2. What is the greatest common factor of all integers of the form $p^{4}-1$, where $p$ is a prime number greater than 5 ?
2. What is the greatest common factor of all integers of the form $p^{4}-1$, where $p$ is a prime number greater than 5 ?
Solution: Let $f(p)=p^{4}-1=(p-1)(p+1)\left(p^{2}+1\right)$. Note that $f(7)=2^{5} \cdot 3 \cdot 5^{2}$ and $f(11)=2^{4} \cdot 3 \cdot 5 \cdot 61$. We now show that their greatest common factor, $2^{4} \cdot 3 \cdot 5$, is actually the greatest common factor of all numbers $p^{4}-1$ so described.
- Since $p$ is odd, then $p^{2}+1$ is even. Both $p-1$ and $p+1$ are even, and since they are consecutive even integers, one is actually divisible by 4 . Thus, $f(p)$ is always divisible by $2^{4}$.
- When divided by $3, p$ has remainder either 1 or 2 .
- If $p \equiv 1$, then $3 \mid p-1$.
- If $p \equiv 2$, then $3 \mid p+1$.
Thus, $f(p)$ is always divisible by 3 .
- When divided by $5, p$ has remainder $1,2,3$ or 4 .
- If $p \equiv 1$, then $5 \mid p-1$.
- If $p \equiv 2$, then $p^{2}+1 \equiv 2^{2}+1=5 \equiv 0$.
- If $p \equiv 3$, then $p^{2}+1 \equiv 3^{2}+1=10 \equiv 0$.
- If $p \equiv 4$, then $5 \mid p+1$.
Thus, $f(p)$ is always divisible by 5 .

Therefore, the greatest common factor is $2^{4} \cdot 3 \cdot 5=240$.