Floored by Primes

By taking modulo $5$ , it's easy to see that $N^2 \equiv 0, 1, -1 \pmod{5}$ . Then , since $\lfloor \frac{N^2}{5} \rfloor = p$ , where p is prime number. Implies : $p \le \frac{N^2}{5} < p+1 \implies 5p \le N^2 < 5p + 5$ . So, we have 3 cases for N. $$ Case 1 : $N^2 = 5p$ Since $p$ is prime number. And $5|N^2 \implies 25|N^2 \implies 25|5p \implies 5|p .$ So, $p = 5$ . For this case, we obtain 1 solution. $$ Case 2 : $N^2 = 5p + 1$ Rewrite them as $N^2 - 1 = (N - 1)(N+1) = 5p$ . So, the possible value of $N - 1$ are $1, 5, p,$ and $5p$ . We divide into $4$ subcase.

Subcase 1 : $N - 1 = 1 \implies 5p = N + 1 = 3$ , not satisfied Subcase 2 : $N - 1 = 5 \implies p = N+1 = 5 + 2 = 7$ Subcase 3 : $N - 1 = p \implies N + 1 = 5 \implies N = 4$ , and $p = N - 1 = 3$ Subcase 4 : $N - 1 = 5p \implies N + 1 = 1 \implies N = 0$ , not satisfied

So, for this case, there are $2$ solution. That is $p = 7$ and $p = 3$ . $$ Case 3 : $N^2 = 5p + 4$ Rewrite them as $N^2 - 4 = (N - 2)(N+2) = 5p$ . So, the possible value of $N - 2$ are $1, 5, p,$ and $5p$ . We divide into $4$ subcase.

Subcase 1 : $N - 2 = 1 \implies 5p = N + 2 = 5 \implies p = 1$ , not possible Subcase 2 : $N - 2 = 5 \implies p = N+2 = 5 + 2 + 2 = 9$ , not satisfied Subcase 3 : $N - 2 = p \implies N + 2 = 5 \implies N = 3$ , and $p = N - 2 = 1$ , not satisfied Subcase 4 : $N - 2 = 5p \implies N + 2 = 1 \implies N = -1$ , not satisfied

So, for this case, there areno solution.

So, total from case $1, 2,$ and $3$ we have $1 + 2 + 0 = \boxed{3}$ solution.

Suppose $\left\lfloor \dfrac{N^2}{5} \right\rfloor = p$ for some prime $p$ . Then, $p \le \dfrac{N^2}{5} < p+1$ , and so, $5p \le N^2 < 5p+5$ .

Since perfect squares must be $0,1,4 \pmod{5}$ we only have $3$ cases:

Case 1: $N^2 = 5p$ . For $5p$ to be a perfect square, we must have $5 \mid p$ . Since $p$ is prime, we get $p = 5$ . Thus, the only solution in this case is $N = 5$ .

Case 2: $N^2 = 5p+1$ . For $5p = N^2-1 = (N-1)(N+1)$ we must have $N-1 = 1, 5, p, 5p$ and $N+1 = 5p, p, 5, 1$ respectively. So, $N = 2 = 5p-1$ , or $N = 6 = p-1$ , or $N = p+1 = 4$ , or $N = 5p+1 = 0$ . The first and last of these are impossible. The 2nd and 3rd yield $p = 7$ and $p = 3$ which are both prime. Thus, the solutions in this case are $N = 4, 6$ .

Case 3: $N^2 = 5p+4$ . For $5p = N^2-4 = (N-2)(N+2)$ we must have $N-2 = 1, 5, p, 5p$ and $N+2 = 5p, p, 5, 1$ respectively. So, $N = 3 = 5p-2$ , or $N = 7 = p-2$ , or $N = p+2 = 3$ , or $N = 5p+2 = -1$ . The first and last of these are impossible. The 2nd and 3rd yield $p = 9$ and $p = 1$ which are both not prime. Thus, there are no solutions in this case.

Therefore, the possible values of $N$ are $4,5,6$ , a total of $3$ values.

