Difference between revisions of "2015 USAJMO Problems/Problem 2"

Revision as of 14:05, 14 May 2015

Solve in integers the equation $x^2+xy+y^2 = \left(\frac{x+y}{3}+1\right)^3.$

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 $x = - n^3 + 3n + 1$

Using substitution we get the solutions: $(n^3 + 3n^2 - 1, - n^3 + 3n + 1) \cup ( - n^3 + 3n + 1, n^3 + 3n^2 - 1)$

@@ Line 25: / Line 25: @@
 <math>2x - 3n^2 - 3n = \pm (2n^3 + 3n^2 -3n -2)</math>
-<math>x = n^3 + 3n^2 - 1</math> or <math>x = n^3 + 3n - 1</math>
+<math>x = n^3 + 3n^2 - 1</math> or <math>x = - n^3 + 3n + 1</math>
-Using substitution we get the solutions: <math>(n^3 + 3n^2 - 1, n^3 + 3n - 1) \cup (n^3 + 3n - 1, n^3 + 3n^2 - 1)</math>
+Using substitution we get the solutions: <math>(n^3 + 3n^2 - 1, - n^3 + 3n + 1) \cup ( - n^3 + 3n + 1, n^3 + 3n^2 - 1)</math>