Science:Math Exam Resources/Courses/MATH437/December 2006/Question 06

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MATH437 December 2006

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Question 06
Let $\displaystyle d=4k^{2}+k$, where $\displaystyle k>1$ is an integer. Show that $\displaystyle x^{2}-dy^{2}=-1$ has no integer solutions.

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.

Hint 1
This time, your first reaction might be to use the theory of Pell equations and continued fractions on this problem will steer you in the correct direction.

Hint 2
Recall from the theory of Pell equations that $\displaystyle x^{2}-dy^{2}=-1$ has a solution if and only if the period of $\displaystyle {\sqrt {d}}$ is even.

Hint 3
Computing the period can be done abstractly by hand.

Hint 4
As a final learning moment, it should be noted that the sufficiency in the previous problem is not true and this problem gives a counter example when say $\displaystyle k=5$

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.

Solution

Proceeding as in the hints, we see it suffices to show that the period of $\displaystyle {\sqrt {4k^{2}+k}}$ is even. First notice that $\displaystyle (2k)^{2}<4k^{2}+k<4k^{2}+4k+1=(2k+1)^{2}$ and thus $\displaystyle \left\lfloor {\sqrt {4k^{2}+k}}\right\rfloor =2k=a_{0}$ in the continued fraction expansion. As $\displaystyle {\sqrt {4k^{2}+k}}=a_{0}+{\frac {1}{a_{1}+...}}=2k+{\frac {1}{a_{1}+...}}$ we see that after cross multiplying that ${\begin{aligned}a_{1}&=\left\lfloor {\frac {1}{{\sqrt {4k^{2}+k}}-2k}}\right\rfloor \\&=\left\lfloor {\frac {{\sqrt {4k^{2}+k}}+2k}{({\sqrt {4k^{2}+k}}+2k)({\sqrt {4k^{2}+k}}-2k)}}\right\rfloor \\&=\left\lfloor {\frac {{\sqrt {4k^{2}+k}}+2k}{4k^{2}+k-4k^{2}}}\right\rfloor \\&=\left\lfloor {\frac {{\sqrt {4k^{2}+k}}+2k}{k}}\right\rfloor \\&=\left\lfloor {\sqrt {4+1/k}}+2\right\rfloor \\&=4\end{aligned}}$ As $\displaystyle {\sqrt {4k^{2}+k}}-2k={\frac {1}{a_{1}+{\frac {1}{a_{2}+...}}}}$ cross multiplying gives us that $\displaystyle a_{1}+{\frac {1}{a_{2}+...}}={\frac {1}{{\sqrt {4k^{2}+k}}-2k}}={\sqrt {4+1/k}}+2$ and thus as we know $a_{1}=4$, ${\begin{aligned}a_{2}&=\left\lfloor {\frac {1}{{\sqrt {4+1/k}}+2-a_{1}}}\right\rfloor \\&=\left\lfloor {\frac {1}{{\sqrt {4+1/k}}+2-4}}\right\rfloor \\&=\left\lfloor {\frac {1}{{\sqrt {4+1/k}}-2}}\right\rfloor \\&=\left\lfloor {\frac {{\sqrt {4+1/k}}+2}{({\sqrt {4+1/k}}+2)({\sqrt {4+1/k}}-2)}}\right\rfloor \\&=\left\lfloor {\frac {{\sqrt {4+1/k}}+2}{4+1/k-4}}\right\rfloor \\&=\left\lfloor k({\sqrt {4+1/k}}+2)\right\rfloor \\&=\left\lfloor {\sqrt {4k^{2}+k}}+2k\right\rfloor \\&=4k\end{aligned}}$ Here we reduce back to the case of $a_{1}$. Hence, the continued fraction expansion is given by $\displaystyle {\sqrt {4k^{2}+k}}=[2k;{\overline {4,4k}}]$. Thus the period of the continued fraction expansion is even and the theory of Pell equations states that the equation $x^{2}-dy^{2}=-1$ has no solution.

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