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.

  • $\begingroup$ It is a Mersenne number, you could try to see if that helps... $\endgroup$ – Lessa121 Dec 15 '13 at 18:47

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.

  • 1
    $\begingroup$ Your chain of reasoning looks too long to me! If $2$ is a quadratic residue of $89$, so that $a^2=2 \mod 89$ for some $a$, then we have immediately that $2^{44} = a^{88} = 1 \mod 89$. We don't need to factorise anything. $\endgroup$ – TonyK Dec 15 '13 at 19:13
  • $\begingroup$ Yes, that is more direct. $\endgroup$ – André Nicolas Dec 15 '13 at 19:14

An idea:



since $\;801=9\cdot 89\;$


You can use exponentiating by squaring:
Consider the following numbers mod $89$:
${16}^2={(2^4)}^2=2^8=256\equiv 78(mod89)$

We have $2^{16}\equiv32=2^5$.So,$2^{16}\equiv2^5$ hence $2^{11}\equiv1mod89$.
We can easily see now that $2^{44}={(2^{11})}^4\equiv1^4\equiv1mod89$ .
I believe this is the shortest way.
By the way:You can take square roots with modular arithmetic if the exponents you are dealing with,are even numbers!


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