# Math Help - Finding ordered pairs

1. ## Finding ordered pairs

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.

2. NA

3. 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

4. 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 $(-33,-33)$.