If $\left\lfloor \dfrac{n^2}{5} \right\rfloor = p$ for some prime $p$ , then we must have that $p \leq \dfrac{n^2}{5} < p+1 \implies 5p \leq n^2 < 5p+5$

Then clearly $n^2$ must be one of $5p, 5p+1, 5p+2, 5p+3, 5p+4$ . However, since $2, 3$ are not quadratic residues modulo $5$ , we have only three cases:

Case 1: $n^2 = 5p$ Then we have that $5|n^2 \implies 5|n \implies 25 | n^2 = 5p \implies 5|p \implies 5=p$ . Then $n^2=25 \implies n=\boxed{5}$ .

Case 2: $n^2=5p+1$ Then we have that $n^2-1 = 5p \implies (n-1)(n+1) = 5p$ . Since clearly $p \neq 5$ , we have either $p<5$ or $p>5$ .

If $p<5$ , then $n-1=p, n+1=5 \implies n=\boxed{4}, p=3$ .

If $p>5$ , then $n-1=5, n+1=p \implies n=\boxed{6}, p=7$ .

Case 2: $n^2=5p+4$ Then $n^2-4 = (n-2)(n+2) = 5p$ . Clearly $5 \neq p$ , so we have similar two subcases:

If $p<5$ then $n-2=p, n+2=5 \implies n=3, p=1$ , but $1$ is not a prime so this case fails.

If $p>5$ then $n-2=5, n+2=p \implies n=7, p=9$ but $9$ is not a prime so this case fails.

Thus, the only values of $n$ are $4, 5, 6$ yielding $\boxed{3}$ values of $n$ .

First we consider the case when N=5k. So we have N^2/5=5 k^2 wich is a prime number if and only if k=1. Now let us consider the general case N=5k+t, t={1,2,3,4} N^2/5=5 k^2+2k t+(t^2)/5. If t={1,2} the number what we want will be k (5k+2t) wich will ber prime if and only if k=1 and t=1. If t=3 we have 5 k^2+6k+1=(5k+1)(k+1) wich is never a prime number. If t=4 we have 5 k^2+8k+3=(5k+3)(k+1) wich is a prime number if and only if k=0. Answer:3 (N={4,5,6})

Let M be the integer part of N^2/5. Consider the different remainders N can have modulo 5. (1) If N=5a, then M=5a^2, which is only prime when a=1, so N=5. (2) If N=5a+1, then M=(5a + 2)a, which is only prime when a=1, so N=6. (3) If N=5a+2, then M=(5a + 4)a, which is never prime. (4) If N=5a-1, then M=(5a - 2)a, which is only prime when a=1, so N=4. (5) If N=5a-2, then M=(5a - 4)a, which is never prime. Thus there are 3 values of N - 4,5,6.

Consider $N^2$ is $5 * p + t$ , where t is {0, 1, 2, 3, 4} , and p is prime

first check for p = 2 (only even prime number) $N^2 = 10 + t$ (not possible)

last digit of $N^2$ can be {0,1,4,9,6,5} and last digit of 5*p can be 5 hence there are only three values of t possible, which are {0,1,4}

case 1: t = 0 $N^2 = 5 * p$ $\Rightarrow$ N = 5 and P = 5

case 2: t = 1 $N^2 = 5 * p + 1$ $\Rightarrow$ $(N - 1) * (N + 1) = 5 * p$ $\Rightarrow$ N = 4 , p=3 and N = 6 , p = 7

case 3: t = 4 $N^2 = 5 * p + 5$ $\Rightarrow$ $(N - 2) * (N + 2) = 5 * p$ $\Rightarrow$ N = 7, p not possible and N = 3 , p not possible

Hence there are total 3 solutions for N = 4 , 5 , 6 ans = 3

By taking modulo 5, it's easy to see that $N^{2}\equiv 0,1,-1(mod5)$ . Then , since $\left \lfloor \frac{N^{2}}{5} \right \rfloor$ , where p is prime number. Implies : $p\leq \frac{N^{2}}{5}<p+1 \rightarrow 5p \leq N^{2} < 5p+5$ . So, we have 3 cases for N.

