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There are some cute exercises based on the Chinese remainder theorem, e.g.,

(1) there exist an arbitrarily large number of consecutive integers, none of which is squarefree (1955 Putnam Competition),

(2) there exist an arbitrarily large number of consecutive integers, none of which is powerful ($n$ is powerful if for every prime $p$ dividing $n$, we have $p^2|n$),

(3) there exist an arbitrarily large number of consecutive positive integers, none of which is a sum of two squares,

(4) the number of integers $1\cdot 2, 2\cdot 3, \dots, n\cdot (n+1)$ divisible by $n$ is $2^{\omega(n)}$, where $\omega(n)$ is the number of distinct prime divisors of $n$.

There are some cute exercises based on the Chinese remainder theorem, e.g.,

(1) there exist an arbitrarily large number of consecutive integers, none of which is squarefree (1955 Putnam Competition),

(2) there exist an arbitrarily large number of consecutive integers, none of which is powerful ($n$ is powerful if for every prime $p$ dividing $n$, we have $p^2 \mid n$),

(3) there exist an arbitrarily large number of consecutive positive integers, none of which is a sum of two squares,

(4) the number of integers $1\cdot 2, 2\cdot 3, \dots, n\cdot (n+1)$ divisible by $n$ is $2^{\omega(n)}$, where $\omega(n)$ is the number of distinct prime divisors of $n$.

There are some cute exercises based on the Chinese remainder theorem, e.g.,  (1) there exist an arbitrarily large number of consecutive integers, none of which is squarefree (1955 Putnam Competition),  (2) there exist an arbitrarily large number of consecutive integers, none of which is powerful ($n$ is powerful if for every prime $p$ dividing $n$, we have $p^2|n$),  (3) there exist an arbitrarily large number of consecutive positive integers, none of which is a sum of two squares,  (4) the number of integers $1\cdot 2, 2\cdot 3, \dots, n\cdot (n+1)$ divisible by $n$ is $2^{\omega(n)}$, where $\omega(n)$ is the number of distinct prime divisors of $n$.

There are some cute exercises based on the Chinese remainder theorem, e.g., 

(1) there exist an arbitrarily large number of consecutive integers, none of which is squarefree (1955 Putnam Competition), 

(2) there exist an arbitrarily large number of consecutive integers, none of which is powerful ($n$ is powerful if for every prime $p$ dividing $n$, we have $p^2|n$), 

(3) there exist an arbitrarily large number of consecutive positive integers, none of which is a sum of two squares, 

(4) the number of integers $1\cdot 2, 2\cdot 3, \dots, n\cdot (n+1)$ divisible by $n$ is $2^{\omega(n)}$, where $\omega(n)$ is the number of distinct prime divisors of $n$.

There are some cute exercises based on the Chinese remainder theorem, e.g., (1) there exist an arbitrarily number of consecutive integers, none of which is squarefree (1955 Putnam Competition), (2) there exist an arbitrarily number of consecutive integers, none of which is powerful ($n$ is powerful if for every prime $p$ dividing $n$, we have $p^2|n$), (3) there exist an arbitrarily large number of consecutive positive integers, none of which is a sum of two squares, (4) the number of integers $1\cdot 2, 2\cdot 3, \dots, n\cdot (n+1)$ divisible by $n$ is $2^{\omega(n)}$, where $\omega(n)$ is the number of distinct prime divisors of $n$.

There are some cute exercises based on the Chinese remainder theorem, e.g., (1) there exist an arbitrarily large number of consecutive integers, none of which is squarefree (1955 Putnam Competition), (2) there exist an arbitrarily large number of consecutive integers, none of which is powerful ($n$ is powerful if for every prime $p$ dividing $n$, we have $p^2|n$), (3) there exist an arbitrarily large number of consecutive positive integers, none of which is a sum of two squares, (4) the number of integers $1\cdot 2, 2\cdot 3, \dots, n\cdot (n+1)$ divisible by $n$ is $2^{\omega(n)}$, where $\omega(n)$ is the number of distinct prime divisors of $n$.