Existence of perfect square between the sum of the first $n$ and $n + 1$ prime numbers

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.