Quadratic Diophantine to Pell reduction

Question

http://mathworld.wolfram.com/DiophantineEquation2ndPowers.html says the equation $$ax^2+bxy+cy^2=k$$ can be reduced to Pell equation.

Can someone explain how?

I also do not think that the article says what you claim it does. They are separate lines of though. — Calvin Lin, Sep 11 '13 at 14:55
They are not sequential statements. There is an entire paragraph (line) between them that talks about implementation in Mathematica, and another paragraph (line) which mentions deeper results for more than 2 variables. — Calvin Lin, Sep 11 '13 at 15:02
Well if you know the correct form of the question I am implying, please feel free to modify this question. I am a bit clueless here. — Turbo, Sep 11 '13 at 15:04

Macavity · Answer 1 · 2013-09-11T16:48:22.527

4

To sketch the how, starting with $ax^2+bxy+cy^2=k$, look at it as a quadratic in $x$. We then need
$$x = \frac{-b\pm\sqrt{b^2-4a(cy^2-k)}}{2a}$$ to be an integer. Clearly this needs $\sqrt{b^2-4a(cy^2-k)} = z$ to be an integer.

So we solve $z^2 + (4ac)y^2 = b^2 + 4ak$, which is a generalised Pell equation and I suppose you are familiar with solving.

Of course we need to (in the end) pick solutions only where $2a \mid (-b \pm z)$.

Another approach is to use a general translation $X = px + qy, Y = ry$ and then pick the values such that the cross term disappears, then complete the square etc... If I recall the specific translation, will post it. An old reference I have is http://www.math.niu.edu/~rusin/known-math/95/quadratics (check part 2)

edited Sep 11 '13 at 16:48

answered Sep 11 '13 at 16:37

Macavity

46,381

Great answer. Thankyou. Actually I am unfamiliar with solving Pell as well. Could you write that portion as well? – Turbo Sep 11 '13 at 19:08
1

Describing how to solve the general Pell equation is quite an elaborate job, there are methods using continued fractions, using factoring in a specific ring, algorithms for finding at least one solution etc., including cases where there are no solutions. The good thing is I am sure you should find many references on the web. For e.g. check http://www.ams.org/notices/200202/fea-lenstra.pdf . Most of them boil down to "somehow" finding one solution, if this exists, then using a characteristic recurrence pattern to find infinitely many more solutions. – Macavity Sep 12 '13 at 15:07
Can you explain on factorization in appropriate rings would produce solution of Pell equation? – Turbo Oct 05 '23 at 00:06

Tito Piezas III · Answer 2 · 2014-11-30T21:22:55.130

3

Edit: This is a shorter and clearer answer to the older version.

Actually, Mathworld is correct when it says solving,

$$a x^2 + b x y + c y^2 = k\tag{1}$$

involves the usual Pell equation,

$$x^2-Dy^2 = \pm 1$$

Given an initial solution {$m,n$} to,

$$a m^2 + b mn + c n^2 = k\tag{2}$$

then we have the identity,

$$a x^2 + b x y + c y^2 - k = k(u^2-Dv^2-1)(u^2-Dv^2+1)\tag{3}$$

where,

$$x = m u^2 - 2 (b m + 2 c n) u v + D m v^2$$

$$y = n u^2 + 2 (2 a m + b n) u v + D n v^2$$

$$D = b^2-4ac$$

for arbitrary variables $u,v$. However, if one wishes to set $(3)$ equal to zero, then $u,v$ should be chosen to satisfy the Pell equation,

$$u^2-Dv^2=\pm 1$$

which, for non-square discriminant $D>0$, should provide an infinite number of integer solutions for the $x,y$ of $(3)$.

edited Nov 30 '14 at 21:22

answered Nov 28 '14 at 00:06

Tito Piezas III

54,565

Oh wow. Thanks! – Calvin Lin Nov 28 '14 at 16:39
@CalvinLin I modified the answer to make it clearer. – Tito Piezas III Nov 29 '14 at 20:40
Do you know if this provides all solutions? i.e., letting $u,v$ range over all integers, do you eventually generate all integer solutions of (1)? – Kieren MacMillan Jan 30 '18 at 19:03
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@KierenMacMillan: I'm not sure. Also, while $(3)$ is valid for *any* $u,v$, one has to choose such that $u^2-Dv^2 =\pm1$ to make $(1)$ true, hence $u,v$ does not range over all integers if the objective is $(1)$. – Tito Piezas III Jan 31 '18 at 15:10
Of course; I meant ranging over all integer solutions to the Pell equation. As for the completeness… I smell a theorem I need to prove. ;) – Kieren MacMillan Feb 01 '18 at 02:53
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@TitoPiezasIII: Do you know of a formula that uses the arbitrary Pellian equation $$u^2-Dv^2=\pm t$$ (where the formula here is the special case $t=1$)? – Kieren MacMillan Jan 06 '20 at 17:07
@tito-piezas-iii What to do if $a=b=c=1$ ? It makes $D=b^2-4ac=-3<0$. – yW0K5o Mar 31 '24 at 03:53
@yW0K5o I believe this implies the equation only has finitely many integer solutions (as is the case for many $a,b,c$). – Tito Piezas III Mar 31 '24 at 08:18

Quadratic Diophantine to Pell reduction

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