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In my current number theory work, I'm running into many equations of the form in the title.

Q1: Am I correct in calling this a quartic Thue equation?

Q2: Are there general methods to attacking this special case where the constant is the square $z^2$? In particular, are there any powerful elementary methods?

Q3: What are the best general references/links regarding this equation?

Example: One equation that I'm currently trying to solve is $$x^4-18x^2y^2+40xy^3-27y^4=z^2.$$ (In fact, in many of the equations I am running into, $b=0$ or $d=0$.) I am trying to prove that the only primitive solution in positive integers is $(x,y,z)=(27,10,171)$.

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  • $\begingroup$ aren't those things elliptic curves ? (provided they aren't singular) $\endgroup$ – mercio Mar 17 '16 at 13:44
  • $\begingroup$ @mercio: Maybe…? Why do you ask? $\endgroup$ – Kieren MacMillan Mar 17 '16 at 13:48
  • $\begingroup$ Why do You need it? The answer to this question You will never be satisfied. In addition, the formula is very cumbersome that nobody would use it. Even for simple equations - these equations are necessary to nobody. $\endgroup$ – individ Mar 17 '16 at 14:08
  • $\begingroup$ provided the curve is nonsingular and you find at least one rational point on it, then it is a rational elliptic curve and there are lots of methods and theorems to study its rational points. $\endgroup$ – mercio Mar 17 '16 at 14:16
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These are only termed quartic Thue equations if the values of $z$ is fixed. As noted above, if there is a rational point, then such an equation defines an elliptic curve; these are much studied but, as far as I am aware, there are no particularly useful elementary approaches that lead to reasonable understanding of their rational points in general. Books by Cassels and Silverman are very useful here.

For your example, it is easy to show that it is isomorphic over $\mathbb{Q}$ to Cremona's 1728v1, which has rank $1$ and no nontrivial torsion. It follows that your solution is one of infinitely many (in fact, it corresponds to the generator on the curve). Doubling this point and mapping it back to your quartic model yields the next solution : $$ x=5383, \; y=1710, \; z=12660211. $$

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  • $\begingroup$ Can it be shown easily that, of all possible solutions, $$ \frac{10y^3-9xy^2+x^3 \pm x\sqrt{x^4-18x^2y^2+40xy^3-27y^4}}{2y^3} $$ is only an integer for the generator $(x,y)=(27,10)$, and in that case, only for the $+$ sign? $\endgroup$ – Kieren MacMillan Mar 17 '16 at 22:35
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    $\begingroup$ Actually, I suspect that this would be very hard to show (or even not true -- one could search for counterexamples easily enough). Any solution with $y$ prime would lead to an integer, for example... $\endgroup$ – Mike Bennett Mar 17 '16 at 23:19
  • $\begingroup$ The family of solutions $(27k,10k), k=1,2,\dots$ all give $383/1000$ with the minus sign and $5$ with the plus sign. Your other solution gives $-383/1000$ and $66234835/5000211$ with the minus and plus, respectively. $\endgroup$ – Kieren MacMillan Mar 17 '16 at 23:47
  • $\begingroup$ Yes, but you need to work out multiples of the generator on the elliptic curve (these don't correspond to multiples of the solutions (x,y,z)). $\endgroup$ – Mike Bennett Mar 18 '16 at 1:07
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M. Bennett gave a good answer, but if you want an elementary method, then here is one. Given,

$$ax^4+bx^3y+cx^2y^2+dxy^3+ey^4=z^2\tag1$$

Since the polynomial is homogeneous, let $x=py$ to express it in the simpler form,

$$ap^4+bp^3+cp^2+dp+e = z^2\tag2$$

with your case as,

$$\color{blue}{p^4 - 18p^2 + 40p - 27 = z^2}\tag3$$

Step 1. Finding an initial solution

If $a$ is a square, then divide all coefficients by it. If you are in luck and your leading coefficient is already $a=1$, then an initial solution can easily be attained. It is given by,

$$p = \frac{(b^2 - 4 c)^2 - 64 e}{8 b(b^2 - 4 c) + 64 d}\tag4$$

The formula yields $\color{blue}{p_1=\frac{27}{10}}$ which you have already found.

Step 2. Finding other solutions

What you want is to make the constant term of $(3)$ as a square. Do the substitution $p=q+\frac{27}{10}$ on $(3)$ to get,

$$q^4+\tfrac{54}{5}q^3+\tfrac{1287}{50}q^2+\tfrac{5383}{250}q+\color{red}{\big(\tfrac{171}{100})^2}=z^2\tag5$$

or, in general,

$$aq^4+bq^3+cq^2+dq+\color{red}{r^2}=z^2\tag6$$

which can now be solved.

a) First way:

If $a=1$, then,

$$q_1 = \frac{(b^2 - 4c)^2 - 64r^2}{8b(b^2 - 4c) + 64d}\tag7$$

$$q_2 = \frac{4 (-b r + d)}{b^2 - 4 c + 8 r}\tag8$$

$$q_3 = -\frac{(d^2 - 4 c r^2) + 8 r^3}{4 r (-b r + d)}\tag9$$

Using $(5)$, then $q_i = 0,\;\frac{383}{855},\;-\frac{12660211}{1309860}$, respectively. Since $p=q+\frac{27}{10}$, then,

$$\color{blue}{p_1 =\tfrac{27}{10},\; p_2 = \tfrac{5383}{1710},\;p_3 =-\tfrac{9123589}{1309860}}$$

the first you know, and the second mentioned by Bennett. Thus, if your original equation was square at both ends like $(5)$, then generally you can quickly find three solutions. (Though some coefficients yield $q_1 = q_2 = q_3$, or something similar.)

b) Second way:

For general $a$, then another solution is,

$$q_4=\frac{8 d r^2(d^2 - 4 c r^2) + 64 b r^6}{(d^2 - 4 c r^2)^2 - 64 a r^6}\tag{10}$$

Using $(5)$, then $q_4=\frac{387354562047}{96977216260}$ which yields $\color{blue}{p_4 =\frac{649193045949}{96977216260}}$.

Step 3. Finding more solutions

Simply repeat Step 2, but use the second value $p_2$,

$$p=q+\tfrac{5383}{1710}$$

which should give a third, then a fourth, etc. Do likewise for all $p_i$, and so on.

Disclaimer:. This tangent method, known way back to Fermat, is easy to understand and you can use it to prove that your initial $x,y$ are not the only ones. However, it is not perfect. It usually misses smaller points, and there are variables such that it will just yield $q_1 = q_2$. The method has been subsumed by the study of elliptic curves which give a thorough treatment of the subject.

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  • $\begingroup$ This is a wonderful explanation! Thank you. If I could accept two answers, I would. $\endgroup$ – Kieren MacMillan Mar 23 '16 at 3:44
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    $\begingroup$ @KierenMacMillan: Revision is finished. $\endgroup$ – Tito Piezas III Mar 23 '16 at 4:42
  • $\begingroup$ The revision makes it even better. Thank you! One question: How did you derive (4)? Is that known a priori? $\endgroup$ – Kieren MacMillan Mar 23 '16 at 11:07
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    $\begingroup$ @KierenMacMillan: I've included three formulas which should be useful if your eqn is square at both ends. The derivation was discussed in your last year's question here. $\endgroup$ – Tito Piezas III Mar 23 '16 at 16:36

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