Conjecture 63. Consecutive primes sum to a perfect power.

Anton Vrba sent the following conjecture:

After a short investigation I have observed that:

A) The sum of consecutive primes could be a perfect power.
Example:    3^13 =  5821+5827+…+7879

B) If the sum of consecutive squared primes is a perfect power, then it is a perfect square.
Example:  586^2 = 41^2+43^2+…+173^2

C)  Not observed a perfect power being the sum of consecutive powered primes to a power higher than two. 

I thus conjecture that:

For the equality:  x^s=p(n)^r+p(n+1)^r+p(n+2)^r+…+p(m)^r

if r and s have following relationship

      r=1 then s>0

      r=2 then 0<s<=2

      r>2 then s=1 

Q:  Find counter examples

Contribution came from Luis Rodríguez

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Luis Rodríguez wrote:

I think the conjecture in its item 3 is
false. Based in the well verified K-tuples
Conjecture, applied in this form:

"If there are infinitely many sets of 5
squared primes that sums a cube then also,
there could be many sets of 5 CONSE-
CUTIVE primes whose squares sum a cube."

(I found hundreds of sets of 5 squared primes
less than 1000 that sum 53^3 , 77^3 or 101^3)

It may be very hard to find an example.

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