2025/018) If a , b , c are any three integers then show that $7 | abc(a^3-b^3)(b^3-c^3)(c^3-a^3)$

Le $P(a,b,c) =  abc(a^3-b^3)(b^3-c^3)(c^3-a^3)$

 

Let us first find out  $a^3 \mod 7$

We have $a^3 \mod 7 \in \{1,-1\}$ when $a \mod 7 \ne 0$

if  $a \mod 7 = 0$ or $b \mod 7 =  0$ or $c \mod 7 \ne 0$ we have $7 |  P(a,b,c)$

$a^3 \mod 7 \in \{1,-1\}$

$b^3 \mod 7 \in \{1,-1\}$

$c^3 \mod 7 \in \{1,-1\}$

each can take one of two values and there are 3 differences so one of them has has to be zero

hence  $7 |  P(a,b,c)$

Proved