Case 1 : $N^{2}=5p$ Since $p$ is prime number. And

$5 \mid N^{2} \rightarrow 25\mid N^{2} \rightarrow 25\mid 5p \rightarrow 5\mid p$

So, $p=5$ ,For this case we can get 1 solution

Case 2: $N^{2} =5p+1$ Rewrite them as $N^{2} -1=(N-1)(N+1) =5p$ .So, the possible value of $of N-1$ are 1,5,p and 5p. we divide into 4 subcase. Subcase 1: $N-1=1 \rightarrow 5p=N+1=3$ (not satisfied) subcase 2 : $N-1=5 \rightarrow p=N+1=5+2=7$ subcase 3 : $N-1=p \rightarrow N+1=5 \rightarrow N=4$ ,and $p=N-1=3$ subcase 4: $N-1=5p \rightarrow N+1=1 \rightarrow N=0$ (not satisfied)

For this case there are 2 solution .that is $p=7\quad and\quad p=3$

Case 3 : $N^{2}=5p+4$ Rewrite them as $N^{2}-4=(N-2)(N+2)=5p$ . So, the possible value of $N-2$ are 1,5,p, and 5p. We divide into 4 subcase.

Subcase 1 : $N-2=1\rightarrow 5p=N+2=5\rightarrow p=1$ ,(not possible) Subcase 2 : $N-2=5\rightarrow p=N+2=5+2+2=9$ ,(not satisfied) Subcase 3 : $N-2=p\rightarrow N+2=5\rightarrow N=3,\quad and\quad p=N-2=1$ (not satisfied) Subcase 4 : $N-2=5p\rightarrow N+2=1\rightarrow N=-1$ (not satisfied)

So, for this case, there are no solution.

So, total from case 1,2, and 3 we have 3 solution.

Let $n=5k+i$ , for $i\in\lbrace0,1,2,3,4\rbrace$ . Let $f(n) = \lfloor n^2/5 \rfloor$ .

$f(5k) = 5k^2$ , which is only prime for k=1.

$f(5k+1) = 5k^2+2k = k(5k+2)$ , which is only prime for k=1.

$f(5k+2) = 5k^2+4k = k(5k+4)$ , which is never prime.

$f(5k+3) = 5k^2+6k+1 = (5k+1)(k+1)$ , which is never prime.

$f(5k+4) = 5k^2+8k+3 = (5k+3)(k+1)$ , which is only prime when k=0.

So, there are only 3 values of n for which f(n) is prime: 4, 5, and 6.

A perfect square can only have a remainder of 0,1 or 4 considering modulo 5. Case 1: N^2=5p, p is a prime number. it is easy to see that p is a multiple of 5, which leads to p=5 since p is a prime number. This gives N=5. Case 2: N^2=5p+1, p is a prime number. Then 5p=(N-1)(N+1). Hence (N-1,N+1)=(1,5p),(p,5) or (5,p) This gives p=3 or p=7, then N=4 or N=6. Case 3: N^2=5p+4, p is a prime number. Then 5p=(N-2)(N+2). Hence (N-2,N+2)=(1,5p),(p,5) or (5,p) This gives no prime number solution, so there is no solution this case. Therefore the answer is 3.

Suppose , $N = 5x +k$ where $k∈$ { $0,\pm 1,\pm 2$ }. So, $\left \lfloor \frac{N^2}{5} \right \rfloor = \left \lfloor \frac{25x^2 + 10xk + k^2}{5} \right \rfloor = \left \lfloor 5x^2 + 2xk +\frac{ k^2}{5} \right \rfloor = x(5x +2k)$ (Because $k^2 < 5$ )

Let's assume $\lfloor \frac{N^2}{5}\rfloor=P$ where $P$ is a prime.

Notice that if $N^2=5P+r$ where $0\leq r<5$ [ $r$ is obviously an integer],

$\lfloor \frac{N^2}{5}\rfloor=\lfloor \frac{5P+r}{5}\rfloor=\lfloor \frac{5P}{5}+\frac{r}{5}\rfloor=\lfloor \frac{5P}{5}\rfloor+\lfloor \frac{r}{5}\rfloor$

[Since $\frac {5P}{5}$ is an integer, we can 'break up' the floor function].

$=\lfloor P\rfloor+0=P$ .

