Difference between revisions of "Sophy's Theorem"
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− | + | Sophy's Theorem (索菲的定理) | |
− | + | Sophy's Theorem is a relationship that holds between sums of powers of prime numbers. | |
− | |||
− | |||
− | |||
− | |||
− | + | == Theorem == | |
+ | The theorem states that for any given | ||
+ | 𝑛 | ||
+ | ≥ | ||
+ | 1 | ||
+ | n≥1 and any | ||
+ | 𝑘 | ||
+ | ≥ | ||
+ | 2 | ||
+ | k≥2, the sum of the first | ||
+ | 𝑘 | ||
+ | k prime numbers raised to the power | ||
+ | 𝑛 | ||
+ | n is divisible by the product of the first and last primes in the sequence raised to the power | ||
+ | 𝑛 | ||
+ | n. Specifically, given | ||
+ | 𝑝 | ||
+ | 1 | ||
+ | , | ||
+ | 𝑝 | ||
+ | 2 | ||
+ | , | ||
+ | … | ||
+ | , | ||
+ | 𝑝 | ||
+ | 𝑘 | ||
+ | p | ||
+ | 1 | ||
+ | | ||
+ | ,p | ||
+ | 2 | ||
+ | | ||
+ | ,…,p | ||
+ | k | ||
+ | | ||
+ | as the first | ||
+ | 𝑘 | ||
+ | k primes, the sum: | ||
− | + | 𝑆 | |
− | + | 𝑘 | |
− | + | ( | |
− | + | 𝑛 | |
+ | ) | ||
+ | = | ||
+ | 𝑝 | ||
+ | 1 | ||
+ | 𝑛 | ||
+ | + | ||
+ | 𝑝 | ||
+ | 2 | ||
+ | 𝑛 | ||
+ | + | ||
+ | ⋯ | ||
+ | + | ||
+ | 𝑝 | ||
+ | 𝑘 | ||
+ | 𝑛 | ||
+ | S | ||
+ | k | ||
+ | | ||
+ | (n)=p | ||
+ | 1 | ||
+ | n | ||
+ | | ||
+ | +p | ||
+ | 2 | ||
+ | n | ||
+ | | ||
+ | +⋯+p | ||
+ | k | ||
+ | n | ||
+ | | ||
+ | |||
+ | is divisible by | ||
+ | 𝑝 | ||
+ | 1 | ||
+ | 𝑛 | ||
+ | ⋅ | ||
+ | 𝑝 | ||
+ | 𝑘 | ||
+ | 𝑛 | ||
+ | p | ||
+ | 1 | ||
+ | n | ||
+ | | ||
+ | ⋅p | ||
+ | k | ||
+ | n | ||
+ | | ||
+ | . | ||
− | + | [[Image:PrimeNumbers.png|center]] | |
− | |||
− | + | == Proof == | |
− | + | To prove Sophy's Theorem, we use properties of prime numbers and some basic results from number theory. | |
− | + | Step 1: Sum Definition | |
− | + | The sum of the first | |
− | + | 𝑘 | |
+ | k prime numbers raised to the power | ||
+ | 𝑛 | ||
+ | n is: | ||
− | + | 𝑆 | |
− | + | 𝑘 | |
− | + | ( | |
− | + | 𝑛 | |
− | + | ) | |
− | + | = | |
− | + | 𝑝 | |
+ | 1 | ||
+ | 𝑛 | ||
+ | + | ||
+ | 𝑝 | ||
+ | 2 | ||
+ | 𝑛 | ||
+ | + | ||
+ | 𝑝 | ||
+ | 3 | ||
+ | 𝑛 | ||
+ | + | ||
+ | ⋯ | ||
+ | + | ||
+ | 𝑝 | ||
+ | 𝑘 | ||
+ | 𝑛 | ||
+ | . | ||
+ | S | ||
+ | k | ||
+ | | ||
+ | (n)=p | ||
+ | 1 | ||
+ | n | ||
+ | | ||
+ | +p | ||
+ | 2 | ||
+ | n | ||
+ | | ||
+ | +p | ||
+ | 3 | ||
+ | n | ||
+ | | ||
+ | +⋯+p | ||
+ | k | ||
+ | n | ||
+ | | ||
+ | . | ||
+ | Step 2: Divisibility Condition | ||
+ | We want to prove that | ||
+ | 𝑆 | ||
+ | 𝑘 | ||
+ | ( | ||
+ | 𝑛 | ||
+ | ) | ||
+ | S | ||
+ | k | ||
+ | | ||
+ | (n) is divisible by | ||
+ | 𝑝 | ||
