## For every integer $n>1$, the family of functions $f_{n,a,b}$ is defined

by

$f_{n,a,b}(x)\equiv a x + b \mod n\,\,\, $ for $a,b,x$ integer and $0< a
### To solve the given problem, you must apply number theory and combinatorics. Here is how you can approach the solution:

First, observe that when $a$ is chosen in $f_{n, a, b}$ as a function, no matter which integer $b$ is, the function is always a retraction due to the property of modular arithmetic. This is easy to see, because $f(f(x)) = a(ax + b) + b = ax + b = f(x)$. For every valid choice of $a$ and $b$, there exists a retraction. Therefore, if $n$ is a prime number, the number of retractions is equal to the number of all possible functions which is $(n-1) \times n = n^2 – n$.

For composite numbers, things will be more complex. If $n=ab$ where $a,b > 1$, such that $gcd(a,b) = 1$, then we can choose corresponding $(a_i,b_i)$ pairs $(a,0),\ldots,(a,b-1);\ (0,b),\ldots,(a-1,b)$ that will map all numbers in the range to themselves. This implies that there are less than $(n^2-n)$ retractions for non-prime $n$’s.

After these observations, you can calculate $sum_{p\, prime} (p^2-p)\ (2\le p\le N)$ for prime values up until $N$ with a prime sieve-like algorithm. For the reduction in the total number of retractions, you can use a recursive function with memoization and inclusion-exclusion principle to find the result, where you need to iterate through all the smallest prime factors of every possible composite number less or equal than $N$.

This solution approach does not provide a concrete and exact algorithm due to its complexity.

For the actual value of $F(10^{14}) \mod (10^9+7)$, due to the complexity and the vastness of the operation (which cannot be performed by a human without a computer), this operation is intended to be a task for programming competitions where one can write a program to follow our guide and find out the answer.

Remember to get the results modulo $1\,000\,000\,007$ at each step of the calculation to ensure that you do not store unnecessarily large numbers.

##### More Answers:

The Inverse Summation of Coprime Couples

Eleven-free Integers

Retractions B

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### To solve the given problem, you must apply number theory and combinatorics. Here is how you can approach the solution:

First, observe that when $a$ is chosen in $f_{n, a, b}$ as a function, no matter which integer $b$ is, the function is always a retraction due to the property of modular arithmetic. This is easy to see, because $f(f(x)) = a(ax + b) + b = ax + b = f(x)$. For every valid choice of $a$ and $b$, there exists a retraction. Therefore, if $n$ is a prime number, the number of retractions is equal to the number of all possible functions which is $(n-1) \times n = n^2 – n$.

For composite numbers, things will be more complex. If $n=ab$ where $a,b > 1$, such that $gcd(a,b) = 1$, then we can choose corresponding $(a_i,b_i)$ pairs $(a,0),\ldots,(a,b-1);\ (0,b),\ldots,(a-1,b)$ that will map all numbers in the range to themselves. This implies that there are less than $(n^2-n)$ retractions for non-prime $n$’s.

After these observations, you can calculate $sum_{p\, prime} (p^2-p)\ (2\le p\le N)$ for prime values up until $N$ with a prime sieve-like algorithm. For the reduction in the total number of retractions, you can use a recursive function with memoization and inclusion-exclusion principle to find the result, where you need to iterate through all the smallest prime factors of every possible composite number less or equal than $N$.

This solution approach does not provide a concrete and exact algorithm due to its complexity.

For the actual value of $F(10^{14}) \mod (10^9+7)$, due to the complexity and the vastness of the operation (which cannot be performed by a human without a computer), this operation is intended to be a task for programming competitions where one can write a program to follow our guide and find out the answer.

Remember to get the results modulo $1\,000\,000\,007$ at each step of the calculation to ensure that you do not store unnecessarily large numbers.

##### More Answers:

The Inverse Summation of Coprime CouplesEleven-free Integers

Retractions B