Notice that any perfect square has a units digit of $0, 1, 4, 5, 6$ or $9$ . Another way of saying that is any perfect square can be written as $5x$ , $5x+1$ or $5x+4$ . That means we have three values of $r$ which are $0$ , $1$ & $4$ . That gives us three cases to consider.

Case $1$ : $N^2=5P$

For $5P$ to be a perfect square, $5$ has to be a factor of $P$ . Since $P$ is prime, the only prime number that has $5$ as its factor is $5$ .

That means $P=5$ and $N=\sqrt{25}=5$ .

So here the only possible value of $N$ is $5$ .

Case $2$ : $N^2=5P+1$

That would imply $N^2-1=5P \Rightarrow (N+1)(N-1)=5P$ .

The only factors of $5P$ are $1, 5, P$ and $5P$ . And for any $N$ , $(N+1)>(N-1)$ .

So the possible combinations of $(N+1, N-1)$ are $(5P, 1), (P,5), (5,P)$ . The first combination is impossible because it would mean $3=5P$ .

The second and the third combination imply $P=7$ & $P=3$ respectively. Both of them are primes and therefore valid.

$P=7$ gives us $N=\sqrt{5\times7+1}=6$ .

$P=3$ gives us $N=\sqrt{5 \times 3+1}=4$ .

So Case $2$ gives us two values of $N$ . They are $4$ and $6$

Case $3$ : $N^2=5P+4$

This case tells us $N^2-4=5P \Rightarrow (N+2)(N-2)=5P$ . We can use the similar reasoning from Case $2$ to determine that the possible combinations of $(N+2, N-2)$ are ( $5P, 1),(5,P), (P,5)$ .

The first and second combinations give us $P=1$ . The last one gives $P=9$ . None of these are primes.

There are no more cases to consider. So $N=5, 6, 4$ which gives us a total of $3$ positive integers.

It is always possible to write $N = 5m+k$ with $k \in \{-2,-1,0,1,2\}$ . We find

$N^2 = (5m+k)^2 = 25m^2 + 10mk + k^2$ .

Since $k^2 < 5$ , it follows that

$\lfloor \frac15 N^2 \rfloor = \lfloor 5m^2 + 2mk + \frac15 k^2 \rfloor = 5m^2 + 2mk = m(5m+2k)$ .

It is now easy to see that $\lfloor \frac15 N^2 \rfloor$ can only be prime for $m = 1$ . This leaves us only five numbers to check, of which three ( $N=4,N=5,N=6$ ) turn out to give a prime.

From the definition of the mod operator, we know that $a \bmod b = a - b\lfloor \frac{a}{b}\rfloor$ . Thus, $\lfloor \frac{N^2}{5} \rfloor = \frac{N^2}{5} - \frac{1}{5}({N^2} \bmod 5)$

For $N = 5k+1$ :

$\frac{N^2}{5} - \frac{1}{5}({N^2} \bmod 5) = 5k^2 + 2k = k(5k+2)$ , which is only prime for $k = 1$ , which provides our first solution, $\lfloor \frac{N^2}{5} \rfloor = 7 \Rightarrow N = 6$

For $N = 5k+2$ :

$\frac{N^2}{5} - \frac{1}{5}({N^2} \bmod 5) = 5k^2 + 4k = k(5k+4)$ , which is never prime (even $k = 1$ still provides $\lfloor \frac{N^2}{5} \rfloor = 9$ ).

For $N = 5k+3$ :

$\frac{N^2}{5} - \frac{1}{5}({N^2} \bmod 5) = 5k^2+6x+1 = (k+1)(5k+1)$ , which is also never prime.

For $N = 5k+4$ :

$\frac{N^2}{5} - \frac{1}{5}({N^2} \bmod 5) = 5k^2 + 8k + 3 = (k+1)(5k+3)$ , which is only prime for $k = 0$ , which provides our second solution, $\lfloor \frac{N^2}{5} \rfloor = 3 \Rightarrow N = 4$

For $N = 5k$ :

$\frac{N^2}{5} - \frac{1}{5}({N^2} \bmod 5) = 5k^2$ , which is only prime for $k = 1$ , which provides our third and final solution, $\lfloor \frac{N^2}{5} \rfloor = 5 \Rightarrow N = 5$

Let $N=5q+r$ , $0\le r\le 4$ and $f(N)=\lfloor\frac{N^2}{5}\rfloor$

Thus we have $N^2=25q^2+10qr+r^2$ , where $f(N)=\lfloor\frac{N^2}{5}\rfloor=5q^2+2qr+\lfloor r^2\rfloor$ .

