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?

#NumberTheory

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Comments

$2^8 \equiv 256 \equiv 249 + 7 \equiv 7$ (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 - 5 years, 2 months ago

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

abc xyz - 5 years, 2 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.

Ameya Daigavane - 5 years, 2 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$