$34^{(84\cdot n)} $ is always congruent to $216 \pmod{ 344} $

Question

I wonder if the expression $34^{(84\cdot n)} $ is always congruent to $216 \pmod {344}$ and to $1 \pmod {559}$ for every n? If true, why?

Answer 1

Yes. In modulo $344$ direct calculation gives $34^3 = 88$, $88^4 = 16$ and $16^7 = 216$, so $34^{84}=216$. Finally $216\cdot 216= 216$ and $34^{84\cdot n} = 216$ follows via induction.

Use similar calculation for the second question.

Answer 2

It's true, as shown here with hardly any calculation.

By Fermat's little theorem, $34^{42}\equiv1\bmod43$. Therefore $34^{84n}\equiv1\bmod43.$

Thus $34^{84n}\equiv1, 44, 87, 130, 173, 216, 259, $ or $302 \bmod 43\times8=344$.

But $34^{84n}\equiv0\bmod8$, so it must be that $34^{84n}\equiv216\bmod344$.

Also, $34\equiv8\bmod 13$, and $8^2\equiv-1\bmod 13$, so $8^4\equiv1\bmod 13$,

so $34^{84n}\equiv1\bmod 13$, so $34^{84n}\equiv1\bmod43\times13=559.$

Answer 3

One has to prove $$ 34^{84n}=216+344X\iff(43-9)^{84n}=(5\cdot43+1)+8\cdot43X\tag1$$ $$34^{84n}=1+559Z\iff(43-9)^{84n}=1+13\cdot43Z\tag2$$ So in $(1)$ we have $$3^{168n}\equiv1\pmod{43}\iff(3^6)^{28n}\equiv(-2)^{28n}\equiv(2^7)^{4n}\equiv(-1)^{4n}\equiv1\pmod{43}$$ In $(2)$ we have from $(1)$ $$34^{84n}\equiv1\pmod{43}$$

Answer 4

$8$ divides $\, 216\,$ & $\,34^{\large 84 n}$ (by $\,n\ge 1)$ so they are congruent $\!\bmod 8,\,$ and also $\!\bmod 43\,$ (both $\equiv 1$ by Fermat by $\,42\mid 84).\,$ So their difference is divisible by $8,43$ so also by ${\rm lcm}(8,43) = 8\cdot 43 = 344$.

By Fermat $\,12\mid 84\,\Rightarrow\, 34^{\large 84 n}\equiv 1\pmod{\!13},\,$ so, as by above, $\,34^{\large 84 n}-1\,$ is divisible by $\,13,43\,$ so by $\,13\cdot 43 = 559.\,$ We used modular order reduction with Fermat to reduce the exponents.