Because every prime number $>$ 3 is $+1,-1mod4$ then by PHP at least one of $a-b,b-c,c-a$ is divisible by 4 and the other two are divisible by 2 so 16 divides $(a-b)(b-c)(c-a)$. because every prime number $>$3 is $+1,-1 mod3$ then by PHP at least two of a,b,c are congurent mod 3 so at least one of $a-b,b-c,c-a$ is divisible by 3 . $\rightarrow$ 48 divides $(a-b)(b-c)(c-a)$..

whats php?

achen29 wrote: whats php? Pigeonhole principle

Hint: All primes are the form $6k+1$ or $6k+5$.

Since all primes larger than 3 are either equivalent to 1 or 3 mod 4, so by pigeonhole at least one of $(a-b),(a-c),(b-c)$ will be divisible by 4, and since all three are even $(a - b)(b - c)(c - a)$ will be divisible by 16. Similarly all primes are either equivalent to 1 or 1 mod 3, so we can also show that $(a - b)(b - c)(c - a)$ has at least one factor of three. Now we have shown that $(a - b)(b - c)(c - a)$ is a multiple of $lcm(16,3)=48$ and we are done. This is my 1st post in hso so sorry if my wording is bad

Just work module $2$, $4$ and $3$ and use the pigeonhole principle.

a,b,c are odd,because a,b,c are prime numbers bigger than 3.Then a-b,b-c,c-a are divisible by 2.Suppose that none of a-b,b-c,c-a isn't divisible by 4.If a≡2 (mod 4), then b≡0 and c≡2 mod 2,but (a-c) is divisible by 4.If a ≡0(mod 4),then b≡2(mod 4),c≡0(mod 4),but (a-c) is divisible by 4 .Contradiction!Therefore at least one of a-b,b-c,c-a is divisible by 4 and a-b,b-c,c-a are even.Therefore (a-b)(b-c)(c-a) is divisible by 16.Suppose that none of a-b,b-c,c-a isn't divisible by 3. If a≡2(mod 3),then b≡1(mod 3),c≡2(mod 3),but (a-c) is divisible by 3 . If a≡1(mod 3),then b≡2(mod 3),c≡1(mod 3),but (a-c) is divisible by 3.Contradiction.Therefore (a-b)(b-c)(c-a) is divisible by 16 and (a-b)(b-c)(c-a) is divisible by 3 and then we have (a-b)(b-c)(c-a) is divisible by 48.

Prime numbers $a, b, c$ are bigger that $3$. Show that $(a - b)(b - c)(c - a)$ is divisible by $48$. Note that, since primes $a,b,c>3,$ each bracket of the expression is divisible by $2$ (or even). Thus, the expression will always be divisible by $8.$ Now we show divisibility by $6$. Building off this fact, we note that, since all primes are $1,3 \pmod 4,$ the expression is always divisible by $4.$ Since (from above) the expression is always going to be a multiple of 2, we now show that is will also, always be a multiple of $3.$ Note that $a,b,c \equiv 1 \pmod 3.$ We then see that $3$ divides the given expression, as well. Now, given that$$8|(a - b)(b - c)(c - a),6|(a - b)(b - c)(c - a),3|(a - b)(b - c)(c - a),$$we have shown that $48|(a - b)(b - c)(c - a).$ $~\blacksquare$

$\color{blue} \boxed{\textbf{SOLUTION}}$ Firstly, $8|(a - b)(b - c)(c - a)$ As, $a,b,c \in p > 3 \in$ odd Resude classes of mod $4$, for primes are $(+1,-1)$ We have $3$ numbers but We can just have $4k+1,4k-1$ these forms of numbers, so By $\textbf{PHP},$ two numbers, WLOG, $a,b = 4k+i, i \in (+1,-1)$ So, $(a-b) \equiv 0 \pmod 4$ We get, $16|(a - b)(b - c)(c - a)$ Similarly, In mod $3$ there are $2$ residue classes for primes, $(+1,-1),$ Similar argument gives, there exist, WLOG $a,b=3k'+j, j \in (+1,-1) \implies (a-b) \equiv 0 \pmod 3$ We get, $3|(a - b)(b - c)(c - a)$ So, $48|(a - b)(b - c)(c - a) \blacksquare$

let's look these numbers in mod 3 , we can see that the number gets 1,2 in mod3 then we can easyly see that , there is three numbers , and two different remainder then two of that numbers get same remainder in mod3 , than this number divisible by 3 , and taking mod 4 for this number , the numbers are odd , we get 1,3 remainder in mod 4 , in same princibe at least two number get same remainder in mod 4 , we get that one of the factor get 0 in mod4 , we know that a,b,c are odd other factor get 0 from mod 2 , we get that the number divisible by 48.