- #1
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]
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One guy said that
But, I don't understand him.Complete square divided by 5 can give the remains 0,1 and 4, but not 3 like here.
Sorry for bad English.