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I am trying to solve this problem. http://www.javaist.com/rosecode/problem-527-1-2-3-type-Pythagorean-triangles-askyear-2018

Here we have to find all positive integral solution of $a^2+2b^2=3c^2$ where $a+b+c\le N$. In the question, $N=25000000$.

My first approach was to loop over $a$ and for fixed value of $a$ find the solutions of $3c^2-2b^2=a^2$ using LMM algorithm (https://thilinaatsympy.wordpress.com/2013/07/06/solving-the-generalized-pell-equation/). But this is very slow. I am unable to find a faster method. Any help will be appreciated.

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  • $\begingroup$ math.stackexchange.com/questions/2773097/… $\endgroup$ – individ Dec 10 '18 at 13:14
  • $\begingroup$ This doesn't give all the solutions. $\endgroup$ – Asif Dec 11 '18 at 6:02
  • $\begingroup$ Gives all solutions. It is necessary to accurately and attentively all to write down. Although you can write such a simple solution. $$a^2+2b^2=3c^2$$ $$a=p^2+6ps+3s^2$$ $$b=p^2-3s^2$$ $$c=p^2+2ps+3s^2$$ $\endgroup$ – individ Dec 11 '18 at 6:41
  • $\begingroup$ I am not sure how you arrived at this parametric solution, but it certainly doesn't give all the solutions and also produces lots of negative integer solutions, which aren't necessary. $\endgroup$ – Asif Dec 11 '18 at 16:22
  • $\begingroup$ Rewrite the negative into a positive solution. You want to get a solution what you like, not a solution to the problem. That's not how math works. We may not like the answer, but the solution is the solution. $\endgroup$ – individ Dec 11 '18 at 16:37
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Above equation shown below has parametric solution:

$a^2+2b^2=3c^2$

$(a,b,c)=((3k^2-6k-5),(3k^2+6k-1),(3k^2+2k+3))$

For $k=4$ we get:

$(a,b,c)=(5,13,11)$

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If we use Pythagorean triples we have:

$(m^2-n^2)^2+(2 mn)^2=(m^2-n^2)^2$

$(m^2+n^2)^2-(m^2-n^2)^2=(2mn)^2=2(2m^2n^2)$

Here $a=m^2+n^2$, $c=m^2-n^2$

Now we rearrange the equation as:

$a^2-c^2=2(c^2-b^2)$

Comparing these relations means:

$c^2-b^2=2m^2n^2$

$2m^2n^2=(m^2-n^2)^2-b^2$

$b^2=m^4+n^4-4m^2n^2$

For example with $m=2$ and $n=1$ we have:

$a=2^2+1^2=5$, $c=2^2-1^2=3$ and:

$b^2=2^4+1^4-4\times2^2\times1^2=1$ or $b=1$

and we have:

$5^2+2\times1^2=3\times3^2$

This is primitive solution and subsequent solutions can be found by multiplying both sides of this relation by a number like $n^2$:

$(5n)^2+2(n)^2=(3n)^2$

$a+b+c<=25000000$

$5n+n+3n=9n<=25000000$

$n<=2777777$

This gives a set of solutions. Also $a=b=c=1$ is a primitives solution which gives another set of solutions. It is not known if the parametric formula for $b$ gives more integer solutions for b; if it does there will be more primitive solutions and thereby more solutions.

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$$(2 a^2 - 3 b^2)^2 + 2 (2 a^2 + 3 b^2 - 6 a b)^2 = 3 (2 a^2 + 3 b^2 - 4 a b)^2$$ $$(a^2 - 6 b^2)^2 + 2 (a^2 + 6 b^2 - 6 a b)^2 = 3 (a^2 + 6 b^2 - 4 a b)^2$$

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I don't think there is a solution. I have solved many such problems for $n$ in terms of $m$ and sometimes other variables. Whenever I found an integer value for $n$, I had the $m,n$ for a Pythagorean triple. Caveat: Euclid's formula generates only primitives, doubles an square multiples of primitives without a multiplier $k$, so I could be missing something but here is my attempt.

$$A^2+2B^2=3C^2\implies m^2-n^2+2(2mn)^2=3(m^2+n^2)\implies m^2-n^2+8m^2n^2=3m^2+3n^2$$

$$8m^2n^2-4n^2-2m^2=4(2m^2-1)n^2+0n-2m^2=0$$ $$a=4(2m^2-1)\qquad b=0\qquad c=-2m^2$$

$$n=\frac{-b\pm \sqrt{b^2- 4ac}}{2a}=\frac{\pm \sqrt{-4(4(2m^2))(-2m^2)}}{2(4(2m^2-1))}=\frac{\pm\sqrt{64m^4}}{8(2m^q-1)}=\frac{m^2}{2m^2-1}$$

Putting this final formula into a spreadsheet, the values of $n$, $m=1\rightarrow n=1$ which is trivial triple but, for $m=2$ thru $m=7635, n$ steadily decreased from $0.571428571428571$ to $0.500000004288663.$

Do check my algebra but I don't think there a solution because, always, $m\notin\mathbb{N}$.

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  • $\begingroup$ You misinterpret the problem. (A,B,C) isn't necessarily a Pythagorean triplet. $\endgroup$ – Asif Sep 2 '19 at 5:28

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