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The question is to find the largest integer that divides all $p^4-1$, where p is a prime greater than 5. Being asked this question, I just assume this number exists. Set $p = 7$, then $p^4-1=2400$. I don't have any background in number theory and not sure what to do next. Thank you for your help!

Edward Wang
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    in what context did you find the question ? –  Aug 13 '17 at 01:21
  • $2^4\cdot3\cdot5$ divides $(p-1)(p+1)(p^2+1)$ and is the gcd of $7^4-1$ and $11^4-1$ – uSir470888 Aug 13 '17 at 01:24
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    @RoddyMacPhee GRE Sub Practice exam. Since no calculator is permitted, I really don't know how to solve this in a quick way. – Edward Wang Aug 13 '17 at 01:31
  • what do we know about all primes greater than 3 ? that's a start. –  Aug 13 '17 at 01:44
  • @RoddyMacPhee They are in the form $4k+1$ or $4k+3$. Therefore 16 divides $p^4-1$, thus this number should be greater than 16. The final answer seems to be 240, after I used my calculator. – Edward Wang Aug 13 '17 at 01:57
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    I was pointing to the fact they are all 1 or 5 mod 6 so we know 6 is in the divisors of the number we want because the value of $p^4$ will be 1 mod 6 and subtracting 1 is 0 mod 6. add in fermat's little theorem and you get that it also is divisible by 5 so $5\cdot lcm(16,6)=240$ and that at least shows 240 is a minimum to the value. –  Aug 13 '17 at 01:59
  • Oh I see. Let that number be $x$, then $16|x$. $p$ are all 1 or 5 mod 6, thus $6|p^4-1$. Then $5|x$, thus $lcm(5,16,6)=240$ is the answer. – Edward Wang Aug 13 '17 at 02:08
  • @RoddyMacPhee We can try $p=11$ to find that 240 is the final answer. – Edward Wang Aug 13 '17 at 02:10
  • it could be higher but there's only 3 possibilities to check at that point to confirm 240 as the answer, 480,1200, and 2400. but baring that one of these also divides each one you can then safely conclude 240. –  Aug 13 '17 at 02:11

3 Answers3

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Let $n$ be the largest integer that divides $p^4-1$ for all prime $p\geq 7.$

We have $11^4-1=14640$ and $7^4-1=2400.$ The $gcd$ of $14640$ and $2400$ is $240.$ So $$n\leq 240.$$ If $p$ is odd then modulo $16$ we have $p^4\in \{(\pm 1)^4, (\pm 3)^4,(\pm 5)^4,(\pm 7)^4\}=\{1^2,9^2, 25^2, 49^2\}=\{1^2,9^2,9^2,1^2\}=$ $=\{1,81,81,1\}=\{1\}.$

If $p$ is not divisible by $3$ then modulo $3$ we have $p^4\in \{(\pm 1)^4\}=\{1\}.$

If $p$ is not divisible by $5$ then modulo $5$ we have $p^4\in \{(\pm 1)^4,(\pm 2)^4\}=\{1,16\}=\{1\}.$

So for any integer $p$ that is not divisible by $2,3,$ or $5$ we have $p^4\equiv 1 \pmod {16}$ and $p^4 \equiv 1 \pmod 3$ and $p^4 \equiv 1 \pmod 5;$ and since $16,3,$ and 5 are pair-wise co-prime, therefore $p^4\equiv 1 \pmod {16\cdot 3\cdot 5}=240,$ so $$n\geq 240.$$

DanielWainfleet
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Let $n$ be the number you are trying to find.

One of the main reasons to look at an example $p=7$ is to gain information about $n$. If $n$ divides all $p^4 - 1$, and $7^4 - 1 = 2400$, then $n$ divides $2400$.

Thus, you've gained information about $n$. Follow that up — what information does knowing that $n \mid 2400$ give you?

In my opinion, the next obvious things to do are to try the next few primes and gain more information about $n$.

My expectation is that will be enough to completely determine what $n$ ought to be, or at least narrow it down to a very small number of possibilities, at which point you can turn towards trying to prove that one of the possibilities is actually the value of $n$.

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Since $p^4-1$ is relatively prime to $p$, we know that the common divisor will not have any large prime factors. Indeed, from your calculation, it will have to be a divisor of $2400=2^5*3*5^2$. However, you could make an even better guess by trying out a few more primes and then taking the greatest common divisor. However, that will get you a proof.

A reasonable thing to try is to factor the polynomial $p^4-1=(p^2+1)(p^2-1)=(p^2+1)(p+1)(p-1)$. There are various things you can see from this. For example, since $p$ is odd, $p\pm 1$ are both even, as is $p^2+1$, giving you at least three factors of $2$. Actually, one of $p\pm 1$ has to be multiple of $4$ (being two consecutive even numbers), and therefore you know that $16$ will divide $p^4-1$ for all odd $p$ (even if $p$ isn't prime). But you should gather more data to figure out exactly what you want to prove.

Aaron
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