# Homework Help: Algebra question

1. Dec 29, 2012

### physics kiddy

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

Let Sn = n2+20n+12, n is a positive integer. What is the sum of all possible values of n for which Sn is a perfect square ?

2. Relevant equations

3. The attempt at a solution

Well, I tried to factorise it : n2+20n+12 = n2+20n+100-88
=(n+10)2-88. And I conclude that there are infinitely such values of n...
I also tried to search properties of perfect squares but could find none applicable here.
To my surprise, answer is 16 ....
How is it like that ???

2. Dec 29, 2012

### Dick

Suppose (n+10)^2-88 is a perfect square. Call it k^2. Then (n+10)^2-88=k^2. So (n+10)^2-k^2=88. Start factoring both sides.

3. Dec 29, 2012

### physics kiddy

That's what I can do without much effort ...

it's

(n+10+k)(n+10-k) = 88
What next ???

4. Dec 29, 2012

### Dick

Start listing ways to factor 88 and try to match the factors with factors on the left. Figure out which ones work.

5. Dec 29, 2012

### physics kiddy

I guess next we need to factorize 88. It's 11*2^3 ..

6. Dec 29, 2012

### Dick

How many ways to split that into two factors that you can match with the algebra factorization?

7. Dec 29, 2012

### haruspex

There's a shortcut you can use with these. Consider the difference between the factors. Can you say whether it is even or odd? What does that tell you about the parity of the factors?

8. Dec 30, 2012

### physics kiddy

(n+10+k)(n+10-k) = 88

Since difference of factors is 3 ...
so,
(n+10+k)-(n+10-k)=3
=> 2k=3
=>k=2/3

9. Dec 30, 2012

### HallsofIvy

You have been told, repeatedly, to find all the ways of factoring 88 into two factors. But you still have not done that.

10. Dec 30, 2012

### Mentallic

What? How did you come to that conclusion?

11. Dec 30, 2012

### Dick

You mean k=3/2. And that just shows that 11*8 is not a factorization you want. Try another one.

Last edited: Dec 30, 2012
12. Dec 30, 2012

### Mentallic

Ahh

13. Dec 30, 2012

### physics kiddy

Oh my god! It was awfully easy:

Factors of 88 -->
1*88
2*44
4*22
8*11

(n+10+k)(n+10-k) = 88
Assuming
n+10+k = 44 and n+10-k = 2;
we have 2n+20=46 =>2n=26 => n = 13
And putting it in n2+20n+12, we have 441 whose root is 21 ... 1st number = 13

again taking n+10+k = 22
and n+10-k = 4;
2n+20 = 26
=> 2n = 6
=> n = 3 ..... 2nd number
so
(3)^2+20(3)+12 = 81 root of which is 9...

so the two numbers 13 and 3 add up to 16, that's it.

Thanks everybody a lot.

14. Dec 30, 2012

### haruspex

Just to explain the shortcut I was trying to get you to find, the difference of the factors is 2k, so even. Therefore the two factors are either both even or both odd. Since the product is even, they must both be even. So you can start by assigning a factor of 2 to each, and then you are just left with either 1 x22 or 2 x11 for the remaining 22.

15. Jan 1, 2013

### physics kiddy

Yea, that also works and even better, no need to search other factors. Thanks a lot. I tried it also.