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Show that $a^{16}-b^{16}$ is divisible by $133$ if $a$ and $b$ are both prime to $85$

Since $(85, a)=1(17,5)$ and $(85, b)=(17,5)$ then $a^{16}-1\equiv (mod ~17)$, $a^{4}-1\equiv (mod~ 5)\implies a^{16}-1\equiv (mod~ 5)$, $b^{16}-1\equiv (mod~ 17)$, $b^{4}-1\equiv (mod~ 17)\equiv b^{16}-1\equiv (mod~ 5)$. Then $a^{16}-1\equiv (mod~ 85)$ $a^{16}-1\equiv (mod~ 85)$ Thus $a^{16}-b^{16}\equiv 0 (mod~85)$.

Please suggest me, what to do now.

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  • $\begingroup$ I find this hard to believe that this holds for all pairs $a,b$ because 133 and 85 are totally unrelated. Where does this problem come from? On the other hand, this does seems to hold for many pairs, which is surprising. $\endgroup$ – lhf Jun 28 '15 at 14:21
  • $\begingroup$ I then think that the problem given in the book is wrong. 133 should be replaced by 85. $\endgroup$ – math131 Jun 28 '15 at 14:22
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$2^{16}-1 = 3\cdot 5\cdot 17\cdot 257 $ so there is no way that $133=7\cdot 19$ divides $a^{16}-b^{16}$ if $(a,b)=(2,1)$.

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It is not true. $2^{16} - 1^{16} \equiv 99 \pmod {133}$ An easy way to show it without a calculator is that $7$ divides $133$, so $7^{16}$ is a multiple of $7$ and $1^{16}$ is not.

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