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The point of this "footnote" is to prove the following theorem which we use on our page about Mersenne primes and the historical note "the Largest Known Prime by Year". Fermat discovered and use the first part of this theorem (p = 1 modulo q) and Euler discovered the second.
Theorem.
Let p and q be odd primes. If p divides Mq, then p = 1 (mod q) and p = +/-1 (mod .
Below we give a proof and an example.
Proof.
If p divides Mq, then 2q = 1 (mod p) and the order of 2 (mod p) divides the prime q, so it must be q. By Fermat's Little Theorem the order of 2 also divides p-1, so p-1 = 2kq. This gives
2(p-1)/2 = 2qk = 1 (mod p)
so 2 is a quadratic residue mod p and it follows p = +/-1 (mod , which completes the proof.
Example: Suppose p divides M31, then the two parts of the theorem together show p = 1 or 63 (mod 248). By 1772 Euler had used this to show M31 was prime.

البرهان الاول

البرهان الثاني
The point of this "footnote" is to prove the following theorem stated by Euler in 1750 and proved by Lagrange in 1775. We use this theorem on our page about Mersenne primes.
Theorem.
Let p = 3 (mod 4) be prime. 2p+1 is also prime if and only if 2p+1 divides Mp.
Proof.
Suppose q = 2p+1 is prime. q = 7 (mod so 2 is a quadratic residue modulo q and it follows that there is an integer n such that n2 = 2 (mod q). This shows
2p = 2(q-1)/2 = nq-1 = 1 (mod q),
showing q divides Mp.
Conversely, let 2p+1 be a factor of Mp. Suppose, for proof by contradiction, that 2p+1 is composite and let q be its least prime factor. Then 2p = 1 (mod q) and the order of 2 modulo q divides both p and q-1, hence p divides q-1. This shows q > p and it follows
(2p+1) + 1 > q2 > p2
which is a contradiction since p > 2.
Note.
This means that if p = 3 (mod 4) and 2p+1 are both prime, then either p is 3 or Mp is composite.

البرهان الثالث
To test for divisibility by three (or nine), we often sum the digits and test the resulting sum. If the resulting sum has several digits, then sum again. Edwin O'Sullivan pointed that if we repeatedly sum the digits of an even perfect number (other than six), we always get one. Conrad Curry later poined out that this result can be found in [Gardner68].
Theorem.
If you sum the digits of any even perfect number (except 6), then sum the digits of the resulting number, and repeat this process until you get a single digit, that digit will be one.

Examples.
28  10  1, 496  19  10  1, and 8128  19  10  1

Proof.
Let s(n) be the sum of the digits of n. It is easy to see that s(n) = n (mod 9). So to prove the theorem, we need only show that perfect numbers are congruent to one modulo nine. If n is a perfect number, then n has the form 2p-1(2p-1) where p is prime (see theorem one). So p is either 2, 3, or is congruent to 1 or 5 modulo 6. Note that we have excluded the case p=2 (n=6). Finally, modulo nine, the powers of 2 repeat with period 6 (that is, 26 = 1 (mod 9)), so modulo nine n is congruent to one of the three numbers 21-1(21-1), 23-1(23-1), or 25-1(25-1), which are all 1 (mod 9).

The point of this "footnote" is to prove the following theorem which we use on our page about Mersenne primes and the historical note "the Largest Known Prime by Year". Fermat discovered and use the first part of this theorem (p = 1 modulo q) and Euler discovered the second.
Theorem.
Let p and q be odd primes. If p divides Mq, then p = 1 (mod q) and p = +/-1 (mod .
Below we give a proof and an example.
Proof.
If p divides Mq, then 2q = 1 (mod p) and the order of 2 (mod p) divides the prime q, so it must be q. By Fermat's Little Theorem the order of 2 also divides p-1, so p-1 = 2kq. This gives
2(p-1)/2 = 2qk = 1 (mod p)
so 2 is a quadratic residue mod p and it follows p = +/-1 (mod , which completes the proof.
Example: Suppose p divides M31, then the two parts of the theorem together show p = 1 or 63 (mod 248). By 1772 Euler had used this to show M31 was prime.

البرهان الاول

البرهان الثاني
The point of this "footnote" is to prove the following theorem stated by Euler in 1750 and proved by Lagrange in 1775. We use this theorem on our page about Mersenne primes.
Theorem.
Let p = 3 (mod 4) be prime. 2p+1 is also prime if and only if 2p+1 divides Mp.
Proof.
Suppose q = 2p+1 is prime. q = 7 (mod so 2 is a quadratic residue modulo q and it follows that there is an integer n such that n2 = 2 (mod q). This shows
2p = 2(q-1)/2 = nq-1 = 1 (mod q),
showing q divides Mp.
Conversely, let 2p+1 be a factor of Mp. Suppose, for proof by contradiction, that 2p+1 is composite and let q be its least prime factor. Then 2p = 1 (mod q) and the order of 2 modulo q divides both p and q-1, hence p divides q-1. This shows q > p and it follows
(2p+1) + 1 > q2 > p2
which is a contradiction since p > 2.
Note.
This means that if p = 3 (mod 4) and 2p+1 are both prime, then either p is 3 or Mp is composite.

