Two Number theory questions in GRE Mathematics Subject Test The first one is if $6x+5y$ is a multiple of $13$, what must $b$ be so that $8x+by$ is also a multiple of $13$? The second one is what is the greatest integer divided $p^{4}-1$ for every prime number $p$ greater than $5$? I get the answer $11$ and $240$ respectively by plugging in different numbers. I wonder is there any regular way to solve them.
 A: For the second part :
Using this (For any prime $p > 3$, why is $p^2-1$ always divisible by 24?), we can write $p^2\equiv 1\pmod{24}$ $\forall p\gt 5$.
 Since , $p\gt 5$ is an odd prime we have, $p^2\equiv 1\pmod 4$. Using this , we can prove that $4\not\mid p^2+1$(you should try it out yourself) . Therefore we have, 
 $$2\mid p^2+1$$.
 Therefore, $$48\mid (p^2+1)(p^2-1)=p^4-1$$. Also, note that $p\gt 5 \implies (p,5)=1$. By FLT we have , $$p^4\equiv 1\pmod 5$$. This implies ,$$5\mid p^4-1$$. Since, $(48,5)=1$, this implies $240\mid p^4-1$ for all $p\gt 5$.
For the first part:
Note that $6(3)+5(-1)=13$. Now assume that ,$8x+by=13$ and substitute ,$x=3$ and $y=-1$ in this equation to find the value of $b$.
A: For the first one, you get
$$\begin{equation}\begin{aligned}
6x & \equiv -5y \pmod{13} \\
2x & \equiv -5(3^{-1})y \pmod{13} \\
8x & \equiv -5(9)(4)y \pmod{13} \\
8x & \equiv -11y \pmod{13}
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
Thus, $b = 11 + 13k$ for any integer $k$ gives that $8x + by$ is a multiple of $13$, with $b = 11$ being the smallest positive integer, as you stated.
As for the second part, I started to answer it, but this existing answer already explains it.
A: For the 2nd part: 
First note that $p^4-1$ may be factorised as:
$p^4-1$ = ($p^2+1$)($p+1$)($p-1$).
Since $p$ is always odd, $p$ is either congruent to 1 or 3 modulo 4.
This gives us $p^2+1$ is congruent to 2 modulo 4, i,e, $p^2+1$ is divisible by 2.
If $p$ is 1 modulo 4, then we have $p-1$ congruent to 0 modulo 4, and $p+1$ is at least divisible by 2, since $p+1$ is even. Thus, multiplying the numbers together, we get that $p^4-1$ is at least divisible by $2 \times 2 \times 4$ =16. In a similar vein, the above argument works if p is 3 modulo 4, giving us $p+1$ congruent to 0 modulo 4, and $p-1$ is at least divisible by 2.
Now, to prove that 16 is the largest power of 2 dividing $p^4-1$ for all $p>5$, all we have to do is simply plug in a specific case to verify this, which is easy.
Similarly, to show that 3 divides $p^4-1$ for all $p>5$, first note that $p$ is either congruent to 1 or 2 modulo 3 (otherwise, if it were congruent to 0 modulo 3, it would be composite). If $p$ is 1 modulo 3, then $3 \mid p-1$, and we are done. If p is 2 modulo 3, then $3 \mid p+1 $, and we are again done. Once again, we merely need to find a specific scenario in order to prove $3^1$ is the highest power of 3 dividing $p^4-1$ for all $p>5$.
Finally, to prove 5 divides $p^4-1$ for all $p>5$, again note that $p$ is congruent to 1,2,3, or 4 modulo 5. The case where $p$ is 1 or 4 modulo 5 is trivial. As for the cases where $p$ is congruent to 2 or 3 modulo 5, it is a simple exercise in arithmetic to verify that $p^2$+1 congruent to 0 modulo 5, i.e $ 5 \mid p^2+1$. Again, use a specific example to show that $5^1$ is the highest power of 5 dividing $p^4-1$ for all $p>5$.
Thus,we conclude that 240=$16 \times 3 \times 5$ divides $p^4-1$ for all $p>5$. It is also easy to prove that this is the largest integer, simply by throwing in a few prime numbers to check the case.
