# Find all natural $n$ numbers such that

Find all natural $$n$$ numbers such that $$15(n!)^2+1$$ is divisible by $$2n-3$$. My try: First I assumed $$2n-3$$ is not prime number. Let $$a$$ be divisor of $$2n-3$$. It's clear that $$a, so $$15(n!)^2$$ is divisible by $$a$$. Which means $$15(n!)^2+1$$ is not divisble by $$a$$. But it is given that $$15(n!)^2+1$$is divisble by $$n$$ which means it is divisible by $$a$$ too. But we have already proved it is not. Contradiction! So $$2n-3$$ must be prime number. Now if we change $$2n-3$$ as $$p$$. We can say $$15*((p+3)/2)!*((p+3)/2)!$$ is congruent to $$-1$$ by p module. By Wilson's theorem $$15*((p+3)/2)!*((p+3)/2)!$$ is congruent to $$(p-1)!$$ by p module. From here I don't know how to continue.

• I really like this problem, but your title needs to be more descriptive. Nov 8, 2020 at 8:51

You do need Wilson's Theorem, but observe:

\begin{align}n &\equiv -(n-3) \pmod {2n-3}\\ n-1 &\equiv -(n-2) \pmod {2n-3}\\ &\ \vdots\\ 1 &\equiv -(2n-4)\pmod {2n-3}\end{align}

This gives:

\begin{align}(n!)^2&\equiv n!(-1)^n(2n-4)(2n-5)\dots(n-2)(n-3) \pmod {2n-3}\\&\equiv(-1)^n(2n-4)!(n)(n-1)(n-2)(n-3) \pmod {2n-3}\\\text{(Wilson)} &\equiv(-1)^{n+1}(n)(n-1)(n-2)(n-3)\pmod {2n-3}\\ &\equiv(-1)^{n+1}n^2(n-1)^2 \pmod {2n-3}\end{align}

Now we seperate into cases where $$n$$ is odd or even. For $$n$$ odd satisfying the division condition,

$$15(n!)^2+1\equiv 15n^2(n-1)^2+1 \equiv 0 \pmod {2n-3}$$

$$z = \dfrac {15n^2(n-1)^2+1}{2n-3}$$ is an integer iff $$16z$$ is. Simplifying we have

$$16z=120 n^3 - 60 n^2 + 30 n + 45 + \frac {151} {2 n - 3}$$

So the above is an integer only if $$2n-3$$ divides $$151$$, which is a prime, giving $$2n-3 = 151$$, $$n = 77$$.

The case for $$n$$ even should be similar. Technically, you also need to consider the cases $$2n-3 = \pm 1$$ separately, since $$\pm1$$ are neither prime nor composite.

• From your result of $(n!)^2\equiv (-1)^{n+1}n^2(n-1)^2\pmod {2n-3}$, we can then get that $15(n!)^2+1\equiv 15(-1)^{n+1}n^2(n-1)^2+1\equiv 0\pmod{2n-3}$. From this, we have $15(-1)^{n+1}(2n)^2(2n-2)^2+16\equiv 15(-1)^{n+1}(3)^2(1)^2+16\equiv (-1)^{n+1}135+16\equiv 0\pmod{2n-3}$. With $n$ odd, this gives $151\equiv 0\pmod{2n-3}$. Since $151$ is prime, we have $2n-3=151\;\to\;n=77$, as you've already determined. With $n$ even, we get $-119\equiv 0\pmod{2n-3}$. Since $119=7(17)$, this means $2n-3=7\;\to\;n=5$ or $2n-3=17\;\to\;n=10$. However, $5$ is odd, so only $n=10$ is a solution. Aug 29, 2023 at 21:19