From USAMO 2015

Algebra Level 3

Solve in integers for $x$ :

$\large {x^2+xy+y^2= \left(\frac{x+y}{3}+1\right)^3}$

If you get $x$ as a polynomial in $n$ , where $n$ is any integer (say $x(n)$ ), then find the maximum value of $x(2)$ .

1 solution

Bloons Qoth
Jun 20, 2016

We first notice that both sides must be integers, so $\frac{x+y}{3}$ must be an integer.

We can therefore perform the substitution $x+y = 3t$ where $t$ is an integer.

Then:

${(3t)^2 - xy = (t+1)^3}$

${9t^2 + x (x - 3t) = t^3 + 3t^2 + 3t + 1}$

${4x^2 - 12xt + 9t^2 = 4t^3 - 15t^2 + 12t + 4}$

${(2x - 3t)^2 = (t - 2)^2(4t + 1)}$

$4t+1$ is therefore the square of an odd integer and can be replaced with ${(2n+1)^2 = 4n^2 + 4n +1}$

By substituting using ${t = n^2 + n}$ we get:

${(2x - 3n^2 - 3n)^2 = [(n^2 + n - 2)(2n+1)]^2}$

${2x - 3n^2 - 3n = \pm (2n^3 + 3n^2 -3n -2)}$

${x = n^3 + 3n^2 - 1}$ or ${x = -n^3 + 3n + 1}$

Since there is no good way to type the above answer into the machine, so I asked for you to plug 2 into what $x$ equals to, which is variable $n$ in this case.

${n^3 + 3n^2 - 1}$ or ${-n^3 + 3n + 1}$

${2^3 + 3*2^2 -1}$ or ${-2^3 +3*2 +1}$