1982 IMO Problems/Problem 4

Revision as of 16:59, 21 August 2008 by Cosinator (talk | contribs) (page/ problem/ solution)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

Prove that if $n$ is a positive integer such that the equation $x^3-3xy^2+y^3=n$ has a solution in integers $x,y$, then it has at least three such solutions. Show that the equation has no solutions in integers for $n=2891$.

Suppose the equation $x^3-3xy^2+y^3=n$ has solution in integers $(x,y)$ with $y=x+k$. Then, completing the cube yields $(y-x)^3-3x^2y+2x^3$. Using the substitution $y=x+k$ yields $k^3-3kx^2-x^3=n$. Notice that equality directly implies that $(k,-x)$ is also a solution to the original equation. Applying the transformation again yields that $(-x-k,-k)$ is also a solution. We show that these three solutions are indeed distinct: If $(x,y)=(k,-x)$ then $x=k,x+k=-x$ which only has solution $x=k=0$ which implies that $n$ is not a positive integer, a contradiction. Similarly, since the transformation from $(k,-x)$ to $(-x-k,-k)$ and $(-x-k,-k)$ to $(x,y)$ is the same as the transformation from $(x,y)$ to $(k,-x)$, we have that the three solutions are pairwise distinct.

For the case $n=2891$, notice that $7|n$. Considering all solutions modulo $7$ of the equation $x^3-3xy^2+y^3\equiv0\pmod{7}$ yields only $x\equiv y\equiv 0\pmod{7}$. But, this implies that $7^3$ divides $2891$ which is clearly not true.

Invalid username
Login to AoPS