how do i prove that $17^n-12^n-24^n+19^n \equiv 0 \pmod{35}$

Question

How do i prove that $17^n−12^n−24^n+19^n≡0(\mod35)$ for every possitive integer n. Can anyone give me a hint of how to start?

Answer 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$(5,7)=?$

Answer 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 $35$.

Answer 3

Hint $\ \ \begin{eqnarray}a\equiv c,&&\ b\equiv d\pmod{m}\\ a\equiv d,&&\ b\equiv c\pmod{k}\end{eqnarray} \Rightarrow\ e = a^n+b^n-c^n-d^n\equiv 0\ \pmod{{\rm lcm}(m,k))}$

because $\ \ {\rm mod}\ m\!:\ a\equiv c\,\Rightarrow a^n\equiv c^n,\ \ b\equiv d\,\Rightarrow\ b^n\equiv d^n\,\Rightarrow\,e\equiv 0,\, $ using the Power Rule.

Similarly $\, {\rm mod}\ k\!:\ e\equiv 0.\,$ Thus $\ m,k\mid e\,\Rightarrow\,{\rm lcm}(m,k)\mid e$.

Remark $\ $ More generally, bringing to the fore the innate symmetry yields:

Assume $ \{a,b\}\equiv \{c,d\}\ {\rm mod}\ m,k\ $ and $\,f(x,y)\,$ is symmetric $\,f(x,y)=f(y,x).$

Then $\ \ f(a,b)\equiv f(c,d)\ {\rm mod}\ m,k\,$ so also mod $\,{\rm lcm}(m,k).$

Answer 4

Reduce each of $17, -12, -24,$ and $ 19$ mod $5$, obtaining $$2^n -2^n -4^n + 4^n\equiv 0 \pmod{5}$$ which is immediate. Then do the same mod $7$, obtaining $$3^n -5^n -3^n +5^n\equiv 0\pmod 7$$ which is similarly trivial. This shows that the original expression is a multiple of both 5 and 7, and therefore of 35.