if p is prime then 1920 divides $p^4-10p^2+9$ if $p>6$ is prime then 1920 divides $p^4-10p^2+9$
Take out the factors of 1920 which are $$1920=2^7(3)(5)$$
and factor the expression that remains as
$$p^4-10p^2+9=(p-3)(p-1)(p+1)(p+3)$$
and I assumed as p is prime then p-1 and p + 1 is even then it is divisible by $2^7$
but I think it is not correct, can you help me?

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*Since $p$ is prime greater than $3$ then $\gcd(p,3)=1$ and by little Fermat theorem we have $p^3\equiv p\pmod 3$
This allows us to reduce $p^4-10p^2+9\equiv p^2-10p^2+0\equiv -9p^2\equiv 0\pmod 3$

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*Similarly $p^4\equiv 1\pmod 5$
This allows us to reduce $p^4-10p^2+9\equiv 1-0p^2-1\equiv 0\pmod 5$

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*For the divisibility by $2^7$ your work is a good starting point

Let explicit $p=2k+1$ then $p^4-10p^2+9=16(k-1)k(k+1)(k+2)$
This is a multiple of $4$ consecutive numbers, so they belong to the set $\{4n,\ 4n+1,\ 4n+2,4n+3\}$ not necessarily in that order.
The product of the two $4n(4n+2)=16n^2+8n$ is divisible by $8$.
So overall the expression is divisible by $16\times 8=2^7$
