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?

## 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 \(83 \)) – Ameya Daigavane · 8 months, 3 weeks ago

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– Abc Xyz · 8 months, 3 weeks ago

Wow !! I didn't think of that. Thanks a LOT Sir !!!Log in to reply

– Ameya Daigavane · 8 months, 3 weeks ago

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.Log in to reply