Power Trap

The recursive formulas, defining the pairs, make sense modulo every prime $p$ . Note that if $x_i$ is not zero modulo $p$ , and $x_i$ is not congruent to $y_i$ modulo $p$ , then $\frac{y_{i+1}}{x_{i+1}}=\left( \frac{y_i}{x_i}\right) ^2$ If $x_i\equiv y_i \pmod p$ then $x_{i+1}\equiv y_{i+1}\equiv 0 \pmod p$ and all further numbers are zero modulo $p$ .

Recall that the multiplicative group of residues modulo any prime $p$ is cyclic of order $p-1$ . If $p-1=2^k$ for some $k$ , then after no more than $k$ iterations, starting from any $x_0$ and $y_0$ not divisible by $p$ , we get $\frac{y_i}{x_i}\equiv 1 \pmod p,$ so $x_{i+1}\equiv y_{i+1}\equiv 0\pmod p$ and all subsequent $x_n$ and $y_n$ are zero modulo $p$ . On the other hand, if $p-1$ is not a power of $2$ , then one can find $\frac{y_0}{x_0}$ so that its order in the multiplicative group modulo $p$ is not a power of $2$ , and then $x_n$ and $y_n$ will never be zero modulo $p$ .

There are $5$ primes less than $1000$ of the form $p=2^k+1:$ $2,3,5,17,257.$ For each of them $k\leq 8,$ so $x_{10}$ will be divisible by $p$ for any $x_0$ and $y_0$ . Thus, the answer is $5$ .