Saudi Arabian Mathematical Competitions

Let $d$ be the greatest common divisor of the numbers in $S$. Since $2^{7}-2=126=2 \cdot 3^{2} \cdot 7 \in S$ for $x=y=1$, one gets that $d$ divides $126$. Now, for $x=2$ and $y=1$, one gets $(2+1)^{7}-2^{7}-1=3^{7}-2^{7}-1 \in S$, therefore $d$ will divide the number $(3^{7}-2^{7}-1)+(2^{7}-2)=3^{7}-3$. Since $9$ does not divide $3^{7}-3$, it follows that $d$ divides $126 / 3=42$.

Let us check that $d=42$, i.e. the prime numbers $2$, $3$ and $7$ divide $(x+y)^{7}-x^{7}-y^{7}$, for all integers $x$ and $y$. Notice it is enough to show that $42$ divides $a^{7}-a$ for all integers $a$, since one can write
$$(x+y)^{7}-x^{7}-y^{7}=\left((x+y)^{7}-(x+y)\right)-\left(x^{7}-x\right)-\left(y^{7}-y\right).$$
But $a^{7}-a=a(a-1)(a+1)\left(a^{2}-a+1\right)\left(a^{2}+a+1\right)$. Now, $2$ divides $a(a-1)$, $3$ divides $a(a-1)(a+1)$, while $7$ divides $a^{7}-a$, by Fermat's Little Theorem.