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A problem appeared in an ICAS math competition paper G (for approximately 15 years olds) in 2010.

The problem was:

37.A student uses trial and error to find all solutions to the question $$y^3=129-x^2$$ x and y can only be integers. How many solutions are there for this equation?

This was one of the "free response" questions at the end, with no multiple choice.

Using brute force search by listing (y,x) pairs for all y: $0 \le y \le 5$ (5 because greater than 5 will result in making $x$ complex, but greater or equal to 0 because there isn't any other lower bound!) gives me only 4 solutions.

Same result occurs when you list (x,y) pairs for all $|x|$ less than 12, since 12 and greater causes $y^3$ to be negative, although it can be.

My four solution pairs of $(x,y)$ were $(2,5)$, $(-2,5)$, $(11,2)$ and $(-11,2)$.

The answer sheet says there are 6 solutions, so there must be a solution where y is negative. After searching the internet, I found this (math help forum) and this (yahoo answers), giving another pair of solutions: $(65,-16)$ and $(-65,-16)$.

65 is a long way away from 12, and -16 is still quite far from 0, so this solution is very easy to miss out (who would list values of x all the way to 60?)

Question:

So knowing that most of us sitting this ICAS math competition paper g would not know anything about diophantine equations, and "other advanced" mathematics, how would you arrive to this answer of 6? Is there any reasoning that can give the minimum y value?


Another (but easy) problem in the same paper to give you an idea of the difficulty of this competition:

22.Which statement is true? A) $\frac{6}{7}<\frac{7}{9}<\frac{9}{11}$ B) $\frac{6}{7}<\frac{9}{11}<\frac{7}{9}$ C) $\frac{9}{11}<\frac{7}{9}<\frac{6}{7}$ D) $\frac{7}{9}<\frac{9}{11}<\frac{6}{7}$


Thank you!

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  • $\begingroup$ I don't think there is an easy solution math.uconn.edu/~kconrad/blurbs/gradnumthy/mordelleqn1.pdf $\endgroup$ – nikoma Jul 26 '13 at 23:21
  • $\begingroup$ Well, thank you anyways! $\endgroup$ – ErnWong Jul 26 '13 at 23:31
  • $\begingroup$ One way to look for solutions is using algebraic number theory writing this equation as $y^3 = (\sqrt{129}-x)(\sqrt{129}+x)$ and then looking at GCF and prime factors in the ring $\mathbb{Z}[\sqrt{129}]$. This does in fact have class number 1, so I believe the same technique used in this link will produce solutions. As this technique is way above the apparent level of the competition, I don't consider this an answer. $\endgroup$ – RghtHndSd Jul 26 '13 at 23:49
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The usual way to try a Mordell equation using elementary means is by adding or subtracting a constant from both sides to get the sum or difference of squares on one side, the sum or difference of cubes on the other, and compare factorizations. With different letters, here we could try $$ w^2 - 4 = v^3 + 125 $$ and see how far that takes us. Here you can immediately see $v = -5, w = \pm 2.$

http://en.wikipedia.org/wiki/Mordell_curve

http://tnt.math.se.tmu.ac.jp/simath/MORDELL/

http://tnt.math.se.tmu.ac.jp/simath/MORDELL/MORDELL+

E_+00129: r = 2   t = 1   #III =  1
          E(Q) = <(-2, 11)> x <(-5, 2)>
          R =   3.1316041389
           6 integral points
            1. (-2, 11) = 1 * (-2, 11)
            2. (-2, -11) = -(-2, 11)
            3. (-5, 2) = 1 * (-5, 2)
            4. (-5, -2) = -(-5, 2)
            5. (16, 65) = -1 * (-2, 11) - 1 * (-5, 2)
            6. (16, -65) = -(16, 65) 
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This is only a note on a fact which simplifies in part the checking.

Notice that looking at the equation modulo $2$ we get: $$y^3 + x^2\equiv 1\ \mathrm{mod}\ 2$$ Thus exactly one of $x$ and $y$ is in $2\mathbb{Z}$.

If I get other ideas, I will edit my answer.

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