how do i prove that $17^n-12^n-24^n+19^n \equiv 0 \pmod{35}$
Solution 1:
HINT:
As $(a-b)|(a^n-b^n)$ for integer $n\ge0$ i.e., $\displaystyle a^n-b^n\equiv0\pmod{a-b}$
(See Why $a^n - b^n$ is divisible by $a-b$?)
$\displaystyle17^n-12^n\equiv0\pmod{17-12}$ and $\displaystyle24^n-19^n\equiv0\pmod{24-19}$
and $24^n-17^n\equiv0\pmod{24-17}$ and $\displaystyle19^n-12^n\equiv0\pmod{19-12}$
Finally, $(5,7)=1\implies$lcm$(5,7)=?$
Solution 2:
Hint $17-12=24-19=5$ and $24-17=19-12=7$.
And $35=5\cdot 7$. Choose appropriate moduli to work with instead of $35$.