Summation of primes

Question

Let $P(n)=\displaystyle\sum^{n}_{i=1}p_i$, where $p_i=i^{th}$ prime $\geq 2$. Does exist $k\in \mathbb{N}$, such that, $P(k)$ and $P(k+1)$, both are squares?

Please provide any clue.

Answer 1

No, there is not such $k$.

If such integer $k$ exists then there are positive integers $a$ and $b$ with $a>b$ such that $$p_{k+1}=P(k+1)-P(k)=a^2-b^2=(a+b)(a-b)\implies a-b=1,\; a+b=p_{k+1}.$$ Hence $b=(p_{k+1}-1)/2$ and $$\sum_{i=1}^{k}p_i=P(k)=b^2=\left(\frac{p_{k+1}-1}{2}\right)^2.$$ Now we show that the above equality does not hold for any positive integer $k$.

For $n=1,2,3$, we easily verify that $$\sum_{i=1}^{n}p_i>\left(\frac{p_{n+1}-1}{2}\right)^2.$$ Moreover, we show by induction that for $n\geq 4$, $$\sum_{i=1}^{n}p_i<\left(\frac{p_{n+1}-1}{2}\right)^2.$$ For $n=4$ it holds: $2+3+5+7=17<25=\left(\frac{11-1}{2}\right)^2$.

Inductive step: we have that $$\sum_{i=1}^{n+1}p_i=\sum_{i=1}^{n}p_i+p_{n+1}<\left(\frac{p_{n+1}-1}{2}\right)^2+p_{n+1}=\left(\frac{p_{n+1}+1}{2}\right)^2\leq \left(\frac{p_{n+2}-1}{2}\right)^2$$ where the first (strict) inequality is the inductive hypothesis and the last inequality holds because $p_{n+2}-p_{n+1}\geq 2$. Hence $$\sum_{i=1}^{n+1}p_i< \left(\frac{p_{n+2}-1}{2}\right)^2$$ and we are done.