/2$ and $y_2:=(w-z)/2$ are relatively prime positive integers. Thus

$$x^2=y_1y_2,$$

and therefore $y_1$ and $y_2$ are relatively prime divisors of $x^2$. This implies that $y_1$ and $y_2$ are perfect squares, so write $y_1=a^2$ and $y_2=b^2$. Hence,

$$u=2x=2ab,\quad z=y_1-y_2=a^2-b^2,\qquad\text{and}\qquad w=y_1+y_2=a^2+b^2.$$

This proves that every primitive triple has the form $(a^2-b^2,2ab,a^2+b^2)$, as wanted.

Of course, the condition $\gcd(a,b)=1$ is needed, otherwise $(a^2-b^2,2ab,a^2+b^2)$ won't be primitive.

$\endgroup$
  • $\begingroup$ Why must $w$ be odd? $\endgroup$ – user19405892 Oct 7 '16 at 20:25
  • $\begingroup$ Suppose the $w$ is even. Then $z$ and $u$ have the same parity. If both are even, then the triple is not primitive. If both are odd, then $w^2$ is multiple of $4$, but $z^2+u^2$ is not multiple of $4$ (it is $\equiv2\bmod 4$ since $x^2\equiv1\pmod4$ for any odd $x$), which is a contradiction. Therefore, $w$ must be odd. $\endgroup$ – user246336 Oct 7 '16 at 20:28
  • $\begingroup$ Even with A odd, B even, C odd, in a triple, it may not be primitive as in $(27,36,45), (75,100,125), (147,196,245), (243,324,405)$ with $GCDs$ of $3^2, 5^2, 7^2, 9^2$, respectively. $\endgroup$ – poetasis Jun 2 at 12:48
1
$\begingroup$

Even in the wikipedia article cited by @JeanMarie I could not find the following quick argument which gives all the Pyhagorean triples at one stroke: the Pythagorean equation being homogeneous, it is equivalent to the equation in rational variables $Z^2 + U^2 = 1$, or equivalently $N(Z + iU) = 1$, where $N$ denotes the norm map from $\mathbf Q(i)$ to $\mathbf Q$ , $i^2 = - 1$. The extension $\mathbf Q(i)/\mathbf Q$ being cyclic, with Galois group generated by the complex conjugation, Hilbert's theorem 90 applies, which shows that $N(Z + iU) = 1$ iff $Z + iU$ is of the form $A + iB /A - iB$. It follows from identification that $Z = A^2 - B^2 /A^2 + B^2$ and $U = 2AB / A^2 + B^2$. Clearing denominators immediately produces the usual Pythagorean integral triples $z = a^2 - b^2 , u = 2ab, w= a^2 + b^2$. Clearing common factors plainly gives the primitive triples.

$\endgroup$
1
$\begingroup$

I thought I had a proof that, if $m,n$ are mutually prime, $m,n$ would generate only primitive Pythagorean triples. The $proof$ is between the asterisk below. $$\text{*************************}$$

$\text{We are given }\quad A=m^2n^2\quad B=2mn\quad C=m^2+n^2$

Let $x$ be the GCD of $m,n$ and let $p$ and $q$ be the cofactors of $m$ and $n$ respectively. Then we have

$$A=(xp)^2-(xq)^2\quad B=2xpxq\quad C=(xp)^2+(xq)^2$$ $$A=x^2(p^2-q^2)\quad B=2x^2(pq)\quad C=x^2(p^2+q^2)$$

If $GCD(m,n)=1$, then $GCD(A,B,C)=1$ and $(A,B,C)$ is a primitive triple. This means that $m$ and $n$ must be co-prime to generate a primitive.

$$\text{*************************}$$

However, a counter-example destroys to so-called proof: $\quad\text{Let }m,n=7,3$. $$A=49-9=40\quad B=2*7*3=42\quad C=49+9=58\quad GCD(40,42,58)=2$$ $$\therefore GCD(m,n)=1\neg\implies GCD(A,B,C)=1$$ The only two formulas I know about that will generate only primitive triplets do not generate all of them but $C-B=1$ in the first one and $C-A=2$ in the second one.

$$A=2n^2+1\quad B=2n^2+2n\quad C=2n^2+2n+1$$ $$A=4n^2-1\quad B=4n\quad C=4n^2+1$$

$\endgroup$
  • $\begingroup$ Where can I find more information about Ellingson'sequation? Wikipedia does not give any. $\endgroup$ – RTn Jun 2 at 7:34
  • $\begingroup$ Can we have a chat on this channel (I don't want to compromise the integrity of an unpublished paper)? Because i know of a similar equation first given in 628 CE (about 1500 years ago). So I think it is not very new (I may be wrong, so in that case I apologize) $\endgroup$ – RTn Jun 2 at 8:52
  • $\begingroup$ Let us continue this discussion in chat. $\endgroup$ – RTn Jun 2 at 9:01
  • $\begingroup$ So you have thought up some equations, have not published them and yet decided to just name them ellingson equations, probably using your own name? That is not good scientific practice. $\endgroup$ – Hirshy Jun 2 at 10:34
  • $\begingroup$ I don't understand why there's an argument. You give various formulas which are the well-known formula of Pythagoras. $$x^2+y^2=z^2$$ $$x=2ps$$ $$y=p^2-s^2$$ $$z=p^2+s^2$$ Your whole game... this is a substitution of numbers of another form. Like this. $$p=2n-1+k$$ $$s=k$$ It's not worth the time. Another formula describing all the solutions - can not be obtained... $\endgroup$ – individ Jun 2 at 12:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.