Review question

How many pairs of positive integers $(x,y)$ satisfy this equation? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R6543

Solution

The number of pairs of positive integers $x$, $y$ which solve the equation $x^3 + 6x^2y + 12xy^2 + 8y^3 = 2^{30}$ is

1. $0$,
2. $2^6$,
3. $2^9-1$,
4. $2^{10}+2$.

We will take it the question means ordered pairs, so that $(x,y)$ is counted as different from $(y,x)$ if $x \neq y$.

First we factorise the equation to get: $(x+2y)^3=2^{30},$ which when we cube root both sides gives us $x+2y=2^{10}.$

The function $y=x^3$ is always increasing, so we don’t lose any solutions by doing this.

Let’s try to systematically list the pairs of positive integers ($x$,$y$) that would satisfy this equation.

The value $x=1$ does not work, because $\dfrac{2^{10}-1}{2}$ is not an integer. Clearly we require $x$ to be even for a solution.

So we have $(2,2^9-1), (4,2^9-2), (6,2^9-3),\dots, (2^{10}-4,2),(2^{10}-2,1).$

So $y$ takes every value from $1$ to $2^9-1$, which gives us that there are $2^9-1$ pairs of positive integers that satisfy the equation, and the answer is (c).