Question 29
In the set of integers, solve the equation
$$
x^{2010} - 2006 = 4y^{2009} + 4y^{2008} + 2007y.
$$
$$
x^{2010} - 2006 = 4y^{2009} + 4y^{2008} + 2007y.
$$
MathNet 2009-part-03 — 175 problems (EN–VI) Question 29
Math Olympiad Vault · MathNet 2009-part-03 — 175 problems (EN–VI)