Square in Interval of Primes Denote by $a_n$ the sum of the first $n$ primes. Prove that there is a perfect square between $a_n$ and $a_{n+1}$, inclusive, for all $n$.
The first few sums of primes are $2$, $5$, $10$, $17$, $28$, $41$, $58$, $75$. It seems there is a perfect square between each pair of successive sums. In addition, we can put a bound on $a_n$, namely $a_n \le 2+3+5+7+9+11+...+(2n+1)=n^2+2n+5$.

(waving my hands frantically to keep me aloft:)
$a_n
\approx \sum_{k=1}^n k\ln k
\approx \int_{1}^n k \ln k\ dk
\approx n^2\ln n
$
and
$a_{n+1}-a_n
\approx n\ln n
$
so if
$b_n
= \lfloor \sqrt{a_n}\rfloor
$,
$b_n
\approx n\sqrt{\ln n}
$.
The next square
after $b_n^2$
is 
$(b_n+1)^2
=b_n^2+2b_n+1
\approx b_n^2+2n\sqrt{\ln n}
$
and since
$a_{n+1}-a_n
\approx n\ln n
$,
$(b_n+1)^2
< a_{n+1}
$
for large anough $n$.
Note that
$(b_n+k)^2
=b_n^2+2kb_n+k^2
\approx b_n^2+2kn\sqrt{\ln n}
$,
so,
by choosing
$k \approx \frac12 \sqrt{\ln n}$, 
it looks like
there are about
$\frac12 \sqrt{\ln n}$
squares between
$a_n$ and
$a_{n+1}$.
