Can the Sum of Factorials Ever be a Complete Square?

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SUMMARY

The discussion centers on proving that the sum of factorials, expressed as \(\sum_{i=1}^{n}i!\), cannot be a complete square for \(n \geq 5\). The initial calculation for \(n=5\) yields a sum of 153, which is not a perfect square. A key argument presented is that complete squares modulo 5 can only yield remainders of 0, 1, or 4, but never 3, which is the remainder obtained from the sum of factorials starting from \(n=5\). This establishes a definitive conclusion regarding the impossibility of the sum being a complete square for the specified range.

PREREQUISITES
  • Understanding of factorial notation and properties
  • Familiarity with mathematical induction techniques
  • Knowledge of modular arithmetic, specifically modulo 5
  • Basic concepts of perfect squares and their properties
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  • Study mathematical induction proofs in depth
  • Explore modular arithmetic and its applications in number theory
  • Investigate properties of factorials and their growth rates
  • Learn about perfect squares and their characteristics in various number systems
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Mathematics students, educators, and enthusiasts interested in number theory, particularly those exploring properties of factorials and perfect squares.

Karamata
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Homework Statement


Proof that for [tex]n\geq 5[/tex] expression [tex]\sum_{i=1}^{n}i![/tex] can't be a complete square


The Attempt at a Solution


Mathematical induction maybe?
[tex]n=5[/tex] [tex]\sum_{i=1}^{5}i!=1+2!+3!+4!+5!=152[/tex] OK
[tex]n\rightarrow n+1[/tex] [tex]\sum_{i=1}^{n+1}i!=\sum_{i=1}^{n}i!+(n+1)![/tex]
...
One guy said that
Complete square divided by 5 can give the remains 0,1 and 4, but not 3 like here.
But, I don't understand him.

Sorry for bad English.
 
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1+2+6+24+120 = 153 right?
subsequent factorials are all divisible by 10, so dividing by 5 will always give the remainder 3.
 

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