Question

There exist unique positive integers x and y that satisfy the equation \(x^2 + 84x + 2008 = y^2\). Find x + y.

Answer 1

My work so far:
factors of 2008:
1 x 2008
2 x 1004
4 x 502
8 x 251

Answer 2

or, if we look at values of y^2:
perfect squares over 2008:
45^2 = 2025 --> 2025-2008 = 17
46^2 = 2116 --> 2116-2008 = 108
47^2 = 2209 --> 2209-2008 = 201
48^2 = 2304 --> 2304-2008 = 296
Should I test \(x^2+84x-[17,108,201,296...]=0\) until I find one that works?

Answer 3

@mathmale can you help when you're not busy being awesome?

Answer 4

BT: Flattery will get you nowhere. (Wink).

That requirement that the integers x and y be unique is a tough one, at least at first glance.

I'm thinking out loud, because it's not immediately apparent to me how one could most efficiently going about finding a solution:

1. If x and y are unique and are positive, wouldn't that seem to indicate that both are in QI?

Answer 5

yes, I believe it would.
What if we made it \(x^2+84x+1794 + 212 = y^2 \rightarrow (x+42)^2+212=y^2\)? Would that get us anywhere?

Answer 6

2. Aha! Why not take the given equation and complete the square? Looks like you got that idea just before I did. :)

Answer 7

Now what?

Answer 8

there are four "conic sections," circle, parab., ellipse and hyperbola. Does it seem that your equation might represent any of those?

Answer 9

a hyperbola, right? But I don't really want to go there...

Answer 10

especially without a calculator...

Answer 11

3. suppose you were to solve your latest equation (or the original one) for y and then graph this function y only in the 1st quadrant. You could then use the TRACE feature, move your cursor along the graph, and that way get a better idea of whether there is any pair of integers x, y that would make the given equation true.

Answer 12

But you seem to be implying you can't use a calculator or don't have one.

Answer 13

yes. This is an AIME problem, and there are no calculators allowed.

Answer 14

4. Look carefully at (x^2 + 84x + 2008 = y^2. It seems to me that y would have to be pretty large to counterbalance that square of x and that +2008 on the left side. Would you agree, or say "maybe" to, the idea that x<<y?

Answer 15

Sorry, but what does x<<y mean?

Answer 16

Just a symbolic way of stating, "x must be much smaller than y."

Answer 17

oh. yes, it is definitely so.

Answer 18

\[ (x+42)^2+212=y^2\] seems to confirm that. We have to take x, add 42 to x, square that result, add 212 to that result, before we come up with a number as large as y squared.

Answer 19

If I were x, I'd have an inferiority complex. :)

Answer 20

5. We do seem to be narrowing down the possibilities for x and y. My next suggestion wuld be to ask you to consider domain and range for this relationship. that might further narrow down the possibilities for the x- and y-values, don't you think?

Answer 21

What would happen if you were to solve for y, take only the positive square root value, and then try to figure out what values x and why could possibly have (domain and range)?

Answer 22

Earlier I put the perfect squares. Would those help now? There is a difference of 184 between 45^2 and 47^2. 50^2-48^2 = 2500-2304 = 196. 53^2-51^2 = 208. This difference is increasing; we need it to equal 212.
54^2-52^2 = 2916-2704 = 212. So, y=54, and x=52-42 = 10. So, x+y=64.

Answer 23

6. What is the greatest value that y could have? You could, if you wanted, maximize y through the usual calculus methods. Or you could simply let x=-42. Unfortunately, the resulting y value would not be an integer; still, knowing this y value will give you the upper limit of your range, wouldn't it? Could we take an analogous approach towards finding the min and max values possible for x?

Answer 24

We've kept on boxing in x- and y-values. Pretty soon the poor things may have no escape route left. Many thanks for the medal.

This brainstorming, while difficult and challenging, is the way to go, when it comes to solving much more difficult problems in science and engineering later on.

Answer 25

ok, I just looked at the solution. I did it right (methodology-wise) but said that 42^2 = 1794, not 1764 as it should be. That and a subtraction error made me get the wrong answer. But this is apparently the way to solve it.

Answer 26

Nice work! Hope we can brainstorm on other problems in the future.
What's your first name?

Answer 27

Ben

Answer 28

And my real name is Warren Goldmann.

I'm looking forward to chatting with you again soon!