# Difference between revisions of "2015 USAMO Problems/Problem 1"

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<math>2x - 3n^2 - 3n = \pm (2n^3 + 3n^2 -3n -2)</math> | <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 | + | <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 | + | 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> |

## Revision as of 22:38, 12 May 2015

### Problem

Solve in integers the equation

### Solution

We first notice that both sides must be integers, so must be an integer.

We can therefore perform the substitution where is an integer.

Then:

is therefore the square of an odd integer and can be replaced with

By substituting using we get:

or

Using substitution we get the solutions: