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 7n257n^{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 7n257n^{25} = 7 x 37 mod 83
This is equivalent to n25n^{25} = 37 = 2202^{20} mod 83
By Fermat's Theorem,
282k2^{82k} = 1 mod 83 for all k.
So we need to choose n such that n25n^{25} =282k+202^{82k+20} mod 83
This will be satisfied if k=15
Therefore n25n^{25} = 212502^{1250} mod 83
And so n = 2502^{50}
This gives one value.

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

That is how 37 = 2202^{20} mod 83?

So can someone please explain this?

#NumberTheory

Note by Abc Xyz
3 years, 5 months ago

No vote yet
1 vote

</code>...<code></code> ... <code>.">   Easy Math Editor

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list
  • bulleted
  • list

1. numbered
2. list
  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in </span>...<span></span> ... <span> or </span>...<span></span> ... <span> to ensure proper formatting.
2 \times 3 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai−1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 ∑i=13 \sum_{i=1}^3
\sin \theta sin⁡θ \sin \theta
\boxed{123} 123 \boxed{123}

Comments

Sort by:

Top Newest

28≡256≡249+7≡7 2^8 \equiv 256 \equiv 249 + 7 \equiv 7 (mod 8383 )
⇒216≡72 \Rightarrow 2^{16} \equiv 7^2 (mod 8383 )
⇒220≡49⋅16≡784≡37\Rightarrow 2^{20} \equiv 49 \cdot 16 \equiv 784 \equiv 37 (mod 8383 )

Ameya Daigavane - 3 years, 5 months ago

Log in to reply

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

abc xyz - 3 years, 5 months ago

Log in to reply

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 - 3 years, 5 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...