def get_factors(n): returnlist = [] for i in range(1,n+1): if (n/i)%1 == 0: returnlist.append(i) return(returnlist) for i in range(1,101): if len(get_factors(i**2-1)) == i: print(i**2-1)def get_factors(n): returnlist = [] for i in range(1,n+1): if (n/i)%1 == 0: returnlist.append(i) return(returnlist) for i in range(1,101): if len(get_factors(i**2-1)) == i: print(i**2-1)RunResetPop Out /

ihatemath123 wrote: def get_factors(n): returnlist = [] for i in range(1,n+1): if (n/i)%1 == 0: returnlist.append(i) return(returnlist) for i in range(1,101): if len(get_factors(i**2-1)) == i: print(i**2-1)def get_factors(n): returnlist = [] for i in range(1,n+1): if (n/i)%1 == 0: returnlist.append(i) return(returnlist) for i in range(1,101): if len(get_factors(i**2-1)) == i: print(i**2-1)RunResetPop Out / It looks like you really do hate math

steppewolf wrote: ihatemath123 wrote: def get_factors(n): returnlist = [] for i in range(1,n+1): if (n/i)%1 == 0: returnlist.append(i) return(returnlist) for i in range(1,101): if len(get_factors(i**2-1)) == i: print(i**2-1)def get_factors(n): returnlist = [] for i in range(1,n+1): if (n/i)%1 == 0: returnlist.append(i) return(returnlist) for i in range(1,101): if len(get_factors(i**2-1)) == i: print(i**2-1)RunResetPop Out / It looks like you really do hate math that comment made my day

I like this! First observe that $n = k^2 - 1$ for some positive integer $k$. As $n$ is not a perfect square, $\tau(n) = k$ must be even, ergo $n$ is odd. We now establish a bound on $\tau(n)$. Recall that, since $n$ is not a perfect square, there exists a bijective correspondence between factors of $n$ greater than $\lfloor\sqrt n\rfloor = k-1$ and factors of $n$ at most $\lfloor\sqrt n\rfloor$. Furthermore, since $n$ is odd, all divisors of $n$ must also be odd. Therefore \[ k = \tau(n) \leq 2\cdot |\{1,3,\ldots, k-1\}| = 2\cdot\tfrac k2 = k. \]Hence equality must hold, and so every odd number between $1$ and $k-1$ must divide $n = k^2 - 1$. In particular, $k-3$ must divide $k^2 - 1$; but \[ \frac{k^2 - 1}{k-3} = k + 3 + \frac{8}{k - 3}. \]Therefore, $k - 3$ is either $-1$ or $1$, so $k$ equals either $2$ or $4$ and $n\in\boxed{3,15}$. Both of these work.

Can't you have more like $24, 35, 48, 63, 80, 99?$

@above Those are all integers such that $\sqrt{n+1}$ is an integer, but they do not have $\sqrt{n+1}$ divisors.

if $n=\prod_{i=1}^{r} p_i^{x_i}$ then we have $\prod_{i=1}^{r}(x_i+1)$ divisors of $n$ . now as $n$ has total of $\sqrt{n+1}$ divisors so $(\prod_{i=1}^{r}(x_i+1))^2 -1= \prod_{i=1}^{r} p_i^{x_i}$ first observe if Case 1-: $p_i=2$ then $n=2^y$ for some $y\in N$ then $(y+1)^2-1=2^y$ which does not have solution Case 2-: now if $x_i\geq 2$ and $p_i >2$ then $(\prod_{i=1}^{r}(x_i+1))^2<\prod_{i=1}^{r} p_i^{x_i}$ so no solution in this case also case 3-: if $x_i<2$ or $x_i=1$ then $4^{r} -1=\prod_{i=1}^{r} p_i^{x_i}$ now if $p_i>4$ or $p\neq 3$ then it has no solution. but if $p=3$ then it has solutrion at $r=1$. now taking $p_1=3$ as $x_i=1$ so let $n=3\prod_{i=2}^{r}p_i$ if $r>3$ then $\{n\}_{min}=3*5*7$ and $\{n\}_{min}>4^3-1$ so for $r>3$ there is no solution. if $r=2$ then $4^2-1=3*p_2 \implies p_2=5$ hence at $r=2$ we get one more solution $n=15$ so there are 2 solution $\boxed{n=3,15}$.

Write $n = k^2 - 1$ (since else $\sqrt{n + 1}$ is not an integer). Note that this is not a square, so $k$ must be even, so $n$ is odd. Note that half of its divisors are less than $k > \sqrt{n}$, and all of its divisors are odd, so all the odd positive integers $1, 3, \dots, k - 1$ divide $k^2 - 1$, so $k - 3$ divides $k^2 - 1 = (k - 3)(k + 3) + 8 \implies (k - 3) \mid 8 \implies k \in \{1, 2, 4, 5, 7, 11\}$ - but $k$ is even so $k \in \{2, 4\}$ which can be checked to work.