Based on a few brute-force calculations, I've formulated the following.

Conjecture. Let $x,y,u,v,p,q,a,b,c \ge 2$ be integers such that $$ (x^2+ay^2)(u^2+bv^2) = p^2+cq^2, $$ and write \begin{align} g_1 &= \gcd(x,y)\gcd(u,v)\gcd(p,q), \\ g_2 &= \gcd(a,xy)\gcd(b,uv)\gcd(c,pq), \\ g_3 &= \gcd(a,b)\gcd(a,c)\gcd(b,c). \end{align} Then $g_1g_2g_3>1$.

Is the conjecture true? What would be a good way of trying to prove it? Alternatively, what would be a good way of narrowing the search for counterexamples (e.g., modular considerations, etc.)?

Note that it isn't enough to simply restrict $g_1 = 1$, since $$(x,a,y,u,b,v,p,c,q)=(2, 2, 3, 3, 7, 2, 2, 10, 9)$$ is a solution with $g_1=1$ (but $g_2g_3>1$).


I have an equation of the form $$ (x^2+ay^2)(u^2+bv^2) = p^2+cq^2, $$ where $a,b,c$ are pairwise relatively prime squarefree integers.

Are there any important results regarding this special equation?

EDIT: In Composition of Binary Forms and the Foundation of Mathematics, Harold M. Edwards says that Gauss [ca. Art 235/236 of Disquisitiones Arithmeticae] proved the following theorem:

If $f$ and $\phi$ can be composed, the ratio of their determinants must be a ratio of squares.

Is this really a necessary and sufficient condition? If so, doesn’t the fact that my equation $$ (x^2+ay^2)(u^2+bv^2) = p^2+cq^2 $$ have two forms on the left-hand side being composed into the single form on the right-hand side then demand that the determinants of the two left-hand forms must be a ratio of squares?

EDIT: Cross-posted to MO a more specific question about the existence or derivation of a complete solution.

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    $\begingroup$ Is your meaning that one considers $a,b,c$ as given fixed pairwise coprime, and then trys to show existence or construction of choices for $x,y,u,v,p,q$? [Or maybe are you doing the reverse, and simply asking how to find triples $a,b,c$ for which there are values of the other variables to make the relation hold?] $\endgroup$ – coffeemath Sep 15 '14 at 2:55
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    $\begingroup$ Your conditions preclude the one thing that resembles this, Arndt gave a recipe, pages 129-131 in Buell. $\endgroup$ – Will Jagy Sep 15 '14 at 2:57
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    $\begingroup$ $(x^2+ay^2)(u^2+bv^2)=(tp)^2+c(tq)^2=t^2(p^2+cq^2)$ The solution of the equation: $x^2+ay^2=t^2$ It turns. $x=p^2-as^2$ ; $y=2ps$ ; $t=p^2+as^2$ For the equation: $u^2+bv^2=p^2+cq^2$ There are solutions if the number of $(bc)$ or $(b-c+1)$ square. $\endgroup$ – individ Sep 15 '14 at 5:42
  • $\begingroup$ @coffeemath: The former. I have $a,b,c$, and the others are to be constructed. Almost-best-case scenario: $x,y,u,v,p,q$ are polynomials in $a,b,c$. Best-case scenario: it's "if and only if". $\endgroup$ – Kieren MacMillan Sep 15 '14 at 12:08
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    $\begingroup$ Got your email. The behavior with $a,b,c$ constants has essentially no bearing on the case when $a,b,c$ depend on the other variables. $\endgroup$ – Will Jagy Sep 15 '14 at 18:52

This is an attempt to show that $$(x^2+ay^2)(u^2+bv^2)=p^2+cq^2 \tag{1}$$ has solutions, without conditions on $a,b,c$ beyond their being distinct. We first select $x,y$ to make $x^2+ay^2=m^2.$ If $a=1$ we may use any pythagorean triple, say $(x,y)=(3,4)$ and $m=5.$ Otherwise put $x=a-1,\ y=2,\ m=a+1$ since $(a-1)^2+4a=(a+1)^2.$ [Note: since later we divide by $m^2$ we need $m \neq 0$ so if by chance $a=-1$ we may use $x=5,\ y=4,\ m=3.$] Replacing the first factor of $(1)$ by $m^2$ and multiply through, we have $$(mu)^2+b(mv)^2=p^2+cq^2.$$ In this, put $q=mv,\ p=ms$ (where $s$ is a "new" variable to be determined) and divide by $m^2$ to obtain $$u^2+bv^2=s^2+cv^2.\tag{2}$$ At this point the argument branches on whether $|c-b|$ is a square. First assume it is not. Then there are the two rearrangements $$u^2-(c-b)v^2=s^2,\\ s^2-(b-c)v^2=u^2.\tag{3}$$ Depending on the sign of $c-b$ the left side of one of these is that of a Pell equation $m^2-Dn^2=1$ [with $D>0$ and $D$ not a square], and we may take our right side $s^2$ or $u^2$ to be $1$ to make one of these exactly a Pell equation (with the $1$ on the right). Since the Pell equation has solutions we have a solution to $(1).$

The remaining case, wherein $|c-b|$ is a square, say $k^2,$ is simple; take a 3-4-5 pythagorean triple and multiply it through by $k^2$ depending on which one of $(3)$ is chosen.

