proving that $89 \mid 2^{44}-1$

Question

i tried to prove that $2^{44} \equiv 1\pmod{89}$.

I noticed that by Fermat's little theorem $2^{88} \equiv 2^{44}\cdot 2^{44} \equiv 1\pmod{89}$ which means that $2^{44}$ is the inverse of itself $\rightarrow 2^{44} \equiv 1 \pmod{89}$ or $2^{44}\equiv 88 \pmod{89}$.

how can I rule out the second option?

Answer 1

By Second Supplement to Law of Quadratic Reciprocity, $\left(\dfrac2p\right) = 1$ for $p \equiv 1 \pmod 8$.

i.e. there exists some $a \in \mathbb Z$ such that $a^2 \equiv 2 \pmod {89}$.

$a^{88} \equiv 1 \pmod {89}$ follows from Fermat's Little Theorem.

Hence $2^{44} \equiv a^{88} \equiv 1 \pmod {89}$.

Answer 2

You have $2^{11}= 2048 \equiv 1 \pmod {89} $. Hence, $$2^{44}=(2^{11})^4 \equiv 1 \pmod {89} $$

Answer 3

By Euler's criterion, $2^{44}\equiv\left(\dfrac2{89}\right)\bmod89$, and $\left(\dfrac2{89}\right)=1$ because $89\equiv1\bmod8$,

where $\left(\dfrac2{89}\right)$ is the Legendre symbol.

Answer 4

$2^{44} - 1 = (2^{22})^2 - 1 = (2^{22}+1)(2^{22} - 1) = (2^{22}+1)(2^{11} + 1)(2^{11} - 1) = (2^{22}+1)(2^{11} + 1)\times 2047 = (2^{22}+1)(2^{11} + 1)\times 23 \times 89$