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:$\displaystyle (m+n)^3=m^3+3m^2n+3mn^2+n^3$$\displaystyle =m^3+n^3+3mn(m+n)$ (1)

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

So$\displaystyle m^3+n^3+99mn =33^3$

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

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

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

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

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

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

I'm not yet sure how to proceed with the second one. $\displaystyle m = n = -33$ is a solution:$\displaystyle (-66)^2+33(-66)+33^2-3(-33)(-33)$$\displaystyle =4.33^2-2.33^2+1.33^2-3.33^2$

$\displaystyle =0$

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

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

Grandad