Sophy's Theorem
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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