Problem from a Theorem

If pp is an odd prime number, then prove that ((p−12)!)2+(−1)p−12 \left( \left( \dfrac{p-1}{2} \right)! \right)^2 + (-1)^\frac{p-1}{2} is divisible by pp.

#NumberTheory

Note by Fahim Shahriar Shakkhor
6 years, 11 months ago

No vote yet
1 vote

  Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

  • Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
  • Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
  • Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
  • Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

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 \( ... \) or \[ ... \] 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

From Wilson's Theorem, (p−1)!≡−1(mod  p)(p-1)! \equiv -1 (\mod p)

⇒1×2×........×p−12×p+12×..........×(p−2)×(p−1)≡−1(mod  p)\Rightarrow 1\times2\times........\times \frac{p-1}{2}\times\frac{p+1}{2} \times ..........\times (p-2)\times (p-1) \equiv -1 (\mod p)

⇒1×2×........×p−12×(−p−12)×..........×(−2)×(−1)≡−1(mod  p)\Rightarrow 1\times2\times........\times \frac{p-1}{2}\times(-\frac{p-1}{2}) \times ..........\times (-2)\times (-1) \equiv -1 (\mod p)

⇒((p−12)!)2×(−1)p−12≡−1(mod  p)\Rightarrow ((\frac{p-1}{2})!)^2 \times (-1)^{\frac{p-1}{2}} \equiv -1 (\mod p)

⇒((p−12)!)2≡(−1)×(−1)p−12(mod  p)\Rightarrow ((\frac{p-1}{2})!)^2 \equiv (-1)\times(-1)^{\frac{p-1}{2}} (\mod p)

⇒((p−12)!)2+(−1)p−12≡0(mod  p)\Rightarrow ((\frac{p-1}{2})!)^2 + (-1)^{\frac{p-1}{2}} \equiv 0 (\mod p)

Fahim Shahriar Shakkhor - 6 years, 11 months ago

Nice and tricky!

For convenience, we let p=2n+1p=2n+1 and so we wanna prove that (n!)2=n!⋅n!≡(−1)n+1(modp)(n!)^2=n!\cdot n!\equiv (-1)^{n+1}\pmod p. Now note that n≡n−p=−(p−n)(modp)n\equiv n-p=-(p-n)\pmod p. So

n!=∏k=1nk≡∏k=1n−(p−k)=(−1)n∏k=1n(p−k)=(−1)n∏k=1n(2n+1−k)=(−1)n∏k=n+12nk(modp).n!=\prod_{k=1}^n k \equiv \prod_{k=1}^n -(p-k)=(-1)^n \prod_{k=1}^n (p-k)=(-1)^n \prod_{k=1}^n (2n+1-k)=(-1)^n \prod_{k=n+1}^{2n} k\pmod p.

And hence we have

n!⋅n!≡n!⋅(−1)n∏k=n+12nk=(−1)n⋅(2n)!=(−1)n⋅(p−1)!≡(−1)n⋅(−1)=(−1)n+1(modp)n!\cdot n! \equiv n!\cdot (-1)^n \prod_{k=n+1}^{2n} k=(-1)^n\cdot (2n)!=(-1)^n\cdot (p-1)!\equiv (-1)^n\cdot (-1)=(-1)^{n+1}\pmod p

and the last congruence step follows by Wilson's theorem.

Jubayer Nirjhor - 6 years, 11 months ago
×

Problem Loading...

Note Loading...

Set Loading...