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Here is my proof:

$(p^4 - q^4) \implies (p^2 + q^2)(p-q)(p+q)$

All prime numbers greater than 3 can be expressed as $6x\pm1$, where $x$ is a positive integer.

let $p = 6n\pm1$ and $q = 6m\pm1$, we can show that: $p^2 + q^2 = (6n\pm1)^2 + (6m\pm1)^2$. $p^2 + q^2 = 36n^2 \pm 12n + 36m^2 \pm 12m + 2$, and such, implies $2(\mod6)$, and $2\mid p^2+q^2$.

With regards to (p-q)(p+q),$p$ and $q$ remainders can be $5(\mod6)$ or $1(\mod 6)$ respectively.

Case 1: $p = 5 (\mod 6)$, and $q = (1 \mod6)$

$p+q = 0(\mod6)$, and $p-q = (4 \mod 6)$, therefore:

$6\mid p+q$ and $2\mid p+q$

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Case 2: $p = 1 (\mod 6)$, and $q = (1 \mod6)$

$p+q = 2(\mod6)$, and $p-q = (0 \mod 6)$, therefore:

$2\mid p+q$ and $6\mid p+q$

\

Case 3: $p = 5 (\mod 6)$, and $q = (5 \mod6)$

$p+q = 4(\mod6)$, and $p-q = (0 \mod 6)$, therefore:

$6\mid p+q$ and $2\mid p+q$.

In all cases, both expressions have a factor of 2, but only one has a factor of 6. Therefore, $12\mid(p^2-q^2)$, and $24\mid(p^2-q^2)(p^2+q^2)$.

However, with my proof, I have only proved that p^4 - q^4, is a factor of 24. However, I don't know how to prove that the expression is also a factor of 48.

Any help would be appreciated, thanks!

Arturo Magidin
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Sam
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2 Answers2

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The squares modulo $3$ are $0$ and $1$. In particular, the square of an integer not divisible by $3$ must be congruent to $1$ modulo $3$, and hence so is the 4th power. So $p^4\equiv q^4\equiv 1\pmod{3}$.

If you square an odd number, the remainder modulo $8$ is $1$. This is easy to verify by simply looking at the squares of $1$, $3$, $5$, and $7$. That means that modulo $16$, the squares must be congruent to either $1$ or $9$ . Squaring again gives you $1$ or $81\equiv 1\pmod{16}$ (as $80=5\times 16$). Thus, $p^4\equiv q^4\equiv 1\pmod{16}$.

So we see that $p^4-q^4$ is divisible by $3$ and by $16$, hence by $48$ (their least common multiple) as desired.

Arturo Magidin
  • 398,050
  • Thanks, for the proof! Could you also provide resources for me to learn about modular arithmetic, what would be the rules of multiplying, dividing, exponentiating, etc. – Sam Jan 05 '24 at 02:39
  • Please strive not to post more (dupe) answers to dupes of FAQs. This is enforced site policy, see here. – Bill Dubuque Jan 05 '24 at 02:46
1

First, it's obvious that $p$ and $q$ are both odd numbers as they are primes bigger than $2$. So we can actually deduce that $$8|(p+q)(p-q)$$ This is because the sum and difference are both even and one of them must be a multiple of $4$. (This can be proven by mod $4$)

And, because $p^2$ and $q^2$ are both odd. We have $$2|(p^2+q^2)$$ As we already have $16|(p^4+q^4)$, all we have to do is prove that one term from $(p+q)$, $(p-q)$ and $(p^2+q^2)$ is a multiple of $3$.

  1. Case 1: $p\equiv\pm1\pmod3$ and $q\equiv\pm1\pmod3$, then $p-q$ is a multiple of $3$
  2. Case 2: $p\equiv\pm1\pmod3$ and $q\equiv\mp1\pmod3$, then $p+q$ is a multiple of $3$

As both $p$ and $q$ are primes, they are not multiples of $3$.

Thus, we have proven it.

YANGyu
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