I came across a question which uses Fermat's little theorem:

**The Question:**

Find a positive integer n such that \(7n^{25}\) - 10 is divisible by 83.

**The Solution given in the book:**

Since 7 x 37 = 259 = 10 mod 83

We have to find a value of n such that \(7n^{25}\) = 7 x 37 mod 83

This is equivalent to \(n^{25}\) = 37 = \(2^{20}\) mod 83

By Fermat's Theorem,

\(2^{82k}\) = 1 mod 83 for all k.

So we need to choose n such that \(n^{25}\) =\(2^{82k+20}\) mod 83

This will be satisfied if k=15

Therefore \(n^{25}\) = \(2^{1250}\) mod 83

And so **n = \(2^{50}\)**

This gives one value.

My problem is that I don't understand the fourth line

That is how 37 = \(2^{20}\) mod 83?

So can someone please explain this?

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## Comments

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TopNewest\( 2^8 \equiv 256 \equiv 249 + 7 \equiv 7 \) (mod \(83 \))

\( \Rightarrow 2^{16} \equiv 7^2 \) (mod \(83 \))

\(\Rightarrow 2^{20} \equiv 49 \cdot 16 \equiv 784 \equiv 37 \)(mod \(83 \))

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Wow !! I didn't think of that. Thanks a LOT Sir !!!

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Note that 2 is a primitive root modulo 83, so that even if the remainder weren't 37, we could find another exponent (but finding the exact exponent is difficult, in general.) instead of 20 here.

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