IMO 1982 · Problem 4

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 when \(n=2891\).

Official text of the problem. Official problem paper (PDF)

Solution

Put \(F(x,y)=x^3-3xy^2+y^3\). Direct expansion gives \[ F(-y,x-y)=-y^3+3y(x-y)^2+(x-y)^3=F(x,y). \] Thus the transformation \(T(x,y)=(-y,x-y)\) preserves every solution. Moreover, \[ T^2(x,y)=(y-x,-x),\qquad T^3(x,y)=(x,y). \] Consequently, from any integer solution we obtain the three integer solutions \[ (x,y),\qquad (-y,x-y),\qquad (y-x,-x). \]

These are distinct. Indeed, since \(T^3\) is the identity, any equality between two of these pairs implies \(T(x,y)=(x,y)\). This gives \[ x=-y,\qquad y=x-y, \] so \(3y=0\), and hence \(x=y=0\). But this would give \(n=0\), contrary to \(n>0\). This proves the first assertion.

For the second assertion, we show that \[ 7\mid F(x,y)\quad\Longrightarrow\quad 7\mid x\ \text{and}\ 7\mid y. \] Suppose first that \(7\nmid y\). Dividing the congruence \(F(x,y)\equiv0\pmod7\) by \(y^3\), and writing \(t\equiv xy^{-1}\pmod7\), gives \[ t^3-3t+1\equiv0\pmod7. \] However, checking all residue classes gives \[ \begin{array}{c|rrrrrrr} t&0&1&2&3&4&5&6\\ \hline t^3-3t+1\pmod7&1&6&3&5&4&6&3 \end{array} \] so this congruence is impossible. Therefore \(7\mid y\). The original congruence then reduces to \(x^3\equiv0\pmod7\), which implies \(7\mid x\), proving the implication.

Now suppose \(F(x,y)=2891\). Since \[ 2891=7^2\cdot59, \] the implication above gives \(x=7a\) and \(y=7b\) for integers \(a,b\). Hence \[ 2891=F(7a,7b)=7^3F(a,b), \] which would make \(2891\) divisible by \(7^3\). This contradicts \(2891=7^2\cdot59\), since \(7\nmid59\). Thus there are no integer solutions when \(n=2891\).

The solution was checked against John Scholes' IMO solution archive (Kalva).