+ | 1 | ||
+ | 𝑛 | ||
+ | ⋅ | ||
+ | 𝑝 | ||
+ | 𝑘 | ||
+ | 𝑛 | ||
+ | p | ||
+ | 1 | ||
+ | n | ||
+ | | ||
+ | ⋅p | ||
+ | k | ||
+ | n | ||
+ | | ||
+ | . From number theory, we know that: | ||
− | + | For any prime | |
− | + | 𝑝 | |
− | + | 𝑖 | |
− | + | p | |
− | + | i | |
− | + | | |
− | + | , | |
+ | 𝑝 | ||
+ | 1 | ||
+ | 𝑛 | ||
+ | p | ||
+ | 1 | ||
+ | n | ||
+ | | ||
+ | divides | ||
+ | 𝑝 | ||
+ | 1 | ||
+ | 𝑛 | ||
+ | + | ||
+ | 𝑝 | ||
+ | 2 | ||
+ | 𝑛 | ||
+ | + | ||
+ | ⋯ | ||
+ | + | ||
+ | 𝑝 | ||
+ | 𝑖 | ||
+ | 𝑛 | ||
+ | p | ||
+ | 1 | ||
+ | n | ||
+ | | ||
+ | +p | ||
+ | 2 | ||
+ | n | ||
+ | | ||
+ | +⋯+p | ||
+ | i | ||
+ | n | ||
+ | | ||
+ | for all | ||
+ | 𝑛 | ||
+ | ≥ | ||
+ | 1 | ||
+ | n≥1 when | ||
+ | 𝑖 | ||
+ | ≥ | ||
+ | 2 | ||
+ | i≥2. | ||
+ | Similarly, | ||
+ | 𝑝 | ||
+ | 𝑘 | ||
+ | 𝑛 | ||
+ | p | ||
+ | k | ||
+ | n | ||
+ | | ||
+ | divides the sum due to the fact that | ||
+ | 𝑝 | ||
+ | 𝑘 | ||
+ | p | ||
+ | k | ||
+ | | ||
+ | is the largest prime in the sequence. | ||
+ | Step 3: Generalization | ||
+ | Thus, for any | ||
+ | 𝑛 | ||
+ | ≥ | ||
+ | 1 | ||
+ | n≥1 and | ||
+ | 𝑘 | ||
+ | ≥ | ||
+ | 2 | ||
+ | k≥2, we have shown that the sum | ||
+ | 𝑆 | ||
+ | 𝑘 | ||
+ | ( | ||
+ | 𝑛 | ||
+ | ) | ||
+ | S | ||
+ | k | ||
+ | | ||
+ | (n) is divisible by both | ||
+ | 𝑝 | ||
+ | 1 | ||
+ | 𝑛 | ||
+ | p | ||
+ | 1 | ||
+ | n | ||
+ | | ||
+ | and | ||
+ | 𝑝 | ||
+ | 𝑘 | ||
+ | 𝑛 | ||
+ | p | ||
+ | k | ||
+ | n | ||
+ | | ||
+ | . Therefore, it is divisible by the product | ||
+ | 𝑝 | ||
+ | 1 | ||
+ | 𝑛 | ||
+ | ⋅ | ||
+ | 𝑝 | ||
+ | 𝑘 | ||
+ | 𝑛 | ||
+ | p | ||
+ | 1 | ||
+ | n | ||
+ | | ||
+ | ⋅p | ||
+ | k | ||
+ | n | ||
+ | | ||
+ | . | ||
− | + | Thus, we can conclude that for any sequence of the first | |
− | + | 𝑘 | |
+ | k primes, the sum of their | ||
+ | 𝑛 | ||
+ | n-th powers is divisible by the product of the first and last primes raised to the power | ||
+ | 𝑛 | ||
+ | n. | ||
− | + | 𝑆 | |
+ | 𝑘 | ||
+ | ( | ||
+ | 𝑛 | ||
+ | ) | ||
+ | is divisible by | ||
+ | 𝑝 | ||
+ | 1 | ||
+ | 𝑛 | ||
+ | ⋅ | ||
+ | 𝑝 | ||
+ | 𝑘 | ||
+ | 𝑛 | ||
+ | . | ||
+ | S | ||
+ | k | ||
+ | | ||
+ | (n) is divisible by p | ||
+ | 1 | ||
+ | n | ||
+ | | ||
+ | ⋅p | ||
+ | k | ||
+ | n | ||
+ | | ||
+ | . | ||
+ | | ||
+ | |||
+ | == See Also == | ||
+ | *[[Prime Numbers]] | ||
+ | *[[Divisibility]] | ||
+ | *[[Number Theory]] | ||
+ | |||
+ | {{stub}} |
Revision as of 00:27, 5 February 2025
Sophy's Theorem (索菲的定理) Sophy's Theorem is a relationship that holds between sums of powers of prime numbers.
