proof of Wilson’s theorem using the Wilson quotient

proof of Wilson’s theorem using the Wilson quotient

Theorem

. An

integer

n

>

1

is prime only if the

Wilson quotient

(

n

-

1

)

!

+

1

n

is an integer.

Proof.

If

n

is

composite

, then its

greatest prime factor

is at most

n

2

, and

n

2

<

(

n

-

1

)

as long as

n

>

2

(and the smallest

positive

composite number

is 4). Therefore,

(

n

-

1

)

!

being the

product

of the numbers from 1 to

n

-

1

includes among its

divisors

the greatest prime factor of

n

, and indeed all its divisors. Since two consecutive integers are always

coprime

, it is the case that

gcd

⁡

(

(

n

-

1

)

!

,

(

n

-

1

)

!

+

1

)

=

1

. Therefore

n

will divide

(

n

-

1

)

!

evenly but not

(

n

-

1

)

!

+

1

. So the Wilson quotient will be a

rational number

but not an integer.

If we only meant to prove the

converse

of Wilson’s theorem we’d be done at this point. But we set out to prove that not only is the stated

relation

false for composite numbers, but that it is true for primes. Proving the former is quite easy. Proving the latter is harder, and in fact neither Edward Waring nor John Wilson left a proof. (Koshy, 2007)

If

n

is prime, then obviously its greatest prime factor is itself, and

(

n

-

1

)

!

will have only 1 as a divisor in common with

n

. But how do we prove that the next number after

(

n

-

1

)

!

is a

multiple

of

n

without having to factorize several values of

(

n

-

1

)

!

and hoping the proof makes itself apparent thus? Modular

multiplication

comes to the rescue. Since

n

is prime, we can multiply any number from 2 to

n

-

2

by another of the same range and the product will be

congruent

to 1 modulo

n

.

For example,

n

=

7

. We verify that

2

×

4

=

8

≡

1

mod

7

and

3

×

5

=

15

≡

1

mod

7

. Since

1

×

1

=

1

, multiplying the range 2 to 5 will give a number that

satisfies

the same

congruence

.

So in general multiplying the range 2 to

n

-

2

gives a number that is congruent to 1 modulo

n

if

n

is prime. (With

n

composite, modular multiplication causes a zeroing out of the range’s overall product).

n

-

1

is different, satisfying

(

n

-

1

)

≡

(

n

-

1

)

mod

n

≡

-

1

mod

n

. And since

1

×

-

1

=

-

1

, modular multiplication of the range 2 to

n

-

1

gives

-

1

. Thus

(

n

-

1

)

!

≡

-

1

mod

n

when

n

is prime, so

(

n

-

1

)

!

+

1

≡

0

mod

n

. Therefore,

(

n

-

1

)

!

+

1

, which is greater than

n

, is also

divisible

by it, and thus dividing it by

n

yields an integer.
∎

References

1

Thomas Kochy,

Elementary Number Theory

with Applications

, 2nd Edition. London: Elsevier (2007): 321 - 323

Title

proof of Wilson’s theorem using the Wilson quotient

Canonical name

ProofOfWilsonsTheoremUsingTheWilsonQuotient

Date of creation

2013-03-22 17:58:57

Last modified on

2013-03-22 17:58:57

Owner

PrimeFan (13766)

Last modified by

PrimeFan (13766)

Numerical id

5

Author

PrimeFan (13766)

Entry type

Proof

Classification

msc 11A51

Classification

msc 11A41

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