# I Don't Get It!

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? Note by Abc Xyz
5 years, 4 months ago

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$2^8 \equiv 256 \equiv 249 + 7 \equiv 7$ (mod $83$)
$\Rightarrow 2^{16} \equiv 7^2$ (mod $83$)
$\Rightarrow 2^{20} \equiv 49 \cdot 16 \equiv 784 \equiv 37$(mod $83$)

- 5 years, 4 months ago

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

- 5 years, 4 months 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.

- 5 years, 4 months ago