2

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.

Analyst_311419
  • 730

3 Answers3

3

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 $b$.

  • 1
    This is a good answer, but a minor point I suggest for completeness is that you explicitly state something like $p^2 \equiv 1 \pmod 4$ for all odd integers, so $p^2 + 1$ always has only one factor of $2$, since the question asks for the largest integer which divides $p^4 - 1$. – John Omielan Oct 11 '19 at 07:07
  • @John omielan, Thanks for pointing out. See it it looks good now. –  Oct 11 '19 at 07:13
  • You're welcome. However, a new small issue is you now state $4 \not\mid p^2$, which is true, but I believe you meant to write $4 \not\mid p^2 + 1$ instead. – John Omielan Oct 11 '19 at 07:20
  • @John Omielan, Thanks for patiently reading and pointing out errors in my answer. –  Oct 11 '19 at 07:24
3

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.

John Omielan
  • 47,976
  • i.e. map the $x$ coef from $6$ to $8$ by scaling the congruence by $,\dfrac{8}6\equiv \dfrac{16}{12}\equiv\dfrac{3}{-1}\equiv\color{#c00}{ -3}\pmod{!13}\ $ which yields $\ 8x\equiv \color{#c00}{-3}(-5y)\equiv 2y\ \ $ – Bill Dubuque Oct 11 '19 at 15:11
1

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.

Hello
  • 2,133