If $2p+1$ is a prime number, then $(p!)^2+(-1)^p$ is divisible by $2p+1$.

Question

I have a question which is as following:

If $2p+1$ is a prime number, then $(p!)^2+(-1)^p$ is divisible by $2p+1$.

Whenever I see questions involving factorial sign, I try to use Wilson theorem, but I don't know weather I can use it here or not, If yes then how??

Please help.

Answer 1

Since it is given that $(2p+1)$ is a prime number. So we can use Wilson theorem for $2p+1$ as:

$$ ((2p + 1)-1)! + 1\equiv 0 \mod(2p + 1) $$

$$(2p)!+1\equiv 0 \mod(2p + 1)$$

$$1\times2\times3\times4\times5\times6....(p)\times(p + 1)\times(p + 2)....(2p - 1)\times(2p) + 1\equiv 0 \mod(2p + 1)$$

$$p! [(2p+1-p)]\times [2p+1-(p-1)]\times[2p+1-(p-2)]\times[2p+1-(1)]+1\equiv 0 \mod (2p+1)$$

$$p![M(2p + 1)+(-1)\times(p)\times(p-1)\times(p-2)....3\times2\times1] +1\equiv 0 \mod(2p + 1)$$

$$p![M(2p + 1)+(-1)^p {p!}] + 1\equiv 0 \mod(2p + 1)$$

$$[p!M(2p + 1)]+[(-1)^p {p!}^2] + 1\equiv 0 \mod(2p + 1)$$

Notice that $[p!M(2p + 1)]$ is already divisible by $(2p + 1)$, So,

$$[(-1)^p {p!}^2] + 1\equiv 0 \mod(2p + 1)$$

Which was needed to be proved.

:) :) :) :)

Answer 2

$p=2$ works. Suppose now $p\ge 3$. Then by Wilson's theorem $$ (p!)^2= (-1)^p p!\cdot (-1)(-2)\cdots (-p)\equiv (-1)^p (2p)!\equiv (-1)^{p+1}\bmod{2p+1}. $$

Answer 3

By Wilson's theorem since $2p+1$ is a prime number, $(2p)!+1$ is divisible by $2p+1$. Put $n$ for $2p+1$, then $(2p)!$ may be written as, $$1\cdot (n-1)\cdot 2 \cdot (n-2) 3\cdot(n-3) \cdots p(n-p)$$ if these factors supposed be multiplied out, it is obvious that we obtain $(-1)^p1^22^23^2\cdots p^2$ together with some multiple of $n$.

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Hence, $1+(-1)^p(p!)^2$ must be divisible by $n$ and therefore $(p!)^2 + (-1)^p$ must be divisible by $n$, i.e, $2p+1$. Hope it helps.