# Homework Help: Mathematical Tripos, archived question about squares

1. May 30, 2013

### JanEnClaesen

1. The problem statement, all variables and given/known data
a, b, x, y are rational numbers

(ay - bx)² + 4(a - x)(b - y) = 0 implies that x = a and y = b or 1 - ab and 1 - xy are squares of rational numbers

2. The attempt at a solution
My attempts are hard to transcribe, is this mandatory for getting help? This is my first post, I am not familiar with the spirit of this forum.

2. May 30, 2013

### Simon Bridge

Welcome to PF;
It is very important to describe what you have attempted if you are to get help.
It is also a good idea to make the question explicit - all you have there is a statement. Are you trying to prove the statement is true or something?

We are used to people having trouble describing their efforts - just give it your best shot.
If the problem is writing equations, try LaTeX ... eg:

$\{ a,b,x,y\} \in \mathbb{R}$

$$(ay - bx)^2 + 4(a - x)(b - y) = 0 \Rightarrow ( x = a \wedge y = b ) \vee (\{\sqrt{1 - ab} \wedge \sqrt{1 - xy}\}\in \mathbb{R} )$$

... that's probably not canonical notation - but, if you use the "quote" button bottom-right of this post, you will see how I did it.

3. May 30, 2013

### JanEnClaesen

LaTeX looks very convenient, thank you. You're right, the statement has to be proven true, but not exactly the statement you latexed, 1 - ab and 1 - xy are to be squares of rational numbers.

I tried working backwards: I substituted some numbers ( a = 3, y = 1/9, b = 1/4, x = 8 ) as to get a clue of the workings of the equation, I didn't get it. I tried substitutions similar to 1 - ab = m² and 1 - xy = n² . That didn't work out.

The equation (a + x)(b + y) - 2(ay + bx) = (a - x)(b - y) didn't prove much help either.
Dividing both ends by (a - x) ² or (b - y)² came to a dead end (doesn't matter which one, since it's sort of symmetric in a, b and x, y, as is evident by the commutativity of 1 - ab and 1 - xy).

4. May 30, 2013

### Ray Vickson

The only suggestion I can give is to look at the thing that = 0. It is a sum of two terms, so either both terms = 0 or both are non-zero. Look at what must happen in each case.

5. May 30, 2013

### JanEnClaesen

Thank you, but could you be a little more specific? You sort of repeated the question, sometimes reformulating the question presents the solution of course, but for my part, there's no solution in sight.

6. May 30, 2013

### Simon Bridge

He's saying that in order for $A+B=0$ either $A=0$ and $B=0$ OR $A=-B$.
In your case, $A=(ay-bx)^2$ and $B=4(a-x)(b-y)$
... so what do you need x and y to be to make both these zero?

That's the first part - for the second part, try expanding the brackets and grouping like terms in x and y .. what shape does the resulting equation represent?

BTW: thanks for the redirect - the $\mathbb{R}$s should be $\mathbb{Q}$s.

Last edited: May 30, 2013
7. May 31, 2013

### Ray Vickson

If you view the equation above as a quadratic in y and solve it, you get y in terms of x and a, b. By imposing the conditions that x and y must both be rational (and, of course, assuming a and b are rational) you will find that some other conditions must hold. You can prove all the required results in this way, but it is vital to look very carefully at the formula for y, to see how it can be simplified.

8. May 31, 2013

### JanEnClaesen

Ah, that does the trick, thank you. Yet I wonder, how could I have known?
Should it be a reflex action to calculate the discriminant of something quadratic?

9. May 31, 2013

### Simon Bridge

It was my first reaction. "Oh look, x and y are squared - probably a conic section..."
You would probably have seen it had you expanded the brackets and stood back - a large writing surface is good for this - why so many people I know have whiteboards and/or big sheets of glass in their workspace

Basically you just get used to recognizing patterns that mean a particular class of equation is probably being dealt with ... then you try it out. It's why they make you do all those proofs and formulas and stuff - to get the experience to spot this sort of thing.