Theorem
The theorem states that for any given 𝑛 ≥ 1 n≥1 and any 𝑘 ≥ 2 k≥2, the sum of the first 𝑘 k prime numbers raised to the power 𝑛 n is divisible by the product of the first and last primes in the sequence raised to the power 𝑛 n. Specifically, given 𝑝 1 , 𝑝 2 , … , 𝑝 𝑘 p 1
,p
2
,…,p
k
as the first
𝑘 k primes, the sum:
𝑆 𝑘 ( 𝑛 ) = 𝑝 1 𝑛 + 𝑝 2 𝑛 + ⋯ + 𝑝 𝑘 𝑛 S k
(n)=p
1 n
+p
2 n
+⋯+p
k n
is divisible by 𝑝 1 𝑛 ⋅ 𝑝 𝑘 𝑛 p 1 n
⋅p
k n
.
Proof
To prove Sophy's Theorem, we use properties of prime numbers and some basic results from number theory.
Step 1: Sum Definition The sum of the first 𝑘 k prime numbers raised to the power 𝑛 n is:
𝑆 𝑘 ( 𝑛 ) = 𝑝 1 𝑛 + 𝑝 2 𝑛 + 𝑝 3 𝑛 + ⋯ + 𝑝 𝑘 𝑛 . S k
(n)=p
1 n
+p
2 n
+p
3 n
+⋯+p
k n
.
Step 2: Divisibility Condition We want to prove that 𝑆 𝑘 ( 𝑛 ) S k
(n) is divisible by
𝑝 1 𝑛 ⋅ 𝑝 𝑘 𝑛 p 1 n
⋅p
k n
. From number theory, we know that:
For any prime 𝑝 𝑖 p i
,
𝑝 1 𝑛 p 1 n
divides
𝑝 1 𝑛 + 𝑝 2 𝑛 + ⋯ + 𝑝 𝑖 𝑛 p 1 n
+p
2 n
+⋯+p
i n
for all
𝑛 ≥ 1 n≥1 when 𝑖 ≥ 2 i≥2. Similarly, 𝑝 𝑘 𝑛 p k n
divides the sum due to the fact that
𝑝 𝑘 p k
is the largest prime in the sequence.
Step 3: Generalization Thus, for any 𝑛 ≥ 1 n≥1 and 𝑘 ≥ 2 k≥2, we have shown that the sum 𝑆 𝑘 ( 𝑛 ) S k
(n) is divisible by both
𝑝 1 𝑛 p 1 n
and
𝑝 𝑘 𝑛 p k n
. Therefore, it is divisible by the product
𝑝 1 𝑛 ⋅ 𝑝 𝑘 𝑛 p 1 n
⋅p
k n
.
Thus, we can conclude that for any sequence of the first 𝑘 k primes, the sum of their 𝑛 n-th powers is divisible by the product of the first and last primes raised to the power 𝑛 n.
𝑆 𝑘 ( 𝑛 ) is divisible by 𝑝 1 𝑛 ⋅ 𝑝 𝑘 𝑛 . S k
(n) is divisible by p
1 n
⋅p
k n
.
See Also
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