As the title, I was asked to show that $2^{44}-1$ is divisible by $89$.
My first attempt is to change the expression into $(2^{22}-1)(2^{22}+1)$.
Then further simplified it into $(2^{11}-1)(2^{11}+1)(2^{22}+1)$, I used my calculator and was able to show that $2^{11}-1$ is divisible by $89$ but then I don't know how to show it with modular arithmetic. I do think that it is quite similar to the form where we can use the Fermat's little theorem. $(\sqrt{2})^{88}-1$. (Though I do understand Flt can only be applied to integer.)
Can someone tell me whether I can take square root in modular arithmetic as well? I am still pretty new to the modular arithmetic. Thank you very much.
Answer
Hint: By Fermat's Theorem, we have $2^{88}\equiv 1\pmod{89}$.
So $(2^{44}-1)(2^{44}+1)\equiv 0 \pmod{89}$.
If we can show that $2^{44}+1\not\equiv 0\pmod{89}$ we will be finished.
One way to do this is to use the fact that $2$ is a quadratic residue of $89$, since $89$ is of the shape $8k+1$.
Remark: Your direct computational approach is perfectly fine. However, it may be that you are expected to make use of "theory," as in the approach described above.
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