14.25. How many natural numbers $x$, less than 10000, exist for which $2^{x}-x^{2}$ is divisible by $7 ?$
14.25. Answer: 2857. The remainders of $2^{x}$ and $x^{2}$ when divided by 7 repeat with periods of 3 and 7, respectively, so the remainders of $2^{x}-x^{2}$ when divided by 7 repeat with a period of 21. Among the numbers $x$ from 1 to 21, the remainders of $2^{x}$ and $x^{2}$ when divided by 7 are equal for exactly 6 numbers. Therefore, among the numbers from 1 to $9996=21 \cdot 476$, there are $476 \cdot 6=2856$ required numbers. Direct verification using the obtained sequence of remainders shows that among the remaining numbers 9997, 9998, and 9999, only the number 9998 has the required property.