We have two cases to analyze:

$q=0$ , where $N=r=1, 2, 3, 4$ and it's easy to see that $\lfloor\frac{N^2}{5}\rfloor$ is a prime number only for $N=4$ .

$q>0$

If $r=0$ , we have $f(N)=5q^2$ and $q^2|f(N)$ , a prime number. Then $q=1 \Rightarrow f(N)=5$ and $N=5$ it's a solution.

If $r=1$ , we have $f(N)=5q^2+2q=q(5q+2)$ and if $q>1$ , we have $1<q<5q+2$ , contradiction with $f(N)$ to be a prime number, where $q=1 \Rightarrow f(N)=7$ and $N=6$ it's a solution.

If $r=2$ , we have $f(N)=5q^2+4q$ , and by a argument like the previous, we have $q=1\Rightarrow f(N)=9$ that it's not a prime number. So we don't have any solution.

If $r=3$ , we have $f(N)=5q^2+6q+1=(5q+1)(q+1)$ and $q+1, 5q+1>1$ . So $f(N)$ cannot be a prime number.

If $r=4$ , we have $f(N)=5q^2+8q+3=(5q+3)(q+1)$ and $q+1, 5q+3>1$ . So $f(N)$ cannot be a prime number.

Thus we have $3$ values: $N=4, 5, 6$ .

Note that $\begin{aligned} \left\lfloor\frac{(5n-2)^2}5\right\rfloor&=\left\lfloor\frac{25n^2-20n}5+\frac45\right\rfloor=(5n-4)n\\ \left\lfloor\frac{(5n-1)^2}5\right\rfloor&=\left\lfloor\frac{25n^2-10n}5+\frac15\right\rfloor=(5n-2)n\\ \left\lfloor\frac{(5n)^2}5\right\rfloor&=\left\lfloor\frac{25n^2}5\right\rfloor=(5n)n\\ \left\lfloor\frac{(5n+1)^2}5\right\rfloor&=\left\lfloor\frac{25n^2+10n}5+\frac15\right\rfloor=(5n+2)n\\ \left\lfloor\frac{(5n+2)^2}5\right\rfloor&=\left\lfloor\frac{25n^2+20n}5+\frac45\right\rfloor=(5n+4)n \end{aligned}$ Hence for all $n\geq2$ and $N\in\{5n-2,5n-1,5n,5n+1,5n+2\}$ , $\left\lfloor\frac{N^2}5\right\rfloor$ can be factored as a product of two numbers, each greater than 2 (since $5n-4\geq2$ ). Thus $\left\lfloor\frac{N^2}5\right\rfloor$ is composite for all $N\geq8$ .

Checking for $N\in\{1,2,\ldots,7\}$ , we see that $\left\lfloor\frac{N^2}5\right\rfloor$ is prime if and only if $N=4,5,6$ . Hence there are $\boxed{3}$ values of $N$ such that $\left\lfloor\frac{N^2}5\right\rfloor$ is prime.

Lemma: For all $x \ge 2 )\ , \lfloor{\frac{(5x+m)^2}{5}}\rfloor$ where $m \in {-2, -1, 0, 1, 2}$ is divisible by $k$ .

It suffices to show that $(5x+m)^2 \cong n (mod 5)$ where $m \in {-2, -1, 0, 1, 2}$ and $n \in {0, 1, 2, 3, 4}$ because that would mean that $\lfloor{\frac{(5x+m)^2}{5}}\rfloor \cong 0 (mod n)$

Note that $(5x+m)^2=25x^2+5xm+m^2 \cong 0+0+m^2 \cong m^2 (mod 5x)$ . Since $m \in {-2, -1, 0, 1, 2}$ , $m^2 \cong q (mod 5x)$ where $q \in {0,1,4}$ . Since all $0,1,4$ is a subset of $0, 1, 2, 3, 4$ , $(5x+m)^2 \cong n (mod 5)$ where $m \in {-2, -1, 0, 1, 2}$ and $n \in {0, 1, 2, 3, 4}$ .

