Show that there are two distinct positive integers such that: $1394|2^a-2^b$

I'm sure pigeon hole principle applies here,but don't recognize holes.Another problem statement is: show that there are two positive integers $a,b$ such that: $$2^a\equiv 2^b\pmod {1394}$$
Of course we have $1394$ cases for division mod $1394$,but what are the pigeons?



Consider $2^i \pmod{1394}$ for $1 \leq i \leq \color{blue}{1395}$

  • $\begingroup$ So this problem has been VERY simple!!! I was afraid of that!! $\endgroup$ – Hamid Reza Ebrahimi Dec 10 '17 at 5:30

Note $1394=2 \times 17 \times 41$.

Since $\varphi(17 \times 41) = 16\times 40 = 640$,

then $2^{640} \equiv 1 \pmod{697}$.

Hence $2^{641} \equiv 2 \pmod{1394}$.

In other words, $1394 \mid 2^{641}-2^1$.

Note. Actually, we can do better. Checking the divisors of $640$, we find that $2^{40} \equiv 1 \pmod{697}$. So $1394 \mid 2^{41}-2^1$.

  • $\begingroup$ One question, what did you do in that last step when you say "checking the divisors of 640"? $\endgroup$ – Francisco José Letterio Dec 10 '17 at 17:25
  • $\begingroup$ Since $2^{240} \equiv 1 \pmod {697}$, then $H = \{2^1,2^2,2^3,\dots,2^{240}\}$ is a cyclic subgroup of $\mathbb U_{697}$, a multiplicative group with $240$ elements in it. Hence the order of $2$ in $H$ is a divisor of $240$. $\endgroup$ – steven gregory Dec 10 '17 at 19:33

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