question about sum of primes?

Question

hi there,
i was asked this riddle a dew days ago but could'nt
solve it, can you help:
let P(k) ( K- a natural number) be a function that givven
a natural number k gives the sum of the k-th first primes.
for example: P(3)=2+3+5=10, P(4)=2+3+5+7=17
prove that between the numbers P(k) and P(k+1)
there must be a square integer.
for example: between 10 and 17 there's 16=4^2

Answer 1

How curious, I am creating a program that will test this and will upload it to http://www.poodwaddle.com/primes.html as soon as I am done, about 10 min. OK its uploaded now.

Answer 2

Let's suppose it's false so that we have squares (x+1)^2 at least P(k+1), and x^2 at most P(k). The difference of the squares then 2x+1>= P(k+1)-P(k)=p_(k+1) the (k+1)th prime.
P(k)>=x^2 implies , 2sqrtP(k) +1>=2x+1>=p_(k+1). So
the sum of the first k primes is at least the square of the
quantity, (1/2)(p_(k+1) - 1). Yet by a theorem proven in 1998
by Dusart we have (1/2)(p_k) is at least the average of the first k primes. For k>=7 we actually have,

(p_k -1)^2 / 4 >(kp_k)/2 which is equivalent to p_k>2k+2

true by induction for all k>=7 because once it's true at p_7
it's true from then on with 2k+2 only increasing by 2 for each increase in k. Thus we have a contradiction if k>=7, and the smaller cases of the riddle can be inspected to conclude the proof.

Answer 3

The key to solving this "riddle" is to realise the following fact which is not hard to prove by induction:

The sum of the first n odd integers equals n^2. That is
1=1
1+3 = 2^2
1+3+5 = 3^2, etc.

Now realise that apart from 2 all the primes are odd. So the process of adding primes is similar to just adding odd numbers. Thus there wil be a square integer in between. This is a non rigorous proof. But you should be able to rigorze it with a little thought. Good luck.

Answer 4

look i don't need ur help for math i am so smart math okay
so what ur name? my name is dennis ameyaw