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Question Number 156951 by mr W last updated on 17/Oct/21
solve for n∈N (n−1)!+1=n^2
$${solve}\:{for}\:{n}\in{N} \\ $$$$\left({n}−\mathrm{1}\right)!+\mathrm{1}={n}^{\mathrm{2}} \\ $$
Commented by mr W last updated on 17/Oct/21
thanks sir! if n=1, then (n−2)!−(n+1) is not defined. therefore n≠1.
$${thanks}\:{sir}! \\ $$$${if}\:{n}=\mathrm{1},\:{then}\:\left({n}−\mathrm{2}\right)!−\left({n}+\mathrm{1}\right)\:{is}\:{not} \\ $$$${defined}.\:{therefore}\:{n}\neq\mathrm{1}. \\ $$
Commented by Rasheed.Sindhi last updated on 17/Oct/21
(n−1)!+1=n^2 (n−1)!−(n^2 −1)=0 (n−1){(n−2)!−(n+1)}=0 n=1 ∣ (n−2)!−(n+1)=0 But n=1 is false because it doesn′t fulfill the original.Why is it false? (n−2)!=n+1 ;n≥2 2≤n≤4: (n−2)!<n+1 n=5: (5−2)!=5+1✓ n>5: (n−2)!>n+1 n=5
$$\left({n}−\mathrm{1}\right)!+\mathrm{1}={n}^{\mathrm{2}} \\ $$$$\left({n}−\mathrm{1}\right)!−\left({n}^{\mathrm{2}} −\mathrm{1}\right)=\mathrm{0} \\ $$$$\left({n}−\mathrm{1}\right)\left\{\left({n}−\mathrm{2}\right)!−\left({n}+\mathrm{1}\right)\right\}=\mathrm{0} \\ $$$${n}=\mathrm{1}\:\mid\:\left({n}−\mathrm{2}\right)!−\left({n}+\mathrm{1}\right)=\mathrm{0} \\ $$$${But}\:{n}=\mathrm{1}\:{is}\:{false}\:{because}\:{it}\:{doesn}'{t} \\ $$$${fulfill}\:{the}\:{original}.{Why}\:{is}\:{it}\:{false}? \\ $$$$\left({n}−\mathrm{2}\right)!={n}+\mathrm{1}\:;{n}\geqslant\mathrm{2} \\ $$$$\:\:\:\:\mathrm{2}\leqslant{n}\leqslant\mathrm{4}:\:\left({n}−\mathrm{2}\right)!<{n}+\mathrm{1} \\ $$$$\:\:\:\:{n}=\mathrm{5}:\:\:\:\:\:\:\:\left(\mathrm{5}−\mathrm{2}\right)!=\mathrm{5}+\mathrm{1}\checkmark \\ $$$$\:\:\:\:{n}>\mathrm{5}:\:\:\:\:\:\:\left({n}−\mathrm{2}\right)!>{n}+\mathrm{1}\:\:\:\:\:\:\:\:\:\: \\ $$$${n}=\mathrm{5}\:\:\: \\ $$
Commented by Rasheed.Sindhi last updated on 18/Oct/21
mr W Sir^(Thanks) ! You′ve always helped me. ...I thought it only matter of the definition: 0!=1 If it were defined 0!=0,the solution x=1 were not false, although the other factor be undefined for it. Afterall ′0!=1′ is a definition which is made for some necessities.What′s your openion sir?Your openion matters to me.
