# A problem on algebric expansion and indices.

1. Oct 10, 2011

### sankalpmittal

1. The problem statement, all variables and given/known data

If :
p2+q2+4r2 = pq+2qr+2rp

Prove that :
p=q=2r

2. Relevant equations

I am not sure but I tried using this equation :

(a+b+c)2 = a2+b2+c2+2(ab+bc+ca)

3. The attempt at a solution

p2+q2+4r2 = pq+2qr+2rp
p2+q2+4r2 -pq-2qr-2rp = 0

2. Oct 10, 2011

### LCKurtz

It isn't true. Look at p = q = r = 1.

3. Oct 10, 2011

### SammyS

Staff Emeritus
That does not agree with p=q=2r anyway.

Look at p = q = 2r = 1, which does work.

4. Oct 10, 2011

### LCKurtz

5. Oct 10, 2011

### I like Serena

Hi sankalpmittal!

Can you write your equation in the form a2 + b2 + c2 = 0?
With a, b, c expressions like (p - q), (p - 4r) or such like?

6. Oct 10, 2011

### Staff: Mentor

Hi sankalpmittal!
It's deceptively easy. (Or do I mean, it's deceptively difficult?) I find some of these idioms confusing.

Just take out a single first-degree factor; do this a few times. I'll start you off:

p(p-q) + ......... = 0

7. Oct 10, 2011

### SammyS

Staff Emeritus
The result can also be obtained by writing the equation as a quadratic in r (I think any of the variables will do.). Using the quadratic formula gives a solution for r with √(-(p-q)2 ) which is real only if p=q. The rest follows from there.

8. Oct 10, 2011

### Dickfore

You can use linear algebra. The quadratic form:
$$p^2 + q^2 + 4 r^2 - p q - 2 q r - 2 p r = 0$$
can be written in matrix form:
$$(p, q, r) \cdot \left(\begin{array}{ccc} 1 & -\frac{1}{2} & -1 \\ -\frac{1}{2} & 1 & - 1 \\ -1 & -1 & 4 \end{array}\right) \cdot \left( \begin{array}{c} p \\ q \\ r \end{array}\right) = 0$$
The eigenvalue problem for this matrix is:
$$\left\vert \begin{array}{ccc} 1 - \lambda & -\frac{1}{2} & -1 \\ -\frac{1}{2} & 1 - \lambda & - 1 \\ -1 & -1 & 4 - \lambda \end{array}\right\vert = 0$$

$$(1 - \lambda) \left\vert \begin{array}{cc} 1 - \lambda & -1 \\ -1 & 4 - \lambda \end{array}\right\vert - \left(-\frac{1}{2}\right) \left\vert \begin{array}{cc} -\frac{1}{2} & -1 \\ -1 & 4 - \lambda \end{array}\right\vert + (-1) \left\vert \begin{array}{cc} -\frac{1}{2} & 1 - \lambda \\ -1 & -1 \end{array} \right\vert = 0$$
$$(1 - \lambda) (3 - 5 \lambda + \lambda^2) + \frac{1}{2} ( -3 + \frac{\lambda}{2} ) - (-\lambda + \frac{3}{2}) = 0$$
$$\left. -\lambda^3 + 6 \lambda^2 - 8 \lambda + 3 - \frac{3}{2} + \frac{\lambda}{4} + \lambda - \frac{3}{2} = 0 \right/ \cdot (-4)$$
$$4 \lambda^{3} - 24 \lambda^{2} + 32 \lambda - 12 + 6 - \lambda - 4 \lambda + 6 = 0$$
$$4 \lambda^{3} - 24 \lambda^2 + 27 \lambda = 0$$
$$\lambda (4 \lambda^2 - 24 \lambda + 27) = 0$$

Last edited: Oct 10, 2011
9. Oct 10, 2011

### Dickfore

[STRIKE]None of those rationals seem to be roots of the cubic equation. Are you sure you haven't made a typo? Or, can you see a typo in my calculations?[/STRIKE]

EDIT:

I found one error and corrected my pervious post. Now there are more options and the cubic equation is different. Can you check whether some of these rationals are roots?

Last edited: Oct 10, 2011
10. Oct 10, 2011

### Dickfore

[STRIKE]I found one rational root is $\lambda = \frac{1}{2}$. After doing long division of[/STRIKE]

[STRIKE]you get a quadratic equation whose roots you can find using the quadratic formula instead of going through the other cases.[/STRIKE]

EDIT:

Scratch that! I corrected one more error. The resulting characteristic equation is correct and easy to solve.

Last edited: Oct 10, 2011
11. Oct 10, 2011

### Dickfore

Multiply the expression by 2:
$$2 p^2 + 2 q^2 + 8 r^2 - 2 p q - 4 q r - 4 p r = 0$$
$$(p^2 - 2 p q + q^2) + (p^2 - 4 p r + 4 r^2) + (q^2 - 4 q r + 4 r^2) = 0$$

12. Oct 11, 2011

### sankalpmittal

I don't know which method is correct. Everyone is giving different opinions well. I start off from last post.

