Let $A_n$ be the sum of the first $n$ prime numbers. Prove that there is a perfect square between $A_n$ and $A_{n+1}$.
This is how I prove the conjucture:
Using Proof by contradiction, suppose there is no perfect square between A_n and A_(n+1).
It implies that k^2 < A_n and (〖k+1)〗^2>A_(n+1) for any natural number k.
Let A_n=p_1+p_2+p_3+⋯+p_n
A_(n+1)=p_1+p_2+p_3+⋯+p_(n+1)
The difference between (〖k+1)〗^2and k^2 is 2k+1 and the difference between A_n and A_(n+1)is p_(n+1), thus p_(n+1)<2k+1.
Since k^2= 1+3+5+…+(2k-1), thus p_n ≥ 2k-1 and p_(n+1) ≥ 2k+1. It then contradicts p_(n+1)<2k+1.
Therefore there is a perfect square between A_n and A_(n+1)where A_n is the sum of the first n prime numbers.
Best Answer
This statement appears to be true, and could perhaps be proved with fairly basic analytic estimates, but it is by no means an elementary question. Here is data up to the point where the sum of consecutive primes exceeds 10,000. Note that, when the sum is exactly 100, the square 81 could have been chosen instead. Posted with CW status as I voted to close the question.
The thing would appear to follow from effective bounds on the size of the $n$-th prime in Rosser and Schoenfeld (1962) available online. The resulting inequality confirms the statement for a specific lower bound $n \geq N,$ and a computer run confirms it below $n.$ I expect i will figure out $N$ tomorrow, it is not immediate...