Solving Problem from 2003 Putnam Math Exam

[ehrenfest]

Homework Statement

Here is the problem.
http://www.unl.edu/amc/a-activities/a7-problems/putnam/-pdf/2003.pdf
Here is the solution.
http://www.unl.edu/amc/a-activities/a7-problems/putnam/-pdf/2003s.pdf
How can they assume, WLOG, in case 2 that B = 0 and A => a > 0 ? It seems to me like that kills generality!

Homework Equations

The Attempt at a Solution

 

[Kummer]

Homework Statement

Here is the problem.
http://www.unl.edu/amc/a-activities/a7-problems/putnam/-pdf/2003.pdf
Here is the solution.
http://www.unl.edu/amc/a-activities/a7-problems/putnam/-pdf/2003s.pdf
How can they assume, WLOG, in case 2 that B = 0 and A => a > 0 ? It seems to me like that kills generality!

When they say B = 0 that has nothing to do with a generality argument. That is something you can say. Because if you complete the square on RHS you have just Nx^2+D with no middle term. That is what they mean by "shifting x" so if you shift x by x - B/A then you end up with a quadradic missing a middle term.

The only generality argument is that a>0 or a<0. So say that a>0. In that case it must mean that A>=a>0 because otherwise LHS is a quadradic of a bigger coefficient.

 

[ehrenfest]

I see for the first part. But you mean shift by x - B/2A, right?

But I do not see why you keep generality when you assume a > 0. Why do you not have to consider the case where a<0?

 

[Kummer]

But I do not see why you keep generality when you assume a > 0. Why do you not have to consider the case where a<0?

Because the proof for a>0 and a<0 are almost identical. For definiteness it is easier to solve one of these cases. The other remaining case is similar.
http://en.wikipedia.org/wiki/Without_loss_of_generality

 

[ehrenfest]

On second thought, I do not see why it is readily apparent that we can set B = 0. When you change coordinates to x - B/2A, how do you know that you are still in case 2? Maybe that is true, but it seems like that they would need to prove that...