Summation of primes 
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.
 A: 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.
