Sum of All Possible Values for Sn to be a Perfect Square?

[physics kiddy]

Homework Statement

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

Homework Equations

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 ?

 

[Dick]

Homework Statement

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

Homework Equations

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 ?

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.

 

[physics kiddy]

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

it's

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

 

[Dick]

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

it's

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

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

 

[physics kiddy]

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

 

[Dick]

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

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

 

[haruspex]

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

it's

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

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?

 

[physics kiddy]

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?

(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

 

[HallsofIvy]

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

 

[Mentallic]

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

Since difference of factors is 3 ...

What? How did you come to that conclusion?

 

[Dick]

(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

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

 

[Mentallic]

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

Ahh

 

[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.

 

[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.

 

[physics kiddy]

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