$$\overset{\mathrm{Thanks}} {\mathrm{mr}\:\mathrm{W}\:\mathrm{Sir}}!\:\mathrm{You}'\mathrm{ve}\:\mathrm{always}\:\mathrm{helped}\:\mathrm{me}. \\ $$$$…\mathrm{I}\:\mathrm{thought}\:\mathrm{it}\:\mathrm{only}\:\mathrm{matter}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{definition}:\:\:\:\mathrm{0}!=\mathrm{1} \\ $$$$\mathrm{If}\:\mathrm{it}\:\mathrm{were}\:\mathrm{defined}\:\mathrm{0}!=\mathrm{0},\mathrm{the}\:\mathrm{solution} \\ $$$$\mathrm{x}=\mathrm{1}\:\mathrm{were}\:\mathrm{not}\:\mathrm{false},\:\mathrm{although}\:\mathrm{the} \\ $$$$\mathrm{other}\:\mathrm{factor}\:\mathrm{be}\:\mathrm{undefined}\:\mathrm{for}\:\mathrm{it}. \\ $$$$\mathrm{Afterall}\:'\mathrm{0}!=\mathrm{1}'\:\mathrm{is}\:\mathrm{a}\:\mathrm{definition}\:\mathrm{which} \\ $$$$\mathrm{is}\:\mathrm{made}\:\mathrm{for}\:\mathrm{some}\:\mathrm{necessities}.\mathrm{What}'\mathrm{s} \\ $$$$\mathrm{your}\:\mathrm{openion}\:\mathrm{sir}?\mathrm{Your}\:\mathrm{openion}\:\mathrm{matters} \\ $$$$\mathrm{to}\:\mathrm{me}. \\ $$
Commented by mr W last updated on 18/Oct/21
i think if 0! is defined, then 0! must be equal to 1. we can not further define 0!=0, since otherwise we would get 1!=1×0!=1×0=0, 2!=2×1!=0 etc.
$${i}\:{think}\:{if}\:\mathrm{0}!\:{is}\:{defined},\:{then}\:\mathrm{0}!\:{must} \\ $$$${be}\:{equal}\:{to}\:\mathrm{1}.\:{we}\:{can}\:{not}\:{further} \\ $$$${define}\:\mathrm{0}!=\mathrm{0},\:{since}\:{otherwise}\:{we}\: \\ $$$${would}\:{get}\:\mathrm{1}!=\mathrm{1}×\mathrm{0}!=\mathrm{1}×\mathrm{0}=\mathrm{0}, \\ $$$$\mathrm{2}!=\mathrm{2}×\mathrm{1}!=\mathrm{0}\:{etc}. \\ $$
Commented by Rasheed.Sindhi last updated on 18/Oct/21
Yes Sir,you′re right!
$${Yes}\:{Sir},{you}'{re}\:{right}! \\ $$
Commented by mr W last updated on 18/Oct/21
0! is defined! so it is ok with (n−1)!−(n^2 −1)=0 and you can put n=1 into it. but if you factorise it to (n−1){(n−2)!−(n+1)}=0 (n−2)! must be defined, that means n−2≥0.
$$\mathrm{0}!\:{is}\:{defined}!\:{so}\:{it}\:{is}\:{ok}\:{with} \\ $$$$\left({n}−\mathrm{1}\right)!−\left({n}^{\mathrm{2}} −\mathrm{1}\right)=\mathrm{0} \\ $$$${and}\:{you}\:{can}\:{put}\:{n}=\mathrm{1}\:{into}\:{it}. \\ $$$${but}\:{if}\:{you}\:{factorise}\:{it}\:{to} \\ $$$$\left({n}−\mathrm{1}\right)\left\{\left({n}−\mathrm{2}\right)!−\left({n}+\mathrm{1}\right)\right\}=\mathrm{0} \\ $$$$\left({n}−\mathrm{2}\right)!\:{must}\:{be}\:{defined},\:{that}\:{means} \\ $$$${n}−\mathrm{2}\geqslant\mathrm{0}. \\ $$
Commented by Rasheed.Sindhi last updated on 18/Oct/21
Thanks very much! But I think n≥2 is only for other values of n. Sir I only talk about supposition. If 0!=0 ,then... Then the solution may contain 1 or not?
