- #1

Math100

- 793

- 220

- Homework Statement
- For ## n>1 ##, show that every prime divisor of ## n!+1 ## is an odd integer that is greater than ## n ##.

- Relevant Equations
- None.

Proof:

Suppose for the sake of contradiction that there exists a prime divisor of ## n!+1 ##,

which is an odd integer that is not greater than ## n ##.

Let ## n>1 ## be an integer.

Since ## n! ## is even,

it follows that ## n!+1 ## is odd.

Thus ## 2\nmid (n!+1) ##.

This means every prime factor of ## n!+1 ## is an odd integer.

Let ## p ## be a prime factor of ## n!+1 ## such that ## p\leq n ##.

Then we have ## p\mid n! ## and ## p\mid (n!+1) ##.

Thus ## p\mid 1 ##, which implies that ## p=\pm 1 ##.

This is a contradiction because ## p ## must be a prime number where ## p\neq \pm 1 ##, so ## p\nleq n ##.

Therefore, for ## n>1 ##, every prime divisor of ## n!+1 ## is an odd integer that is greater than ## n ##.

Suppose for the sake of contradiction that there exists a prime divisor of ## n!+1 ##,

which is an odd integer that is not greater than ## n ##.

Let ## n>1 ## be an integer.

Since ## n! ## is even,

it follows that ## n!+1 ## is odd.

Thus ## 2\nmid (n!+1) ##.

This means every prime factor of ## n!+1 ## is an odd integer.

Let ## p ## be a prime factor of ## n!+1 ## such that ## p\leq n ##.

Then we have ## p\mid n! ## and ## p\mid (n!+1) ##.

Thus ## p\mid 1 ##, which implies that ## p=\pm 1 ##.

This is a contradiction because ## p ## must be a prime number where ## p\neq \pm 1 ##, so ## p\nleq n ##.

Therefore, for ## n>1 ##, every prime divisor of ## n!+1 ## is an odd integer that is greater than ## n ##.