البرهان الثالث
To test for divisibility by three (or nine), we often sum the digits and test the resulting sum. If the resulting sum has several digits, then sum again. Edwin O'Sullivan pointed that if we repeatedly sum the digits of an even perfect number (other than six), we always get one. Conrad Curry later poined out that this result can be found in [Gardner68].
Theorem.
If you sum the digits of any even perfect number (except 6), then sum the digits of the resulting number, and repeat this process until you get a single digit, that digit will be one.

Examples.
28  10  1, 496  19  10  1, and 8128  19  10  1

Proof.
Let s(n) be the sum of the digits of n. It is easy to see that s(n) = n (mod 9). So to prove the theorem, we need only show that perfect numbers are congruent to one modulo nine. If n is a perfect number, then n has the form 2p-1(2p-1) where p is prime (see theorem one). So p is either 2, 3, or is congruent to 1 or 5 modulo 6. Note that we have excluded the case p=2 (n=6). Finally, modulo nine, the powers of 2 repeat with period 6 (that is, 26 = 1 (mod 9)), so modulo nine n is congruent to one of the three numbers 21-1(21-1), 23-1(23-1), or 25-1(25-1), which are all 1 (mod 9).

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The point of this "footnote" is to prove the following theorem which we use on our page about Mersenne primes and the historical note "the Largest Known Prime by Year". Fermat discovered and use the first part of this theorem (p = 1 modulo q) and Euler discovered the second.
Theorem.
Let p and q be odd primes. If p divides Mq, then p = 1 (mod q) and p = +/-1 (mod .
Below we give a proof and an example.
Proof.
If p divides Mq, then 2q = 1 (mod p) and the order of 2 (mod p) divides the prime q, so it must be q. By Fermat's Little Theorem the order of 2 also divides p-1, so p-1 = 2kq. This gives
2(p-1)/2 = 2qk = 1 (mod p)
so 2 is a quadratic residue mod p and it follows p = +/-1 (mod , which completes the proof.
Example: Suppose p divides M31, then the two parts of the theorem together show p = 1 or 63 (mod 248). By 1772 Euler had used this to show M31 was prime.

البرهان الاول

البرهان الثاني
The point of this "footnote" is to prove the following theorem stated by Euler in 1750 and proved by Lagrange in 1775. We use this theorem on our page about Mersenne primes.
Theorem.
Let p = 3 (mod 4) be prime. 2p+1 is also prime if and only if 2p+1 divides Mp.
Proof.
Suppose q = 2p+1 is prime. q = 7 (mod so 2 is a quadratic residue modulo q and it follows that there is an integer n such that n2 = 2 (mod q). This shows
2p = 2(q-1)/2 = nq-1 = 1 (mod q),
showing q divides Mp.
Conversely, let 2p+1 be a factor of Mp. Suppose, for proof by contradiction, that 2p+1 is composite and let q be its least prime factor. Then 2p = 1 (mod q) and the order of 2 modulo q divides both p and q-1, hence p divides q-1. This shows q > p and it follows
(2p+1) + 1 > q2 > p2
which is a contradiction since p > 2.
Note.
This means that if p = 3 (mod 4) and 2p+1 are both prime, then either p is 3 or Mp is composite.

البرهان الثالث
To test for divisibility by three (or nine), we often sum the digits and test the resulting sum. If the resulting sum has several digits, then sum again. Edwin O'Sullivan pointed that if we repeatedly sum the digits of an even perfect number (other than six), we always get one. Conrad Curry later poined out that this result can be found in [Gardner68].
Theorem.
If you sum the digits of any even perfect number (except 6), then sum the digits of the resulting number, and repeat this process until you get a single digit, that digit will be one.

Examples.
28  10  1, 496  19  10  1, and 8128  19  10  1

Proof.
Let s(n) be the sum of the digits of n. It is easy to see that s(n) = n (mod 9). So to prove the theorem, we need only show that perfect numbers are congruent to one modulo nine. If n is a perfect number, then n has the form 2p-1(2p-1) where p is prime (see theorem one). So p is either 2, 3, or is congruent to 1 or 5 modulo 6. Note that we have excluded the case p=2 (n=6). Finally, modulo nine, the powers of 2 repeat with period 6 (that is, 26 = 1 (mod 9)), so modulo nine n is congruent to one of the three numbers 21-1(21-1), 23-1(23-1), or 25-1(25-1), which are all 1 (mod 9).

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