If $p$ is an odd prime, prove that $1^2\cdot3^2\cdot5^2\cdots(p-2)^2 \equiv (-1)^{(p+1)/2} \pmod p$
Then $$1^2\cdot3^2\cdot5^2\cdots(p-2)^2={1/16} (p-2)^2(3p-1)^2\tag{1}$$
Now since $p$ is odd then $(p+1)/2$ is even and $(-1)^{(p+1)/2}=1$. Now I tried substract $1$ from $(1)$ to find out if the result is a multiple of $p$. But I can't get the answer any suggestions.
On
The statement is false. For example, if $p = 5$ then $\frac{p+1}{2} = 3$ and $(-1)^3 = -1$, however $1^2 \times 2^2 \times 3^2 = 36$, and it is not true that $36 \equiv -1 \mod 5$.
For the true statement, consider the group of natural numbers less than or equal to $p$ modulo $p$ by multiplication i.e. $\{1,2,...,p-1\}$ under multiplication modulo $p$.
If we take the product of all the elements, then we see that every element is paired with its inverse except $1$ and $p-1$ (not very difficult to prove), whose product is $-1$ modulo $p$, giving Wilson's theorem.
However, if we exclude $p-1$, then every element is paired with its inverse, except the identity $1$ which does not contribute. Consequently, we get $(p-2)! \equiv 1 \mod p$ and so $((p-2)!)^2 \equiv 1 \mod p$.
For example, take $p=13$,then $11!$ leaves a remainder of $1$, therefore so does its square.
The right statement is that $(p-2)! \equiv 1 \mod p$ always, for $p$ a prime. Note that $0! = 1$ so this applies for $2$ as well.
Multiply both sides by $(p-1)^2$ for the equivalent:
$$1 = (p-1)^2 (-1)^{(p+1)/2} \text{ mod } p$$
where the left hand side was (the product of all nonzero elements modulo $p$)$^2 \equiv^{\star} (-1)^2 = 1$, and the right hand side expands to the product of: $p^2 - 2p + 1$ mod $p \equiv 1$, multiplied by $(-1)^{(p+1)/2}$.
And so we have the equivalent assertion:
$$1 = (-1)^{(p+1)/2} \text{ mod } p$$
which is, as pointed out already, false when $\frac{p+1}{2}$ is odd: for example, when $p=5$.
$^\star$: The asterisk for that particular equivalence is an application of Wilson's Theorem, for which you can find more discussion in (e.g.) MSE 307.