Saudi Arabia 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.