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I stumbled upon this equation: $$\exp_a\left\{\sum_{k=1}^n \gcd(k,n)\cos\left(\frac{2\pi k}{n}\right)\right\}\equiv 1\pmod n\tag*{$\big(\forall a\in\mathbb{Z}\land n\in\mathbb{Z}^+\big)$}$$ such that $\exp_a\left\{b\right\} = a^b$ for all $a,b\in\mathbb{R}$ and $e=2.7182818\ldots$ is the natural base. Is this equation true at all? How can it be proven? I tried looking at similar equations, and doing some research and some exploration through memory, the following theorem looks fairly similar:

Fermat's Little Theorem:

$$a^{p-1}\equiv 1\pmod p.\tag*{$\big(\forall a\in\mathbb{Z}$ and prime $p\big)$}$$

However, the modulus is a prime $p$ in this theorem, but in the upmost equation, the modulus is a positive integer $n$. Could I turn this around somehow, or am I not looking in the right direction?

I found this odd equation in images by looking at math equations to set as my desktop background, and I am now sharing this to you in just different notation (as it looks prettier in my opinion). I cannot find the actual image unfortunately, but it and the equation above is nearly identical.

Nota Bene (Please Note): My skill level is not that great when solving/proving such equations...

Thank you in advance.

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Mr Pie
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Because $\; \phi(n) = \sum_{k=1} ^n \gcd(k, n) \cos(2 \pi k / n)\; $ and $\; a^{\phi(n)} \equiv 1 \pmod n \;$ your equation follows. The congruence is the natural generalization of Fermat's little theorem. The sum equation for Euler's totient function is mentioned in the Wikipedia article and in MSE question 2757228 "Derivation of the gcd formula for the nth cyclotomic polynomial".

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  • How is $a^{\varphi(n)}\equiv 1\pmod n$? Is this a generalisation of Fermat's Last Theorem? – Mr Pie May 13 '18 at 11:48
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    Nevermind. I went to this post and learnt more. Congratulations! $$(+1) \ \ \color{green}{\checkmark}$$ – Mr Pie May 16 '18 at 04:09
  • Ok well you should probably reference your source since we stumble on garbage left in the street, as a result of inattention during a leisurely stroll, but mathematics is the result of someone doing negation to the utmost just saying bud – Adam Ledger Oct 26 '18 at 03:24
  • but I did "bit bash" out a discrete Fourier transform for the modulo function quite a few years ago with little to no real comprehension of where to take it, but it can obviously be taken into consideration for a corresponding congruence relation https://math.stackexchange.com/questions/2753303/discrete-fourier-transform-for-the-modulo-function – Adam Ledger Oct 26 '18 at 03:27
  • note that it can easily be rearranged to be the DFT of the floor of a rational number, which might make it easier to arrive at something that's a function of the greatest common divisor, but it is going to be very long winded – Adam Ledger Oct 26 '18 at 03:30
  • also I realise it may look prettier in your view, but for others that have tried their best to select the most easily understandable notation for which to express something, it just looks like total trash – Adam Ledger Oct 26 '18 at 03:33