Simple proof regarding integers

  • Thread starter Thread starter Yosty22
  • Start date Start date
  • Tags Tags
    Integers Proof
AI Thread Summary
To prove that if 4 divides m² + n², then 4 also divides mn, one should start from the equation m² + n² = 4k for some integer k. The discussion emphasizes the importance of avoiding division when working with integers, as it may not yield integer results. Participants suggest exploring concrete examples to build intuition about the problem. The goal is to manipulate the equation logically to express mn as an integer multiple of 4. Ultimately, the discussion encourages independent discovery and understanding of the proof process.
Yosty22
Messages
182
Reaction score
4

Homework Statement



Show that if m and n are integers such that 4|m2+n2, then 4|mn

Homework Equations

The Attempt at a Solution



Since 4 divides m2+n2, then we can say that m2+n2 = 4k, where k is an integer. I haven't done any mathematical proofs of any kind yet, but we were supposed to see if we could do this. I am kind of stuck as to where to go from here. The only thing I could think of is to try to introduce a m*n term into the equation m2+n2 = 4k. To do this, I multiplied and divided the left-hand side of the equation by mn to get: mn(m2+n2/mn) = 4k. Since I am looking to show that 4|mn, or mn = 4*(some integer), I tried to isolate the mn term. That means 4k/(m2+n2/mn) = mn, or 4kmn/(m2+n2) = mn. However, I don't think this proves anything. Any advice on how to proceed (or even where to start)?

Thanks.
 
Physics news on Phys.org
First, when working with integers, in general, dividing is not a good idea. Since you know nothing about ##m^2+n^2##, ##m^2+n^2/mn## isn't necessarily an integer.

First, think about how you need you're proof to end. You want to show that mn can be written as an integer multiple of 4.
After that, where you go next really depends on what you are allowed to assume about integers. Are there any results that may be relevant that you've proven in class?
 
There is nothing specifically we talked about in class (We haven't really done any proofs at all - we have just talked about Sentential Logic, truth tables, and basic set theory stuff). Although we haven't really covered proofs in class yet, it is considered a "proofs class" so he wanted us to be able to work this out. I know in the end, I want to find that mn = 4l, where l is another integer. However, I can't see how I can isolate a (m*n) term in m2+n2 = 4k (especially without dividing). To step through the proof in a logical way, I would assume you have to start with (m2+n2) = 4k and do things to the equation until you get something of the form m*n = 4l. I just have no idea how I would go about doing it at all.
 
Try some concrete examples and see if you can't convince yourself why its true.
 
It sounds like your prof doesn't really expect an elegant solution, and he really just wants to what you can discover on your own. I wouldn't want to deprive you of that.
 
MostlyHarmless said:
First, when working with integers, in general, dividing is not a good idea. Since you know nothing about ##m^2+n^2##, ##m^2+n^2/mn## isn't necessarily an integer.
If you meant ##\frac{m^2 + n^2}{mn}##, you need parentheses in what you wrote. Without them, what you wrote is ##m^2 + \frac{n^2}{mn}##. Also m2/mn is ambiguous, as it could be interpreted as m2/(mn) or as (m2/m) * n.
MostlyHarmless said:
First, think about how you need you're proof to end. You want to show that mn can be written as an integer multiple of 4.
After that, where you go next really depends on what you are allowed to assume about integers. Are there any results that may be relevant that you've proven in class?
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
Back
Top