0

I want to parameterize the equation $c^2 = a^2 + b^2 = d^2 + 4 b^2$ but every avenue I go down leads to an equation I can’t solve.

I know $a^2 + b^2 = c^2$ is parameterized by $(2uvk, (u^2 - v^2)k, (u^2 + v^2)k)$ so it should be as simple as:

$(u^2 + v^2)k = (p^2 + q^2)n$

$(uv)k = (2pq)n$

Or some other combination of terms, but I’m still having trouble actually solving it.

Could anyone provide some guidance?

ServingSpy
  • 45
  • 1
    Note that that's not the parametrization for all pythagorean triples, but only for PRIMITIVE triples. You need to include a factor of $k$. – Calvin Lin Jun 16 '23 at 16:15
  • @CalvinLin Thank you! I fixed that in the prompt – ServingSpy Jun 16 '23 at 16:47
  • You fixed that in one place. Can you follow through all of the changes needed? – Calvin Lin Jun 16 '23 at 16:48
  • Hint:2^2=4 – Safal Das Biswas Jun 16 '23 at 18:36
  • You already made it discussing triplets just bashing out that 2^2 out of $2uvk$ – Safal Das Biswas Jun 16 '23 at 18:37
  • see if you can prove that there are no such examples with integers $a,b,c,d \neq 0$ – Will Jagy Jun 16 '23 at 19:17
  • @Will Jagy if I construct an example then? – Safal Das Biswas Jun 16 '23 at 20:52
  • @HappieHappieHappie that would be surprising – Will Jagy Jun 16 '23 at 21:00
  • 13^2=12^2+5^2=4.6^2+5^2 anything else? – Safal Das Biswas Jun 16 '23 at 21:38
  • Many such exist because in PT there are 2uv factor infact c^2 has two different representation in PT and hence it works. – Safal Das Biswas Jun 16 '23 at 21:40
  • Suprise Suprise Suprise @WillJagy Sir Lol!!! Like a Villan I am talking. But What's correct is correct may be it's given by a Villan – Safal Das Biswas Jun 16 '23 at 21:42
  • 1
    @HappieHappieHappie In the original $c^2 = a^2 + b^2 = d^2 + 4 b^2$ what are your values for $a,b,c,d$ exactly? – Will Jagy Jun 16 '23 at 21:49
  • @HappieHappieHappie Your example requires $c=13$ and the $4 \cdot 6^2$ says that $b = 6.$ Then $d=5.$ The thing that fails is that $c^2 - b^2 = 13^2 - 6^2 = 133. $ For your example to be legitimate you would need $a^2 = 133,$ but $133$ is not a perfect square and the resulting $a$ is not an integer. – Will Jagy Jun 16 '23 at 22:07
  • This with the comments indicates that there are no solutions in one case. Unfortunately, it's not an elementary solution. (It's also missing another case, due to forgetting about primitive solutions) – Calvin Lin Jun 16 '23 at 22:12
  • @WillJagy Can You review you comment well, There exist infinitely many primitive solution to the above Given Equation Not explicitly but by using the fact that every composite number pqr of the form 4k+1 can be written as a^2+b^2 – Safal Das Biswas Jun 16 '23 at 22:19
  • And Why d=5? I am really confused what you said – Safal Das Biswas Jun 16 '23 at 22:20
  • Fact I used in $pqr$ is that $p\equiv q\equiv r\equiv 1 \pmod 4$ where all are primes. Now use fact of famous identity $(a^2+b^2)(u^2+v^2)=h^2+f^2$ which can be obtained by Gaussian integer split. – Safal Das Biswas Jun 16 '23 at 22:24
  • @HappieHappieHappie let me not confuse you further. If you have an example, please display it by writing $a=$ and giving the specific value for $a,$ then same for $b,c,d$ – Will Jagy Jun 16 '23 at 22:27
  • @CalvinLin thanks. I would guess the descent can be done in the manner of Mordell's book, the chapter on quartic equations with only trivial solutions. However, cannot be sure until all the details work out. – Will Jagy Jun 16 '23 at 22:30
  • Actually values are critical to find But I have construction that told us that there are infinitely many values because for example $13\times 17\times 29=c$ is very large like that I can construct large solutions.Then there exist different $a, b, e, d$ – Safal Das Biswas Jun 16 '23 at 22:32
  • 1
    @HappieHappieHappie I'm inclined to agree with Will here. Just state any solution that you've found, even if it's very large. $ c= 6409$ isn't too bad. $\quad$ However, I don't think there are corresponding integers $a, b, d$ that satisfy $6409^2 = a^2 +b^2 = d^2 + 4b^2$. – Calvin Lin Jun 16 '23 at 22:36
  • Okay I see Op Want's $b$ in 2 equations That I missed I thought it was, $c^2=a^2+b^2=d^2+4e^2$ that I missed. that there are two b's – Safal Das Biswas Jun 16 '23 at 22:36
  • My solution works if there are five distinct variable and not 4 – Safal Das Biswas Jun 16 '23 at 22:38
  • I hope You got my point What I am trying to say. It's my careless ness to not look for equations properly I am sorry. – Safal Das Biswas Jun 16 '23 at 22:40
  • $a^2=u^2+3v^2$ which has A parameterization $a=m^2+3n^2$ $u=m^2-3n^2$ and $v=2mn$ there after we need $m^2+3n^2=g^2+f^2$ I think from here we can proceed. – Safal Das Biswas Jun 16 '23 at 22:46
  • What about $\pmod 3$ on last equation now? – Safal Das Biswas Jun 16 '23 at 22:51
  • It' has Solutions in $\mathbb{F}_3$ for sure – Safal Das Biswas Jun 16 '23 at 22:52
  • 1
    @HappieHappieHappie - the point you need to get is that if someone starts questioning your comment, it is wise to look deeply at what you've written before giving a flippant answer. It is even more wise when the person who is questioning is of high reputation (which you can see in comments by hovering the mouse over their signature) because they didn't get that reputation by being careless. And if their name is Will Jagy, it wiser still. To be sure, "look carefully, and correct if wrong" is all you should do, not back down when you are right. We all make mistakes. But tread with care. – Paul Sinclair Jun 17 '23 at 19:22
  • Thank you, for letting me know, I will follow. – Safal Das Biswas Jun 18 '23 at 09:33

0 Answers0