# Homework Help: Prove that 11^2 does not devide n^2+3n+5

1. Jun 13, 2009

### rbetan

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

prove that 11^2 does not devide n^2+3n+5; for any n.

In order for this to make sense n must be an integer.

2. Relevant equations

3. The attempt at a solution

want to show that n^2+3n+5 is not congruent to 0(mod 121)

Assume towards a contradiction that 11^2 divides n^2+3n+5

we can rewrite n^2+3n+5=(n+7)(n-4)+33.

since i assumed that 11^2|n^2+3n+5 then 11^2|(n+7)(n-4)+33

(now is the part that i am not sure about)

but 11^2 does not divide 33. Therefore 11^2 does not divide n^2+3n+5.

thanks for any help.

2. Jun 13, 2009

### snipez90

I don't think you need any clever factorizations to do this problem. The simplest factorization possible will give you a lot of insight. Assuming that my reasoning was correct, working in mod 3 is a good way to approach this type of problem (in fact, when dealing with squares, working in mod 3 or 9 is helpful; for instance, squares are congruent to either 0 or 1 in mod 3).

*EDIT* Actually, I think I made a mistake somewhere. Your argument can work but I think you are missing a step.

Last edited: Jun 13, 2009
3. Jun 13, 2009

### gunch

Considering the whole thing mod 121 quickly gets messy, and we see that 11|33 so we should instead try to see what we get if we consider it modulo 11 instead. From your factorization you should be able to see that,
11|(n+7)(n-4)+33
What can you say about n modulo 11 from this?

4. Jun 13, 2009

### snipez90

Ah thanks gunch. rbetan, you last implication is not justified, but gunch has provided the missing step.

5. Jun 14, 2009

### rbetan

hey gunch: I do not see how does it follow that 11|(n+7)(n-4)+33; can any one explain this step.

Now taking for granted that 11|(n+7)(n-4)+33. then (n+7)(n-4)+33= 0(mod 11)

so (n+7)(n-4)=-33(mod11); but -33(mod11)=0; so (n+7)(n-4)=0(mod11)

so it follows that 11|(n+7)(n-7) then 11^2|(n+7)(n-7)

but 11^2 does not divide 33. Therefore, 11^2 cannot divide (n+7)(n-4)+33.

so 11^2 does not divide n^2+3n+5.

is this correct.

6. Jun 14, 2009

### snipez90

It doesn't follow, but it patches your original argument. The whole point is that we know 11 divides 33 but if 11 doesn't even divide (n+7)(n-4), then there is nothing left to prove because clearly 11^2 won't divide (n+7)(n-4) if 11 does not. Thus, we assume that 11 does divide (n+7)(n-4). Noting that n + 7 = n - 4 (mod 11), we know that 11^2 divides (n+7)(n-4) if 11 does, but now 11^2 does not divide 33, so 11^2 cannot divide (n+7)(n-4) + 33.

7. Jun 15, 2009

### faith003

Using the quadratic formula and quadratic reciprocity you can show that 11^2 indeed divides (n+7)(n-4). Therefore, it divides (n+7)(n-4) but not 33. Hence it's not divisible by 11^2