Hi Dickfore ,

(p-q)2 + (p-2r)2 + (q-2r)2 = 0

Now ?

ok !

(p-q)2 = 0
p=q

(p-2r)2 = 0
p=2r

(q-2r)2 = 0

so , q=2r

p=q=2r
QED

That was simple haha.
Thanks all :)

Hiii SammyS ,
Your relation worked though. How did you get it ? By observation ?
Well by the method which Dickfore suggested ,
p=q=2r=n , where n can be any number.

Last edited: Oct 11, 2011
13. Oct 11, 2011

### SammyS

Staff Emeritus
Well, all the methods work.

I really liked the one by Dickfore.

I had to fiddle around with WolframAlpha to get it to recognize the expression. Then it gave lots of stuff. It wasn't until I messed around with the radical expression that I saw that it gave the solution. Similarly, with Dick's solution, where you have the sum of three squares is equal to zero, the only way to get a real solution is for each of them to be zero.

$4 r^2-2(p+q)r+p^2-p q+q^2 = 0$

$\displaystyle r=\frac{2(p+q)\pm\sqrt{12} \sqrt{-p^2+2 p q-q^2}}{8}$

$\displaystyle r=\frac{(p+q)\pm\sqrt{3} \sqrt{-(p-q)^2}}{4}$

So for a real solution we need p=q. Then r = 2p/4 = p/2 = q/2 .

NascentOxygen's also works. p(p-q) + q(q-2r) + 2r(2r-p) = 0 .

Last edited: Oct 11, 2011
14. Oct 11, 2011

### I like Serena

Errr... how does this work?
I don't get it.

15. Oct 11, 2011

### SammyS

Staff Emeritus
If p-q = 0 , q-2r = 0 and p-2r = 0, then the expression is zero. But it's not so obvious that this is the only solution.

16. Oct 11, 2011

### sankalpmittal

This method of NascentOxygen is also good.

p(p-q) + q(q-2r) + 2r(2r-p) = 0 .

Here p(p-q) , q(q-2r) , 2r(2r-p) are zero , but its not obvious so ,

let

p(p-q) = 1

so p2 -pq -1 = 0

q(q-2r) = 1
q2 - 2rq - 1 = 0

2r(2r-p) = -2

2r2 - pr +1 = 0

I think solving and combining these equation can help the case.
Well thanks all for taking effort !

Yet to try :
p(p-q) = 1
Thinking the other way
p=1
or p-q = 1

q(q-2r) = 1
q=1
q-2r = 1

2r(2r-p) = -2
2r = -2
2r-p = 1

Here we can obviously get : p=q=2r = 1 or we can try simultaneously solving equations p-q = 1 , q-2r = 1 , 2r-p = 1.

Well we can use other combination in p(p-q) as p=2 and p-q = 1/2

Thanks all of you :)

Last edited: Oct 11, 2011
17. Oct 12, 2011

### Staff: Mentor

There's a trivial solution: p=q=r=0

and a non-trivial solution: p=q=2r

though the trivial solution is contained within the non-trivial.

18. Oct 12, 2011

### I like Serena

Well... I still don't get it.
I see no reason why there couldn't be more solutions.
And the problem asks to proof that p=q=2r is the only solution.

19. Oct 12, 2011

### sankalpmittal

Hiii ILS !!!

This method of NascentOxygen is also good.

I Method : is a trivial solution

p=q=r=0

II Method is given below :

p(p-q) + q(q-2r) + 2r(2r-p) = 0 .

Here p(p-q) , q(q-2r) , 2r(2r-p) are zero , but its not obvious so .

Then p = q = 2q = 0

III Method

p(p-q) = 1

so p2 -pq -1 = 0

q(q-2r) = 1
q2 - 2rq - 1 = 0

2r(2r-p) = -2

2r2 - pr +1 = 0

I think solving and combining these equation can help the case.
Well thanks all for taking effort !

IV Method :

In p(p-q) + q(q-2r) + 2r(2r-p) = 0 .

Yet to try :
p(p-q) = 1
Thinking the other way
p=1
p-q = 1

q(q-2r) = 1
q=1
q-2r = 1

2r(2r-p) = -2
2r = 1
2r-p = -2

Since 1+1-2=0

Here we can obviously get : p=q=2r = 1 or we can try simultaneously solving equations p-q = 1 , q-2r = 1 , 2r-p = -2.

Well we can use other combination in p(p-q) as p=2 and p-q = 1/2

p-q = 1 , q-2r = 1 , 2r-p = -2.

Now try solving these equations simultaneously , you will get a non trivial solution of p=q=2r

q=2r+1
p= 2r+2

So 2r+2 - 2r - 1 = 1
1=1 Hmm This question is irritating me now !
Hmm

??????

20. Oct 12, 2011

### I like Serena

To be clear, Dickfore's method is a (the?) proper solution:

(p-q)2 + (p-2r)2 + (q-2r)2 = 0

Since a square is always at least zero, each of the 3 terms must be zero.
From there it follows that p=q=2r exactly represents all possible solutions.