Square roots in square brackets

There are $2n+1$ terms equal to $n$ for each $n$ , because there are $2n+1$ integers in $[n^2,(n+1)^2)$ . So $y= \sum_{n=1}^{x-1} n(2n+1) = \frac{(x-1)x(2x-1)}3 + \frac{x(x-1)}2 = \frac{x(x-1)(4x+1)}6.$
Clearly $(2,3)$ and $(3,13)$ are solutions, so we need only show that for $x \ge 4$ , $y$ cannot be prime.

Well, at least one of the three terms in product in the numerator is even, and at least one is divisible by 3. Suppose $x \ge 4$ . Call the terms $f_1, f_2, f_3$ . Note that all the $f_i$ are at least $3$ .

If $2|f_1$ and $3|f_1$ , then we can write $y = (f_1/6)f_2f_3$ as a product of three numbers, at least two of which ( $f_2$ and $f_3$ ) are strictly greater than $1$ , so it can't be prime.

If $2|f_1$ and $3|f_2$ , then we can still write $y$ in that way, because we can write $y = (f_1/2)(f_2/3)f_3$ and both $f_1/2$ and $f_3$ are strictly greater than $1$ .

Those are the only two cases (after we reorder the terms appropriately), so $y$ cannot be prime for $x \ge 4$ . So the answer is $2 \cdot 3 + 3 \cdot 13 = \fbox{45}$ .

Note that $\left\lfloor \sqrt{1} \right\rfloor + \cdots + \left\lfloor \sqrt{x^2 -1} \right\rfloor =$ $1 + 1 + 1 + 2 +2 +2 +2+2 + \cdots + \overbrace{(x-1) + (x-1) + \cdots + (x-1)}^{2x+1 \mbox{ times}}$

hence the sum defining $y$ can be easily written as

$y = \sum_{i=1}^{x-1} (2i+1)\cdot i= \sum_{i=1}^{x-1} 2i^2 + i =$ $2\sum_{i=1}^{x-1} i^2 + \sum_{i=1}^{x-1}i = \dfrac{(x-1)x(2x-1)}{3} + \dfrac{(x-1)x}{2}$

that leads to

$6y = x(x-1)(4x+1)$

Now $x$ is prime, and it divides $6y$ so either $x$ divides $6$ , or $x$ divides $y$ .

In the first case $x$ can be $2$ or $3$ . Both values, put in the equation, give a prime number for $y$ .

Namely we have $x=2\ \Rightarrow y = 3$ and $x = 3 \Rightarrow y= 13$ .

In the second case $x$ divides $y$ . But $y$ is prime and this means $x = y$ .

Putting $x = y$ in the equation we have $(x-1)(4x+1) = 6$ that has no integer solutions.

So the only pair of prime solutions are $(2;3)$ and $(3;13)$ and the answer is $45$ .

Terms from floor(x-1) up to floor (sqrt(x^2 -1 ) all have value (x-1)

Such (2x-1) terms are there.

So the expression of y is sum of terms (x-1)(2x-1) for all prime values of x.

For x=2, y=3

For x=3, y=13

All higher values of y are found composite.

So answer is 2x3+3x13 = 45

Someone please provide complete analytical solution.