Problem - 3192
Show that $5\mid 4^{2n}-1$ for $n\ge 1$.
Because $4^{2n}-1\equiv (-1)^{2n} -1 \equiv 1^n -1 \equiv 0\pmod{5}$, therefore the claim holds.
Show that $5\mid 4^{2n}-1$ for $n\ge 1$.
Because $4^{2n}-1\equiv (-1)^{2n} -1 \equiv 1^n -1 \equiv 0\pmod{5}$, therefore the claim holds.