Huge Primes

Number Theory Level 2

$97!\; n + d$ Find the smallest integer $d > 1$ , such that the number above is prime for an infinite number of integers $n$ .

Notation :
$!$ denotes the factorial notation. For example, $10! = 1\times2\times3\times\cdots\times10$ .

The answer is 101.

2 solutions

Jesse Nieminen
Mar 11, 2016

Dirichlet's theorem tells us that there exists an infinite amount of positive integers $n$ , for which $an + d$ is a prime number if $a$ and $d$ are coprime positive integers.

Using Dirichlet's theorem we know that $97!$ and $d$ must be coprime, in order to have an infinite amount of solutions for $n$ .

Since $97!$ is clearly divisible by all primes less than or equal to $97$ and since $d$ cannot be $1$ , $d$ must be the smallest prime number greater than $97$ .

By trial and error we find that $98$ , $99$ and $100$ are not prime numbers, but $101$ is a prime number.

Hence, $\boxed{101}$ is the solution.

The wiki tells everything :-P

Akshat Sharda - 5 years, 3 months ago

But how do we know that a and d have to be coprime does the theorem state that they must be coprime for this to be true?

Colin Carmody - 5 years, 3 months ago

If $a$ and $d$ aren't coprime their greatest common divisor is not 1. Thus, they are both divisible by some integer $x > 1$ , and therefore $an + d$ is divisible by $x > 1$ but because $x \leq d$ , $an + d$ can be prime only if $x = d$ is prime and $n = 0$ . Hence, there cannot be an infinite amount of solutions for $n$ if $a$ and $d$ aren't coprime.

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Jesse Nieminen - 5 years, 3 months ago

Saurabh Mittal
Apr 19, 2018

Here 97! contains all the factor till 97. so we have to look for a number that is prime as well as greater than 97 . that is 101 .

Yes, this shows that $101$ is the smallest candidate, but you didn't show that there is an infinite number of primes of form $97! n + 101$ .

Jesse Nieminen - 3 years, 1 month ago

Huge Primes

The answer is 101.

2 solutions

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