Relationship between factorials and squares of natural numbers

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SUMMARY

The discussion centers on the relationship between factorials and the squares of natural numbers, specifically through two equations involving factorials: \( n^2 = \frac{(n + 1)! - n!}{(n - 1)!} \) and \( (n + 1)^2 = \frac{(n + 1)! + n!}{(n - 1)!} + 1 \). Participants clarify terminology, confirming that "faculty" refers to "factorial" and "distribution law" refers to "distributive law." Additionally, the conversation touches on the conjecture that \( n! \) is never a perfect square for \( n > 1 \), suggesting a deeper exploration of prime factors and their coefficients.

PREREQUISITES
  • Understanding of factorial notation and properties
  • Familiarity with basic algebraic manipulation
  • Knowledge of prime factorization
  • Awareness of Bertrand's postulate
NEXT STEPS
  • Research the Gamma function and its relation to factorials
  • Explore the properties of perfect squares in relation to factorials
  • Study the distributive law in algebra
  • Investigate Bertrand's postulate and its implications in number theory
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Mathematicians, educators, and students interested in number theory, particularly those exploring the properties of factorials and their relationships to natural numbers.

Prez Cannady
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TL;DR
Two equations relating factorials with squares of natural numbers. They seem to work.
Was fooling around and wrote down these two equations today that appear to work. I'm not all that bright and I'm positive these either have some proof or restate some conjecture--probably something in a textbook. Could somebody help me out?

<br /> \forall n \in \mathbb{N}_0\smallsetminus\{0\}<br />
<br /> n^2 = \frac{\left(n + 1 \right)! - n!}{\left(n - 1 \right)!} \\<br />
<br /> \left(n + 1 \right)^2 = \frac{\left(n + 1 \right)! + n!}{\left(n - 1 \right)!} + 1 \\<br />
 
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The right hand side simplifies in each case as the denominator is a factor of the numerator. E.g ##\frac{n!}{(n-1)!}=n##
 
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This is basically the definition of the faculty and the distribution law. If you want to read more about the faculty, which formulas hold, and which generalizations exist, look up the Gamma function, and the Beta function.
 
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Is faculty an autocorrect for factorial?
 
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PeroK said:
Is faculty an autocorrect for factorial?
Lost in translation, sorry. (We use the same word for both meanings.)
 
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PeroK said:
The right hand side simplifies in each case as the denominator is a factor of the numerator. E.g ##\frac{n!}{(n-1)!}=n##

Indeed. Was just curious if there was a name for it or if I'm just writing down n^2 and (n + 1)^2 in a needlessly complicated fashion.
 
fresh_42 said:
Lost in translation, sorry. (We use the same word for both meanings.)
Just so I'm clear:

1. faculty -> factorial
2. distribution law -> distributive law

Is that correct?
 
Prez Cannady said:
Just so I'm clear:

1. faculty -> factorial
2. distribution law -> distributive law

Is that correct?
Yes.
 
@Prez Cannady :
This may interest you: Can you prove## n! ## is never a perfect square for ##n >1 ##?
 
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Prez Cannady said:
Indeed. Was just curious if there was a name for it or if I'm just writing down n^2 and (n + 1)^2 in a needlessly complicated fashion.
It's so easy to show that I would be surprised if it has gotten more attention.
##\frac{\left(n + 1 \right)! - n!}{\left(n - 1 \right)!} = \frac{(n+1)n(n-1)! - n(n-1)!}{(n-1)!} = \frac{n^2 (n-1)!}{(n-1)!} = n^2##

Factorial and faculty are both "Fakultät" in German.
 
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  • #11
Interesting... Would one start by separating the expansion to the product of composite factors multiplied by the product of prime factors (the latter can never be a perfect square)?
 
  • #12
valenumr said:
Interesting... Would one start by separating the expansion to the product of composite factors multiplied by the product of prime factors (the latter can never be a perfect square)?
After having chased that rabbit hole and having thought about this more, is it not sufficient to say simply that there is a maximal prime factor with coefficient (edit) exponent 1?
 
Last edited:
  • #13
That's the easiest way to show it, using Bertrand's postulate and checking the first few cases separately.
 
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