I came across with the following question:
is the following correct: $$34 | (39^{16}-1)$$
After checking the calculator I've concluded it's true, but I was wondering if there is any other way doing it using some basic number theory skills.
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3This should be your first reflex. – TheSilverDoe Jul 04 '23 at 19:00
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1Is it obvious to you that $2\mid39^{16}-1$, and do you know Fermat's little theorem? – J. W. Tanner Jul 04 '23 at 19:07
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@J.W.Tanner Yes I am familiar with fermat's little theorem, but how is it helpful here? – YahavB Jul 04 '23 at 19:13
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3Since $39$ is coprime to $17$, that says $39^{17-1}\equiv1\bmod17$ – J. W. Tanner Jul 04 '23 at 19:14
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@J.W.Tanner So the answer is incorrect because it isnt divisible by 17 right? – YahavB Jul 04 '23 at 19:18
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For a compendium of techniques for answering questions like this please check out this as well as hundreds of others, most but not all of them linked to that one. – Jyrki Lahtonen Jul 04 '23 at 19:29
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No, $39^{16}-1$ is divisible by $17$ – J. W. Tanner Jul 04 '23 at 20:24
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THis is a special case of Euler's Theorem since $\phi(34) = 16,$ and $\gcd(39,34)=1.\ $ More generally see mod order reduction in the linked dupe, and the links there. – Bill Dubuque Jul 05 '23 at 00:50
1 Answers
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Note that $34=2\cdot 17$ and $2\mid (39^{16}-1)$. Next we need to show $17\mid (39^{16}-1)$. Fermat's little theorem is a shortcut. Or without applying Fermat's little theorem, we can do a direct computation, which gives
$$39^{16}-1\equiv5^{16}-1=25^8-1\equiv8^8-1\equiv(-4)^4-1\equiv(-1)^2-1=0\pmod {17}$$
MathFail
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