Let $f(x)=\lfloor{\frac{N^2}{5}}\rfloor$

Thus, $f(8)$ , $f(9)$ , $f(10)$ , $f(11)$ and $f(12)$ are divisible by $2$ ; $f(13)$ , $f(14)$ , $f(15)$ , $f(16)$ , $f(17)$ are divisible by $3$ , and so on. Note also that it is obvious that f(5x-2)>x, so none of f(x) where $x \ge 8$ is prime. Hence we just have to test the cases where $x \le 7$

Here are the values of $x$ and $f(x)$ where $x \le 7$ .

For $x=1, 2, 3, 4, 5, 6, 7$ , $f(x)=0, 0, 1, 3, 5, 7, 9$ respectively.

Out of the $f(x)$ , only $f(4)=3$ , $f(5)=5$ , and $f(6)=7$ are prime.

Hence, for $\boxed{3}$ positive integers $N$ , $\lfloor{\frac{N^2}{5}}\rfloor$ is prime.

The given equation is $p=\left\lfloor\frac{N^2}{5}\right\rfloor$ To remove the floor (greatest integer) function, we can make it so that $\frac{N^2}{5}$ is always an integer, and we can do that by subtracting a variable, say $\alpha$ , from the numerator. We know that $\alpha\in\{0,1,2,3,4\}$ since the denominator is 5. After multiplying by 5, our equation becomes $5p=N^2-\alpha$ We can further shrink the possibilities for $\alpha$ by finding the quadratic residues modulo 5. They are 0,1, and 4, so the possibilities for $\alpha$ are 0, 1, and 4. Now, we have 3 cases:

Case 1: $\alpha=0$

We have $5p=N^2$ This implies $5|p$ , and since $p$ is a prime, $p=5$ . Therefore, $N=5$ is our only possibility in this case.

Case 2: $\alpha=1$

We have $5p=N^2-1=(N-1)(N+1)$ The factors of $5p$ are 1,5, $p$ , and $5p$ . Either $N-1$ is 1, or the lesser of 5 and $p$ . If $N-1=1$ , $N=2$ and that doesn't work. Therefore, $N-1=5$ or $N+1=5$ . Checking, we find that both $N=4$ and $N=6$ work, for two possibilities in this case.

Case 3: $\alpha=4$

We have $5p=N^2-4=(N-2)(N+2)$ Similarly to Case 2, $N-2$ is either 1 or the lesser of 5 and $p$ . If $N-2=1$ , $p=1$ , so that doesn't work. Therefore, $N-2=5$ or $N+2=5$ . Checking, neither give $p$ prime, so there are no solutions in this case.

The 3 solutions are $N=4$ , $N=5$ , and $N=6$ .

F = floor( N^2 / 5 )

N = {3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18}

F = {1 3 5 7 9 12 16 20 24 28 33 39 45 51 57 64}

difference between F terms is 2 2 2 2 3 4 4 4 4 5 6 6 6 6 7 8 8 8 8 ... and so on. There are a few cool things about this sequence of differences.

First, after all the 2 2 2 2, we are at 9. We add 3. That is divisible by 3.

Next, we add 1 1 1 1 mod3, then add 2 mod3. That is divisible by 3.

Next, we add 2 2 2 2 mod, then add 1mod3. That is divisible by 3.

Every time we add an odd step up, the result is divisible by 3. Now, let's look at the even patterns.

The 4 4 4 4 just make even numbers. So divisible by 2 The 6 6 6 6 just make numbers divisible by 3. That's still good. The 8 8 8 8 just make even numbers. So divisible by 2 The 10 10 10 10 make numbers divisible by 5 The 12 12 12 12 make numbers divisible by 2. The 14 14 14 14 make numbers divisible by 3.

And the pattern continues. Therefore, the only prime ones occur at n = 4,5,6. All above these are divisible by 2 or 3 or 5.

This is an extremely brute force way of doing it. Sorry, I'm not very good at proving things. I just like patterns. I also think you might want to take away the solutions given at the bottom of the page in the discussion if you ask someone for a submission for points, because it would be very easy to just copy an elegant solution below.

