Finding ordered pairs

**Vicky1997** · January 5th 2010, 04:41 PM

The number of the ordered pairs of intergers (m,n) for which mn> or = 0

and m cubed + n cubed +99mn = 33 cubed is equal to ?

a. 2 b. 3 c. 33 d. 35 e. 99

Initially my answer was a. 2 ( 0, 33 ) (33, 0 ) but I know there would be more pairs. How do I find the other pairs without randomly plugging in numbers?

Vicky.

**bigwave** · January 7th 2010, 09:44 AM

NA

**Vicky1997** · January 7th 2010, 04:29 PM

Originally Posted by bigwave

not sure of the background of this question so this is just my observation on this.

since the equation $= 33^3$ which is a fixed number and positive
then m and n will have be both positive. if you plug any number other than 0 you cannot get the 33^3 only m=0 and n=0 will work
$0^3 + n^3 = 33^3$ so if $m=0\ n=33$
$m^3 + 0^3 = 33^3$ so if $n=0\ m = 33$
so your answer of #2 (0,33)(33,0) appears to be correct

the answer to this seems to be just looking at the equation and seeing what really is possible given the constraint was a big hint.

again this is just my observation. there may be something too that I don't see in this, (my disclaimer anyway)

hope this helps. (some)

Thanks for your help.
But I did find one more pair (-33. -33 ). It was just a good guess though.

I'm still trying to figure out the rest because the answer is 35 pairs.

Vicky

**Grandad** · January 7th 2010, 11:27 PM

Hello Vicky

Originally Posted by Vicky1997

The number of the ordered pairs of intergers (m,n) for which mn> or = 0

and m cubed + n cubed +99mn = 33 cubed is equal to ?

a. 2 b. 3 c. 33 d. 35 e. 99

Initially my answer was a. 2 ( 0, 33 ) (33, 0 ) but I know there would be more pairs. How do I find the other pairs without randomly plugging in numbers?

Vicky.

This is quite tricky. I have part of the solution for you - I'm not quite sure how to prove the final part. Look at these expressions:

$(m+n)^3=m^3+3m^2n+3mn^2+n^3$

$=m^3+n^3+3mn(m+n)$ (1)

and

$p^3-q^3=(p-q)(p^2+pq+q^2)$ (2)

So

$m^3+n^3+99mn =33^3$

$\Rightarrow (m+n)^3 -3mn(m+n)+99mn = 33^3$ from (1)

$\Rightarrow (m+n)^3-33^3-3mn(m+n-33)=0$

$\Rightarrow (m+n-33)([m+n]^2+33[m+n]+33^2)-3mn(m+n-33)=0$ from (2)

$\Rightarrow (m+n-33)([m+n]^2+33[m+n]+33^2-3mn)=0$

$\Rightarrow m+n-33=0$ or $(m+n)^2+33(m+n)+33^2-3mn=0$

The first of these possibilities, $m+n=33$ , gives $34$ integer solutions for $mn\ge0$ ; namely $(0,33), (1,32), (2, 31), ...,(33,0)$ .

I'm not yet sure how to proceed with the second one. $m = n = -33$ is a solution:

$(-66)^2+33(-66)+33^2-3(-33)(-33)$

$=4.33^2-2.33^2+1.33^2-3.33^2$

$=0$

but I'm not sure how to show that this is the only one.

Anyway, there are your 35 solutions: $(0,33), (1,32), (2, 31), ...,(33,0)$ and $(-33,-33)$ .

Grandad

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