Science:Math Exam Resources/Courses/MATH437/December 2006/Question 01

MATH437 December 2006

Work in progress: this question page is incomplete, there might be mistakes in the material you are seeing here.

  •  Q1  •  Q2  •  Q3  •  Q4  •  Q5  •  Q6  •  Q7  •  Q8  •  

   Other MATH437 Exams

  •  December 2006  •  December 2011  •  

Question 01
Suppose that x,y,z are positive integers satisfying $\displaystyle (x,y)=1$ and $\displaystyle xy=z^{2}$. Show that there are integers u and v for which $\displaystyle x=u^{2}$ and $\displaystyle y=v^{2}$.

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 hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
The Fundamental Theorem of Arithmetic is the major idea in this proof. Start with the $\displaystyle z^{2}$ term and proceed from there.

Checking a solution serves two purposes: helping you if, after having used the hint, 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

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. By the fundamental theorem of arithmetic, let $\displaystyle z^{2}=\prod _{i=1}^{n}p_{i}^{2\alpha _{i}}$ be the prime factorization. Suppose that $\displaystyle p_{i}\mid x$ for any i. Then since x and y are coprime, this means that $\displaystyle p_{i}\nmid y$. Thus, we must have that $\displaystyle p_{i}^{2\alpha _{i}}\parallel x$. If one of x or y (or both) has no prime factors, then since these numbers are positive, they must be equal to 1 which is a square. Repeating this argument for all such i and noticing that this argument is symmetric in x and y, we see that $\displaystyle x=\prod _{p_{i}\mid x}p_{i}^{2\alpha _{i}}=\left(\prod _{p_{i}\mid x}p_{i}^{\alpha _{i}}\right)^{2}=u^{2}$ and $\displaystyle y=\prod _{p_{i}\mid y}p_{i}^{2\alpha _{i}}=\left(\prod _{p_{i}\mid y}p_{i}^{\alpha _{i}}\right)^{2}=v^{2}$ completing the proof.

Click here for similar questions

MER QGQ flag, MER RH flag, MER RS flag, MER RT flag, MER Tag Fundamental theorem of arithmetic, Pages using DynamicPageList3 parser function, Pages using DynamicPageList3 parser tag