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 $a<n-1$, 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.
Find all natural $n$ numbers such that
divisibilitynumber theory
Best Answer
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.