# Putnam 2003 A4

1. Oct 26, 2007

### ehrenfest

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

2. Relevant equations

3. The attempt at a solution

Last edited by a moderator: May 3, 2017
2. Oct 26, 2007

### Kummer

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.

Last edited by a moderator: May 3, 2017
3. Oct 26, 2007

### 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?

4. Oct 26, 2007

### Kummer

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

5. Oct 26, 2007

### ehrenfest

On second thought, I do not see why it is readily apparant 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...