# Arithmetic/Divisibility homework problem

1. Apr 23, 2012

### mtayab1994

1. The problem statement, all variables and given/known data

Solve for n in the following equations:

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

2. Relevant equations

3. 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.

Last edited: Apr 23, 2012
2. Apr 23, 2012

### scurty

Re: Arithmetic/Divisibility

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!

3. Apr 23, 2012

### mtayab1994

Re: Arithmetic/Divisibility

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

4. Apr 23, 2012

### Staff: Mentor

Re: Arithmetic/Divisibility

Right.

5. Apr 23, 2012

### mtayab1994

Re: Arithmetic/Divisibility

Yes now how can i continue from there?

6. Apr 23, 2012

### Whovian

Re: Arithmetic/Divisibility

7. Apr 23, 2012

### mtayab1994

Re: Arithmetic/Divisibility

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

8. Apr 23, 2012

### Whovian

Re: Arithmetic/Divisibility

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

9. Apr 23, 2012

### mtayab1994

Re: Arithmetic/Divisibility

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)?

10. Apr 23, 2012

### Whovian

Re: Arithmetic/Divisibility

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.

11. Apr 23, 2012

### mtayab1994

Re: Arithmetic/Divisibility

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?

Last edited: Apr 23, 2012
12. Apr 23, 2012

### Staff: Mentor

Re: Arithmetic/Divisibility

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

13. Apr 23, 2012

### mtayab1994

Re: Arithmetic/Divisibility

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

14. Apr 23, 2012

### Staff: Mentor

Re: Arithmetic/Divisibility

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

15. Apr 23, 2012

### mtayab1994

Re: Arithmetic/Divisibility

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

16. Apr 23, 2012

### Staff: Mentor

Re: Arithmetic/Divisibility

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.

17. Apr 23, 2012

### mtayab1994

Re: Arithmetic/Divisibility

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.

18. Apr 23, 2012

### Staff: Mentor

Re: Arithmetic/Divisibility

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

19. Apr 23, 2012

### mtayab1994

Re: Arithmetic/Divisibility

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.

20. Apr 23, 2012

### Staff: Mentor

Re: Arithmetic/Divisibility

Solve n^2-(k+1)n+(2k+3)=0 for n.