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 conjecture:
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 $$A_n = p_1+p_2+\ldots+p_n$$ $$A_{n+1} = p_1+p_2+\ldots+p_{n+1}$$
The difference between $(k+1)^2$ and $k^2$ is $2k+1$ and the difference between $A_{n+1}$ and $A_n$ is $p_{n+1}$, thus $p_{n+1}<2k+1$.
Since $k^2 = 1+3+5+\ldots+(2k-1)$, we have $p_n ≥ 2k-1$ 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.
