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I have the equation: $$y^2=x^3-2$$ It seems to be deceivingly simple, yet I simply cannot crack it. It is obviously equivalent to finding a perfect cube that is two more than a perfect square, and a brute force check shows no solutions other than $y=5$ and $x=3$ under 10,000. However, I can't prove it.

Are there other integer solutions to this equation? If so, how many? If not, can you prove that there aren't?

Bonus: What about the more general equation" $$y^2=x^3-c$$ Where $c$ is a positive integer?

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    $\begingroup$ The specific equation here is an exercise in Chapter 2 of M. Ram Murty, J. Esmonde, Problems in Algebraic Number Theory. As André implied it depends on unique factorization in $\Bbb{Z}[\sqrt{-2}]$. The general case (IIRC AKA Mordell Curves) is much more difficult. $\endgroup$
    – Jyrki Lahtonen
    Jul 18, 2016 at 15:30
  • $\begingroup$ @AndréNicolas Could you possibly link me to some pages about unique factorization? The Wolfram Mathworld page isn't that helpful. $\endgroup$
    – Nico A
    Jul 18, 2016 at 15:31
  • $\begingroup$ @TreFox: Unfortunately, for generational reasons and because I have not taught number theory for a while, I am not familiar with online stuff. $\endgroup$
    – André Nicolas
    Jul 18, 2016 at 16:05
  • $\begingroup$ @AndréNicolas No problem! Could you perhaps explain a bit about what $Z[\sqrt{-2}]$ means? I know Z is the set of all integers, but I'm not sure what the square root of -2 means in this context. $\endgroup$
    – Nico A
    Jul 18, 2016 at 16:07
  • $\begingroup$ It is the collection of all numbers of the form $a+b\sqrt{-2}$, where $a$ and $b$ range over the ordinary integers. As to $\sqrt{-2}$, if you are not yet acquainted with complex numbers, it would take a while to explain. Such extensions of the ordinary integers are surprisingly often useful in solving problems about ordinary integers. This kind of number theory is usually not done (in North America at least) until the second or third year of university. By the way, Wikipedia is often less terse than MathWorld. $\endgroup$
    – André Nicolas
    Jul 18, 2016 at 16:12

2 Answers 2

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The only integral solutions to your first problem are $(3, \pm 5)$. The general class of equations are known as Mordell's equation. A fairly elaborate discussion and case by case analysis is provided here.

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  • $\begingroup$ So am I right in thinking this is probably a typo if it appeared on an Algebra 1 math worksheet? $\endgroup$
    – Nico A
    Jul 18, 2016 at 15:19
  • $\begingroup$ No. Your thinking is not right. Solutions to $y^2=x^3-2$ can be done in Algebra $1$. $\endgroup$
    – Adhvaitha
    Jul 18, 2016 at 15:22
  • $\begingroup$ How would one go about solving that equation in Algebra 1 - other than just trial and error? $\endgroup$
    – Nico A
    Jul 18, 2016 at 15:39
  • $\begingroup$ @TreFox As mentioned in the article on page $7$. $\endgroup$
    – Adhvaitha
    Jul 18, 2016 at 15:41
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    $\begingroup$ The factorization $x^2+2=(x-\sqrt{-2})(x+\sqrt{-2})$ is already not Grade 7 stuff, and neither is the ring $\mathbb{Z}[\sqrt{-2}]$. Perhaps you were just supposed to play around and find the solutions $x=\pm 5$, $y=3$ and report that you had found no others. $\endgroup$
    – André Nicolas
    Jul 18, 2016 at 16:11
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Fact : $\mathbb{Z}[\sqrt{-2}]$ is Unique factorization domain. lemma : in every UFD, if the product of two numbers, which are relatively prime is a cube, then each of them must be a cube. There is no any solution

$$x^3 = (y+\sqrt{-2})\times(y-\sqrt{-2})$$

the greatest common divisor of these factors will divide 2-times the $\sqrt{-2}$, which is lead to only finitly many cases. (some cases can be shown impossible, only by the modular an congrunce arithmetic.)

finally we have: $$y+\sqrt{-2}=(a+b\sqrt{-2})^3$$ which lead us to the system of equations as follows: $$a^3-6ab^2=y$$ and $$3a^2b-2b^3=1$$ then $$b(3a^2-2b^2)=1$$ which implies the assertion.

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  • $\begingroup$ Welcome to MathSE! Your post will be easier to read if you format it. Please see this informative tutorial on formatting. $\endgroup$
    – Théophile
    Jul 18, 2016 at 15:32
  • $\begingroup$ I can't type in that way, I can't understand what is the proble. for example when I want to write " a-sub-1" .... I type a_1 but the result does not differ, i can't understand what should I do. $\endgroup$
    – Davood
    Jul 18, 2016 at 19:40
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    $\begingroup$ You need to surround the expression with dollar signs. So to get $a_1$, type $a_1$. If you use two dollar signs, then it will appear centred on a new line. For example, $$a_1$$ produces $$a_1$$ $\endgroup$
    – Théophile
    Jul 18, 2016 at 20:37

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