Note that we must have $\left\lfloor \dfrac{N^2}{5} \right \rfloor \ge 2$ or $N \ge 4.$ Now we take cases based on the residue of $N \pmod 5.$

---

• Case 1: $N \equiv 0 \pmod 5$

Let $N = 5a$ for some integer $a.$ Because $N \ge 4,$ $a$ is also positive. Then we have $\left\lfloor \frac{N^2}{5} \right \rfloor = \left\lfloor \frac{25a^2}{5} \right \rfloor = 5a^2.$ This means that $a = 1$ and $N = 5,$ which is the only value found here.

• Case 2: $N \equiv 1 \pmod 5$

Let $N = 5a+1$ (again for some positive integer $a$ ). Then $\left\lfloor \frac{N^2}{5} \right \rfloor = \left\lfloor \frac{25a^2+10a+1}{5} \right \rfloor = 5a^2+2a = a(5a+2).$ For this expression to be prime, one of the factors in the product must equal $1.$ If $a = 1,$ then $N = 6,$ which indeed works. However, $5a+2$ cannot equal $1$ because then $a(5a+2)$ would not be an integer (it is the floor of a real number). So, the only value from this case is $N = 6.$

• Case 3: $N \equiv 2 \pmod 5$

Let $N = 5a+2,$ so that $\left\lfloor \frac{N^2}{5} \right \rfloor = \left\lfloor \frac{25a^2+20a+4}{5} \right \rfloor = 5a^2+4a = a(5a+4).$ $a = 1, N = 7$ doesn't work (it gives $9$ ). Taking $5a+4 = 1$ makes $a(5a+4)$ non-integral, so it doesn't work. Thus, there are no values of $N$ in this case.

• Case 4: $N \equiv 3 \equiv -2 \pmod 5$

Let $N = 5a-2,$ which gives $\left\lfloor \frac{N^2}{5} \right \rfloor = \left\lfloor \frac{25a^2-20a+4}{5} \right \rfloor = 5a^2-4a = a(5a-4).$ Setting either factor equal to $1$ gives $N = 3,$ which contradicts our statement that $N \ge 4.$ So, there are no values in this case either.

• Case 5: $N \equiv 4 \equiv -1 \pmod 5$

If $N = 5a-1,$ then $\left\lfloor \frac{N^2}{5} \right \rfloor = \left\lfloor \frac{25a^2-10a+1}{5} \right \rfloor = 5a^2-2a=a(5a-2).$ If $a = 1,$ then $N = 4,$ which works. It $5a-2 = 1,$ then $a(5a-2)$ is not an integer. Therefore, $N = 4$ is our last value of $N.$

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In conclusion, the only possible values of $N$ are $N = 4, 5, 6,$ so there are $\boxed{3}$ positive integers such that $\left\lfloor \dfrac{N^2}{5} \right \rfloor$ is prime.

Use this simple code in mathematica we see that only numbers 4;5;6 only satisfy the condition. u can change 1000 to any number for your own choice......result would be same For[i = 2, i < 1000, i++, Print[{Element[Floor[i^2/5], Primes], i}]] so answer is 3

We can show that every floor of $\frac{n^{2}}{5}$ is not a prime except for n = 4, 5, and 6.

Of course the most convenient technique used is none other then Trial and Error

We make cases, let N be **Case 1: (5k \Rightarrow

Let $N = 5k + q$ ; $q = 0,1,2,3,4$ ; $k >= 0$ .

If $N = 5k$ , $\frac(N^{2})(5) = 5k^{2}$ . Hence, $k = 1$ or $N = 5$ is an answer. If $N = 5k + 1$ , $\frac(N^{2})(5) = k(5k+2)$ . Hence, $k = 1$ or $N = 6$ is an answer. If $N = 5k + 2$ , $\frac(N^{2})(5) = k(5k+4)$ . Hence, no answer. If $N = 5k + 3$ , $\frac(N^{2})(5) = (k+1)(5k+1)$ . Hence, no answer. If $N = 5k + 4$ , $\frac(N^{2})(5) = (k+1)(5k+3)$ . Hence, $k = 0$ or $N = 3$ is an answer.