IMO 1995 · Problem 4

Find the maximum value of \(x_0\) for which there exists a sequence \(x_0,x_1\ldots,x_{1995}\) of positive reals with \(x_0=x_{1995}\), such that for \(i=1,\ldots,1995\), \[ x_{i-1}+\frac{2}{x_{i-1}}=2x_i+\frac{1}{x_i}. \]

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

Answer

\[ \boxed{2^{997}}. \]

Solution

Multiplying the given equation by \(x_{i-1}x_i\) and factoring gives \[ (x_{i-1}-2x_i)(x_{i-1}x_i-1)=0. \] Thus each step satisfies at least one of \[ x_i=\frac{x_{i-1}}2,\qquad x_i=\frac1{x_{i-1}}. \]

Put \(y_i=\log_2 x_i\). The two possible steps become \[ y_i=y_{i-1}-1,\qquad y_i=-y_{i-1}. \] Choose one valid type for each step, and let \(s\) be the number of subtractions and \(r\) the number of negations. Then \[ s+r=1995. \]

Composing all the steps gives \[ y_{1995}=(-1)^r y_0+c, \] where \(c\) is a sum of \(s\) terms, each equal to \(1\) or \(-1\): each subtraction contributes \(-1\), whose sign is reversed by every subsequent negation. Consequently, \[ |c|\le s,\qquad c\equiv s\pmod2. \]

If \(r\) were even, the condition \(y_{1995}=y_0\) would force \(c=0\), so \(s\) would also be even. This contradicts \(s+r=1995\). Therefore \(r\) is odd, and hence \(r\ge1\) and \(s\le1994\). The closing condition now gives \[ y_0=-y_0+c, \] so \[ 2y_0=c\le s\le1994. \] Thus \(y_0\le997\), proving \[ x_0\le2^{997}. \]

To attain this bound, set \[ x_i=2^{997-i}\quad(0\le i\le1994), \qquad x_{1995}=2^{997}. \] All terms are positive. The first \(1994\) steps satisfy \(x_i=x_{i-1}/2\), while the last satisfies \[ x_{1995}=\frac1{x_{1994}}=2^{997}=x_0. \] By the factorization above, every required equation holds. Therefore the maximum is \(\boxed{2^{997}}\).

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