Arithmetic/Divisibility homework problem

[mtayab1994]

Homework Statement

Solve for n in the following equations:

$$n-2\mid n^{2}-n+3$$

Homework Equations

The Attempt at a Solution

Should i use the euclidean division to solve it? Thank you before hand.

Edit: When i used euclidean division i got n+1 and a remainder of 5 but i don't know how to get n.

 

[scurty]

Start by using the definition of divides to produce an equation. This equation will introduce a new variable, say, k. Solve for n using the quadratic formula (which will involve k) and then choose values of k so that n is an integer (I assume n is an integer in this problem).

See what you come up with!

 

[mtayab1994]

So that means you should get: n^2-n+3=k(n-2) am i right?

 

[Mark44]

So that means you should get: n^2-n+3=k(n-2) am i right?

Right.

 

[mtayab1994]

Right.

Yes now how can i continue from there?

 

[Whovian]

Ever heard of quadratic equations?

 

[mtayab1994]

Ever heard of quadratic equations?

Yes, but n^2-n+3 doesn't have any real roots, if that's what you're talking about.

 

[Whovian]

No, we group everything on one side of the equation so we have coefficients in terms of k and solve THAT.

 

[mtayab1994]

No, we group everything on one side of the equation so we have coefficients in terms of k and solve THAT.

So we get: n^2-k(n-2)-n+3=0, but where are we supposed to get b^2-4ac. Is b=k(n-2)?

 

[Whovian]

Not quite. We get n^2-(k+1)n+(2k+3)=0 after multiplying out and grouping like terms, and it should be obvious how to proceed from here.

 

[mtayab1994]

So for Δ i got k^2-6k-11 and then i counted the discriminant again and i got 80. Is what I've done so far correct?

 

[Mark44]

Reread what scurty wrote in post #2. He laid out pretty much what you need to do.

 

[mtayab1994]

Reread what scurty wrote in post #2. He laid out pretty much what you need to do.

So there is no need to find the discriminant of the equation : n^2-(k+1)n+(2k+3)=0?

 

[Mark44]

I don't see any need for finding the discriminant in this problem.

 

[mtayab1994]

I don't see any need for finding the discriminant in this problem.

But the quadratic formula involves finding the discriminant, or do i just use random values of k?

 

[Mark44]

Of course you find the discriminant in the quadratic formula. What I meant was that there's no reason to find it and stop there.

Again, take a look at what scurty said in post #2. You are asking questions that suggest you haven't read it or didn't understand it.

 

[mtayab1994]

Of course you find the discriminant in the quadratic formula. What I meant was that there's no reason to find it and stop there.

Again, take a look at what scurty said in post #2. You are asking questions that suggest you haven't read it or didn't understand it.

Well, he said to produce a new equation, which I did and then he said to find n using the quadratic formula, which will involve k, and that's what I did. I'm confused because we have just started this lesson, and we haven't done any problem like this so far.

 

[Mark44]

Did you solve the quadratic equation for n? If so, what did you get?

 

[mtayab1994]

Did you solve the quadratic equation for n? If so, what did you get?

For n^2-(k+1)n+(2k+3)=0 I got a discriminant of k^2-6k-11 then i solved that equation and I got Δ=80 so i got k1=3+2√5 and k2=3-2√5.

 

[Mark44]

Solve n^2-(k+1)n+(2k+3)=0 for n[/color].

 

[mtayab1994]

Solve n^2-(k+1)n+(2k+3)=0 for n[/color].

Ok when solving for n i got : $$n=\frac{k+\sqrt{k^{2}-8k+4}}{2},\frac{k-\sqrt{k^{2}-8k+4}}{2}$$

 

[Mark44]

Ok when solving for n i got : $$n=\frac{k+\sqrt{k^{2}-8k+4}}{2},\frac{k-\sqrt{k^{2}-8k+4}}{2}$$

No.

Do you know how to use the quadratic formula? You said before that the discriminant was k² - 6k -11, but that's not what you show above.

 

[mtayab1994]

No.

Do you know how to use the quadratic formula? You said before that the discriminant was k² - 6k -11, but that's not what you show above.

yes instead of k^2-8k+4 should be k^2-6k-11

 

[mtayab1994]

I think i got it for k^2-6k-11=0 we get n =(k+1)/2 right?

 

[Mark44]

After you make that correction, there is still another one you need to make.

Then READ POST #2!

 

[scurty]

For what it's worth, when I solved this, I got 4 answers for n. I'm not sure if there is a theorem that's states a maximum number for n based on certain criteria, but I tested out about 30 values for k and only two of them worked. Maybe someone else can chime in.

 

[Mark44]

I think i got it for k^2-6k-11=0 we get n =(k+1)/2 right?

Why do you think that k^2-6k-11=0?

 

[mtayab1994]

Well after all you really don't need to form and equation and give values for k all you have to do is:
$$n-2\mid n^{2}-n+3\Longleftrightarrow\frac{(n-2)(n+1)+5}{n-2}\Longleftrightarrow n+1+\frac{5}{n-2}$$

and since: $$(n+1)\in Z$$ therefore: $$\frac{5}{n-2}\in Z$$

thus: $$n-2\mid5$$Then you form a table with 2 rows one row contains the divisors of '5' and the other row contains n the number that we have to plug into n-2 to get that divisor of 5. Like the following.

(n-2) 1 -1 5 -5

(n) 3 1 7 -3

And then the solutions are: $$S=\{-3,1,3,7\}$$

So how's that? I think my way is a lot easier then solving all of those equations.

 

[scurty]

Those are the answers I got. There are often more ways than one to solve a problem, in this case, that method would probably be the easiest way to solve the problem. However, you seemed to struggle with simple algebra involving solving for a quadratic equation.. You might want to touch up on that a bit for the future. Algebra techniques are common in number theory, I remember using the quadratic formula numerous times when I took the course, it would be wise to understand why the other method works as well.

Edit: I just noticed that you don't have k anywhere in your equation. The definition of divides introduces a new variable, you can't just assume k=1, you have to explain why certain values of k work and others don't. It works in this case but not all cases.

 

[skiller]

Sorry for butting in, but I like to think of myself as a decent mathematician who has terrible difficulty with number theory.

I couldn't solve this at all. I had to write a program to solve for all cases for 1 to 1000 and I got the same answers: 3 & 7

Could someone please tell me where I'm going wrong analytically?

PS I've read post #2.