Note that, as pointed out by @individ, one really does not need Pell to solve equation $(2).$

  • $\begingroup$ Above given formula for the equation; $u^2+bv^2=s^2+cv^2$ always gives a solution. And the Pell equation is not necessary. $\endgroup$ – individ Sep 16 '14 at 11:41
  • $\begingroup$ @coffeemath: Thanks! I'm trying to put together a "complete solution", and this will definitely help. $\endgroup$ – Kieren MacMillan Sep 16 '14 at 12:19
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    $\begingroup$ @coffeemath: When I attack a Diophantine equation, I always start by hoping I can find a "complete solution", which is an integer parameterization if it admits one, or a characterization of all possible cases otherwise. Sometimes, this task is relatively easy (e.g., equal sums of squares); in other cases it is more difficult; in many (most?) cases, I have to give up fighting that particular windmill. I didn't mean to say that $\gcd(p,q)>1$ is a drawback: it’s a [useful] restriction. If I could prove that $\gcd(p,q)>1$ is a necessary condition, that would be great. $\endgroup$ – Kieren MacMillan Sep 16 '14 at 13:27
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    $\begingroup$ It's just. The formula holds when $u^2+(b-c)v^2=s^2$ When $(b-c)$ is greater than zero. If less redraw slightly different form. $\endgroup$ – individ Sep 16 '14 at 13:47
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    $\begingroup$ @individ I agree no Pell needed, as one can do a rational parametrization of $u^2+(b-v)v^2=1$ based on the initial solution $u=1,v=0$ and then multiplying through by a denominator. I'll throw that in, citing your input. $\endgroup$ – coffeemath Sep 16 '14 at 13:55

Never mind: I found a counterexample…

$$(x,a,y,u,b,v,p,c,q)=(1471528, 9777203, 9668027, 5703851, 3364185, 2582134, 16, 819951918602068063799187294877418918951, 5)$$

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    $\begingroup$ The formula in the General form for this equation is very long. What's the point of wasting time and to write it? $\endgroup$ – individ Sep 18 '14 at 5:20
  • $\begingroup$ If you have a general formula for ALL solutions, it could be used to prove some very powerful theorems very simply. That's the point. $\endgroup$ – Kieren MacMillan Sep 18 '14 at 11:26
  • $\begingroup$ Is your general formula somehow derived from my counterexample? If so, I'm sure we can find a much smaller counterexample as a "seed value". $\endgroup$ – Kieren MacMillan Sep 18 '14 at 11:27

It is necessary to say a few words about the formula about which I have spoken. For the equation: $$(x^2+ay^2)(u^2+bv^2)=p^2+cq^2$$

Need to write this simple formula.







I think that this formula gives all solutions. Mutually simple solution obtained after reduction to common divisor.

For example there was a similar situation with the equation: $X^2+Y^2=Z^{n}$

It is enough to write the formula generates an endless series of decisions in all degrees. For this we use the Pythagorean triple. And the number of their sets.

$$a^2+b^2=c^2$$ $$a=2ps$$ $$b=p^2-s^2$$ $$c=p^2+s^2$$

$p,s$ - what some integers. Then the solution can be written.
$$X=2psc^{n-1}$$ $$Y=(p^2-s^2)c^{n-1}$$ $$Z=c^2$$

And mutually simple solutions can get if you cut down on common divisor. Although there will be not one simple solution.

  • $\begingroup$ As a set of formulas, this is rather amazing. I will need to prove whether it is the complete solution, before I can wield it like the Odinsword… but thank you for even putting me on the right path. $\endgroup$ – Kieren MacMillan Sep 18 '14 at 20:58
  • $\begingroup$ This can't be the complete solution: considering my counterexample (see other answer), we have $\gcd(u,v)=1$. Hence $n^2+cj^2=1$, and since $c>1$, we must have $j=0$ and $n=\pm1$. But now $q=0$, which is not true of my counterexample. $\endgroup$ – Kieren MacMillan Sep 18 '14 at 21:14
  • $\begingroup$ I realise you may have meant that the complete solution with no common factors would be: \begin{align} x &= r^2-as^2 \\ y & = 2rs \\ u &= k^2-bt^2 \\ v &= 2kt \\ p &= n^2-cj^2 \\ q &= 2nj, \end{align} where all of the common polynomial factors have been removed. But this also can't be the complete solution, since my counterexample has $y$ odd, and your solution forces $y$ to be even. $\endgroup$ – Kieren MacMillan Sep 18 '14 at 23:01
  • $\begingroup$ @KierenMacMillan I meant that after substitution of different factors. Then reduce by common divisor. In this case, will mutually simple solutions? $\endgroup$ – individ Sep 19 '14 at 4:45

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