If $p,q$ are primes, and $p>q>5$, prove that $240\vert p^4 - q^4$

Question

The source of the question is from 2019 UKMT Senior Maths Challenge, Further Investigation 1.6 :

"If $p,q$ are primes, and $p>q>5$, prove that $240\vert p^4 - q^4$"

I am struggling with the question , as I’m not too sure on the proof for this statement.

My attempts :

I do know that for when $p>q>2$, you can express the primes as $2x + 1$, where $x$ is a positive integer.
I also know that when $p>q>3$, you can express the primes as $6x\pm1$. However I’m unaware of a similar general formula for $p>q>5$ which is the reason I can not complete my work.

$(p^4 - q^4) \implies (p+q)(p-q)(p^2+q^2)$,
Consider the case, where $p>q>2$
All numbers $p$ and $q$, can be expressed in the form 2n + 1.
$(p^2 + q^2) \pmod 2 \implies 0\pmod2$, $2\vert(p^2+q^2)$
With regards to $(p+q)(p-q)$, p and q can be $1\pmod 4$ or $3\pmod4$, therefore $8\vert p^2 - q^2$.
This implies $16\vert p^4 - q^4$, when $p>q>2$

Now consider the where, $p>q>3$
All prime numbers can be expressed in the form $6n \pm 1$, following from the first case, we know $2\vert p^2 + q^2$
With regards to $(p+q)(p-q)$, p and q can $1\pmod 6$ or $5\pmod6$, therefore $12\vert (p+q)(p-q)$.
Knowing that $16\vert p^4 - q^4$ for $p>q>2$ we can take the lowest common denominator of 16 and 12, to get $48\vert p^4 - q^4$, when $p>q>3$

The reason why I don't know how to prove p>q>5, because I'm not aware of a standard form for all prime numbers greater than 5, (I used 2n+1, for primes greater than 2, and $6n\pm1$ for primes greater than 3).

Answer 1

We have $240=2^4\cdot 3\cdot 5$, by CRT we only need to check that $p^4-q^4$ is divisible by each prime power.

Modulo $2^4$

We have $x^8\equiv_{16}1$ for all $x\in\Bbb Z_{16}^\times$ since $\phi(16)=8$. However, $\Bbb Z_{16}^\times\cong\Bbb Z_2\times\Bbb Z_4$ so we have $$x^4\equiv_{16}1$$ Therefore $p^4-q^4\equiv_{16}0$

Modulo $3$

We have $x^2\equiv_31$ for all $x\in\Bbb Z_3^\times$, so $$p^4-q^4\equiv_3(p^2)^2-(q^2)^2\equiv_30$$

Modulo $5$

We have $x^4\equiv_5 1$ for all $x\in\Bbb Z_5^\times$, so $$p^4-q^4\equiv_5 1-1\equiv_5 0$$

Answer 2

Sufficient to show that $~p^4 - q^4~$ is divisible by each of $~16, ~5,~$ and $~3.$

From the constraints :

• $~p = q + 2k ~: ~k \in \Bbb{Z^+} \implies $
  $p^4 - q^4 = 8kq^3 + 24k^2q^2 + 32k^3q + 16k^4.$
  So, divisibility by $~16~$ is determined by whether
  $~kq^3 + k^2q^2~$ must be divisible by $~2.~$
  If $~k~$ is even, both numbers are even, and if $~k~$ is odd, both numbers are odd.

• From Fermat's little theorem, $~p^4 \equiv 1 \equiv q^4 \pmod{5} \implies 5 ~| ~(p^4 - q^4).$

• From Fermat's little theorem, $~p^2 \equiv 1 \equiv q^2 \pmod{3} \implies p^4 \equiv 1 \equiv q^4 \pmod{3} \implies $
  $3 ~| ~(p^4 - q^4).$

Answer 3

Elementary Way :

(A) Consider that all Primes larger than $5$ are ODD
Hence , each of the $3$ factors $(P+Q)$ , $(P-Q)$ , $(P^2+Q^2)$ are EVEN , hence , we have got $2 \times 2 \times 2 = 2^3$

Now , $P$ can be $4X+1$ or $4X-1$ , while $Q$ can be $4Y+1$ or $4Y-1$. When both $P$ & $Q$ have the $+1$ form or both $P$ & $Q$ have the $-1$ form, then $(P-Q)$ is not just EVEN , it is a multiple of $4$.
When one has the $+1$ form while the other has the $-1$ form , then $(P+Q)$ is a multiple of $4$.
Hence , we have got one more $2$ factor to get $2^4$

(B) Consider that all Primes larger than $5$ are of the form $3X+1$ or of the form $3X-1$
When both $P$ & $Q$ have the $+1$ form or both $P$ & $Q$ have the $-1$ form, then $(P-Q)$ is a multiple of $3$.
When one has the $+1$ form while the other has the $-1$ form , then $(P+Q)$ is a multiple of $3$.
Hence , we have got $3$ factor

(C1) Similarly , when we take the Primes larger than $5$ to be like $5X+1$ or $5X-1$ , then either $(P+Q)$ or $(P-Q)$ will be a multiple of $5$.

(C2) When we take the Primes larger than $5$ to be like $5X+2$ or $5X-2$ , then either $(P+Q)$ or $(P-Q)$ will be a multiple of $5$.

(C3) When $P$ is like $5X+1$ or $5X-1$ , while $Q$ is like $5X+2$ or $5X-2$ (or vice-versa) , then $(P^2+Q^2)$ will be a multiple of $5$.

All Cases , we have got factor of $5$.

SUMMARY :

Over-all , we have the factors $2^4,3,5$
Hence , $240$ is a factor.

ADDENDUM :

We do not have to use the $6N+1$ form , though that will give slight variation in the Proof , which could get a little longer when we want the factor of $5$.
It is easier when we use multiple forms , which are suitable to get each factor.