Science:Math Exam Resources/Courses/MATH437/December 2006/Question 07

MATH437 December 2006

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[hide] Question 07
Show that the only integer solutions to the equation $\displaystyle y^{2}+11=x^{3}$ are $\displaystyle (3,\pm 4)$ and $\displaystyle (15,\pm 58)$

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

[show] Hint 1
Keep in mind that because there are solutions to this equation, trying anything locally (that is, reducing modulo a prime) will not help you.

[show] Hint 2
With a computer, it would be easy to solve for the integer points on an elliptic curve. Given that this is an exam question, algebraic number theory is probably your saving grace.

[show] Hint 3
Factor the left hand side over $\displaystyle \mathbb {Q} ({\sqrt {-11}})$ .

[show] Hint 4
Dirichlet's unit theorem says that the unit group of this field's ring of integers is trivial (that is, it is only $\pm 1$ ). Keep in mind that since $\displaystyle -11\equiv 1\mod {4}$ , the ring of integers here is given by $\mathbb {Z} \left[{\frac {1+{\sqrt {-11}}}{2}}\right]$ . Also, this is one of the special fields with class number 1 (or alternatively notice that this is a Euclidean domain) and so its ring of integers in particular is a unique factorization domain.

[show] Hint 5
Don't forget to show that the factors on the left are coprime. For this you might want to refer back to the original equation.

Checking a solution serves two purposes: helping you if, after having used all the hints, you still are stuck on the problem; or if you have solved the problem and would like to check your work.

If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work.
If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

[show] Solution
We proceed as in the hints. We factor the left hand side in $\displaystyle \mathbb {Q} ({\sqrt {-11}})$ as $\displaystyle (y+{\sqrt {-11}})(y-{\sqrt {-11}})=x^{3}$ . Since $\displaystyle -11\equiv 1\mod {4}$ , the ring of integers here is given by $\mathbb {Z} \left[{\frac {1+{\sqrt {-11}}}{2}}\right]$ which is a unique factorization domain. We compute the greatest common divisor of the terms on the left hand side. Let $\displaystyle I=(y+{\sqrt {-11}},y-{\sqrt {-11}})$ , the ideal generated by these two elements. Now notice that in the equation $\displaystyle y^{2}=x^{3}-11$ , if x is even, then modulo 8 considerations show that $\displaystyle y^{2}\equiv 5\mod {8}$ and this is a contradiction since the only odd square is 1 modulo 8. Thus, x is odd and so y is even. We may further suppose that these factors are coprime for otherwise, if a prime p divides both x and y, then it must divide 11 and hence must equal 11. Rewriting the equation would then give $\displaystyle (11y')^{2}+11=(11x')^{3}$ or simplified $\displaystyle 11(y')^{2}+1=121(x')^{3}$ and hence 11 divides 1 which is a contradiction. So y is even and x and y are coprime. Further, we have that $\displaystyle (y+{\sqrt {-11}})(y-{\sqrt {-11}})=x^{3}\in I$ and that $\displaystyle y+{\sqrt {-11}}+y-{\sqrt {-11}}=2y\in I$ Hence $1=\gcd(x,y)\in I$ and thus, the two elements $y+{\sqrt {-11}},y-{\sqrt {-11}}$ are coprime. Thus, we may write $\displaystyle y+{\sqrt {-11}}=\pm \left({\frac {\alpha '+\beta '{\sqrt {-11}}}{2}}\right)^{3}=\left({\frac {\alpha +\beta {\sqrt {-11}}}{2}}\right)^{3}$ where above, the right hand side absorbs the units since $\pm 1=(\pm 1)^{3}$ (and we relabeled the $\displaystyle \alpha ',\beta '$ values) where $\displaystyle \alpha ,\beta \in \mathbb {Z}$ . Expanding and comparing coefficients yields ${\begin{aligned}8y&=\alpha ^{3}-33\alpha \beta ^{2}\\8{\sqrt {-11}}&=(3\alpha ^{2}\beta -11\beta ^{3}){\sqrt {-11}}\end{aligned}}$ or simplified ${\begin{aligned}8y&=\alpha ^{3}-33\alpha \beta ^{2}\\8&=\beta (3\alpha ^{2}-11\beta ^{2})\end{aligned}}$ The second equation tells us that $\displaystyle \beta \in \{\pm 1,\pm 2,\pm 4,\pm 8\}$ since it must be a factor of 8. Checking these 8 values shows that only $\displaystyle \beta =-1$ or $\displaystyle \beta =2$ gives admissable values. This leads to the points $\displaystyle (\alpha ,\beta )=(\pm 1,-1)$ or $\displaystyle (\alpha ,\beta )=(\pm 4,2)$ The first point gives $y=\mp 4$ and the second point yields $y=\mp 29$ . These correspond to the given points and thus complete the problem.