1
$\begingroup$

I have attempted by using the fact that the even numbers $b$ in a PPT $(a,b,c)$ can be generated using $ b = \frac{s^2 - t^2}{2} $ for any odd integers $s >t \ge 1$ that share no common factors.

We can express any odd numbers as $2k+1$ for some non-negative integers $k$, so:

$$\frac{(2n+1)^2 - (2m^2 + 1)}{2} = \frac{4(n^2 + n - m^2 - m)}{2} \\(2n+1)^2 - (2m^2 + 1)= 2(n(n+1)-m(m+1)) \\ \text{So }\frac{ s^2 - t^2}{2} = 2(2k)=4k \text{ for some integer k }$$

But I think all I did was that I proved that all even numbers $b$ in PPT must be a multiple of 4, rather than proving that all multiples of 4 can appear as b, how would I proceed from this?

$\endgroup$
  • 2
    $\begingroup$ Try using the generating form for pythagorean triples. $\endgroup$ – Don Thousand Jul 10 '19 at 4:08
3
$\begingroup$

What you have to prove is that if I give you a $b$ that is a multiple of $4$, you can express it as $\frac {s^2-t^2}2$ with $s,t$ odd and coprime. As you did, write $b=4k$, then we are looking to have $$8k=s^2-t^2=(s+t)(s-t)$$ One way is just to let $s=t+2$. Then we get $$s-t=2\\s+t=4k\\s=2k+1\\t=2k-1$$ And we know $s,t$ are odd and coprime because if they have a common factor it must divide $s-t=2$

$\endgroup$
  • $\begingroup$ Not sure if I understood it correctly, but is it supposed to be s = 2k + 1 and t = 2k - 1 instead? Because if it's 4k then wouldn't s+t gives 8k instead of 4k? $\endgroup$ – Premsupapong Vanitcharenthum Jul 10 '19 at 5:00
  • $\begingroup$ @PremsupapongVanitcharenthum: Yes, that is correct. Fixed. $\endgroup$ – Ross Millikan Jul 10 '19 at 5:01
  • $\begingroup$ The first equation in DanielWainfleet's answer is all you need for proof. $\endgroup$ – poetasis Jul 21 '19 at 12:29
1
$\begingroup$

$(4n^2-1)^2+(4n)^2=(4n^2+1)^2.$

All Pyth. triplets are of the form $(\,k(a^2-b^2),\,2kab,\, k(a^2+b^2)\,)$ with positive integers $k,a,b.$

All primitive triplets have $k=1$, with $a,b$ co-prime and such that $a-b$ is odd. These conditions are necessary but are also sufficient. (Along with, obviously, $a>b.)$

If $2ab=4n$ then $ab=2n,$ and we see that the above conditions on $a,b$ for a primitive triplet are met if $a=2n$ and $b=1.$

$\endgroup$
  • $\begingroup$ A form of your first equation was enough. $\endgroup$ – poetasis Jul 21 '19 at 12:27
  • $\begingroup$ @poetasis. The rest was to show how you might use the formula for triplets to find the 1st line. And the second "if" in my last line is not an "iff",... which is a clue that the forms of the 1st line are not the only primitive Pyth. triplets with $2ab=4n.$ $\endgroup$ – DanielWainfleet Jul 21 '19 at 16:56
0
$\begingroup$

All non-trivial triplets have values of $B$ that are multiples of $4$ where $B=2mn$. If $m,n$ are of like parity, Euclid's formula generates trivial triples, e.g. $f(1,1)=(0,2,2)$ or $f(2,2)=(0,8,8)$ but if they are opposite parity, i.e. if $m=2x,x\in \mathbb{N}\text{ and }n=(2y-1),y\in\mathbb{N}$ or the opposite, then $4|2(2x)(2y-1)$ or $4|(2x-1)(2y)$.

$\endgroup$

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.