 Solution

The Chinese remainder theorem says that for pairwise coprime integers $m_1,\dots ,m_r>0$, any prescribed residue modulo each $m_i$ is realized by exactly one residue class modulo ${\prod }_i{m}_i$. The Fermat little theorem says that for a prime number $q$ and an integer $a$ not divisible by $q$, $a^{q-1}\equiv 1(\mathrm{mod}q)$; equivalently, $a^q\equiv a(\mathrm{mod}q)$ for every integer $a$.

For the given prime number, write $p=2k+1$. Its square is $p^2=1+4k(k+1)=1+8t$ for some integer $t$, since one of two consecutive integers is even. Therefore $p^4=(1+8t{)}^2\equiv 1(\mathrm{mod}16)$. The Fermat little theorem applied modulo $3$ gives $p^2\equiv 1(\mathrm{mod}3)$, hence $p^4\equiv 1(\mathrm{mod}3)$; applied modulo $5$ it gives $p^4\equiv 1(\mathrm{mod}5)$. These applications are valid because $p>5$.

The three moduli $16,3,5$ are pairwise coprime integers and have product $240$. By the Chinese remainder theorem,