(√(1! + 2! + 3! + ... + n!)) ∈ N n = ?


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Question Number 148154 by mathdanisur last updated on 25/Jul/21

$\sqrt{1! + 2! + 3! + . . . + n!} \in N$ $n = ?$
	Answered by puissant last updated on 25/Jul/21

	$n = 1$
	Commented by Olaf_Thorendsen last updated on 25/Jul/21

	$and n = 3$
	Commented by mathdanisur last updated on 25/Jul/21

	$Y e s S e r, b u t h o w$
	Commented by ajfour last updated on 25/Jul/21

	$y e s s i r, t r u l y, h o w ?$
	Answered by mindispower last updated on 25/Jul/21
	$Let n∈N suche Σ_(k=1) ^n k!=m^2 for n=1,n=3 we get solution ∀n≥4 Σ_(k=1) ^n k!=1+2!+3!+Σ_(k≥4) k!≡(1+2+6+24)[5] ≡3[5] n=5k,n^2 ≡0[5] n=(5k+_− 1)^2 ≡1[5] n=(5k+_− 2)≡4[5] so impossibl to get 3[5] ⇒∀n≥4; (√(Σ_(k=1) ^n k!))∉N n∈{1,3}$
	$L e t n \in N s u c h e$ $\underset{k = 1}{\sum^{n}} k! = m^{2}$ $f o r n = 1, n = 3 w e g e t s o l u t i o n$ $\forall n ⩾ 4$ $\underset{k = 1}{\sum^{n}} k! = 1 + 2! + 3! + \sum_{k ⩾ 4} k! \equiv (1 + 2 + 6 + 24) [5]$ $\equiv 3 [5]$ $n = 5 k, n^{2} \equiv 0 [5]$ $n = {(5 k \underline{+} 1)}^{2} \equiv 1 [5]$ $n = (5 k \underline{+} 2) \equiv 4 [5]$ $s o i m p o s s i b l t o g e t 3 [5]$ $\Rightarrow \forall n ⩾ 4; \sqrt{\underset{k = 1}{\sum^{n}} k!} \notin N$ $n \in {1, 3}$
	Commented by mathdanisur last updated on 26/Jul/21

	$T h a n k y o u S e r c o o l$
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