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 $a_n \le 2+3+5+7+9+11+...+(2n+1)=n^2+2n+5$.
 A: Let $p$ be the $(n+1)$st prime. Then $a_n\le (-8)+(1+3+5+7+9+\ldots +p-2)=(\frac{p-1}{2})^2-8$ provided $p\ge 11$ (the smaller cases can be dealt with by checking manually). So with $b_n:=\lfloor \sqrt {a_n}\rfloor $ we have $b_n< \frac{p-1}{2}$. Then $a_n<(b_n+1)^2=b_n^2+2b_n+1< a_n+p=a_{n+1}$ as was to be shown. 
A: (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
$a_{n+1}$.
A: This result follows from the well established fact that all primes numbers, with the exception of the first one which is$~2$, are odd.
Observe that for any positive integer$~n$, (a) the sum of all positive odd numbers less than $2n$ is $n^2$, and (b) the first gap of $2n$ successive natural numbers containing no perfect square is the range $n^2+1,\ldots,n^2+2n$.
I prove a stronger statement: for any set $S$ of positive odd numbers that includes $1$, if $2n-1$ is its greatest element, there is a perfect square among the $2n$ numbers $a-(2n-1),\ldots,a$ where $a=1+\sum S$.
(Apply this to the sequence of prime numbers after subtracting $1$ from the initial prime number $2$, which subtraction is compensated by the additional term $1$ in the expression for $a$.)
By the above observation (b) it suffices to show that $a-(2n-1)\leq n^2$, and since $n\geq1$ follows from $\sum S\leq n^2$ obtained from observation (a) (the sum of a subset of positive numbers is at most the sum of all).