$$\mathbb{T}\mathrm{han}\Bbbk\mathrm{s}\:\mathrm{very}\:\mathrm{much}! \\ $$$$\mathcal{B}{ut}\:{I}\:{think}\:{n}\geqslant\mathrm{2}\:{is}\:{only}\:{for}\:{other} \\ $$$${values}\:{of}\:{n}. \\ $$$${Sir}\:{I}\:{only}\:{talk}\:{about}\:{supposition}. \\ $$$${If}\:\mathrm{0}!=\mathrm{0}\:,{then}… \\ $$$$\mathcal{T}{hen}\:{the}\:{solution}\:{may}\:{contain}\:\mathrm{1} \\ $$$${or}\:{not}? \\ $$
Answered by mindispower last updated on 17/Oct/21
⇔n^2 −1−(n−1)!=0,n≥1,n=1not solution ∀n,N−{0,1}=E ⇔(n−1)(n+1−(n−2)!)=0 (n+1)=(n−2)! n≥5 (n−2)!≥n+1...? n=5 3!≥6 true suppose ∀n≥5 (n−2)!≥n+1,⇒(n−1)!≥n+2 (n−1)!=(n−2)!.(n−1)≥(n+1)(n−1)=n^2 −1 n.n−1≥5n−1=4n+n−1≥20−1+n=n+19>2n+2 ⇒(n−1)!>n+2 ⇒∀n≥5 (n−2)!≥(n+1),∀n>6 (n−2)!>n+1 just try n∈{2,3,4,5) 2⇒1+1=2^2 ,n=3⇒3=9,n=4⇒7=16 n=5⇒25=5^2 ⇒n=5
$$\Leftrightarrow{n}^{\mathrm{2}} −\mathrm{1}−\left({n}−\mathrm{1}\right)!=\mathrm{0},{n}\geqslant\mathrm{1},{n}=\mathrm{1}{not}\:{solution} \\ $$$$\forall{n},\mathbb{N}−\left\{\mathrm{0},\mathrm{1}\right\}={E} \\ $$$$\Leftrightarrow\left({n}−\mathrm{1}\right)\left({n}+\mathrm{1}−\left({n}−\mathrm{2}\right)!\right)=\mathrm{0} \\ $$$$\left({n}+\mathrm{1}\right)=\left({n}−\mathrm{2}\right)! \\ $$$${n}\geqslant\mathrm{5} \\ $$$$\left({n}−\mathrm{2}\right)!\geqslant{n}+\mathrm{1}…? \\ $$$${n}=\mathrm{5}\:\mathrm{3}!\geqslant\mathrm{6}\:{true} \\ $$$${suppose}\:\forall{n}\geqslant\mathrm{5}\:\left({n}−\mathrm{2}\right)!\geqslant{n}+\mathrm{1},\Rightarrow\left({n}−\mathrm{1}\right)!\geqslant{n}+\mathrm{2} \\ $$$$\left({n}−\mathrm{1}\right)!=\left({n}−\mathrm{2}\right)!.\left({n}−\mathrm{1}\right)\geqslant\left({n}+\mathrm{1}\right)\left({n}−\mathrm{1}\right)={n}^{\mathrm{2}} −\mathrm{1} \\ $$$${n}.{n}−\mathrm{1}\geqslant\mathrm{5}{n}−\mathrm{1}=\mathrm{4}{n}+{n}−\mathrm{1}\geqslant\mathrm{20}−\mathrm{1}+{n}={n}+\mathrm{19}>\mathrm{2}{n}+\mathrm{2} \\ $$$$\Rightarrow\left({n}−\mathrm{1}\right)!>{n}+\mathrm{2} \\ $$$$\Rightarrow\forall{n}\geqslant\mathrm{5}\:\left({n}−\mathrm{2}\right)!\geqslant\left({n}+\mathrm{1}\right),\forall{n}>\mathrm{6}\:\left({n}−\mathrm{2}\right)!>{n}+\mathrm{1} \\ $$$${just}\:{try}\:{n}\in\left\{\mathrm{2},\mathrm{3},\mathrm{4},\mathrm{5}\right) \\ $$$$\mathrm{2}\Rightarrow\mathrm{1}+\mathrm{1}=\mathrm{2}^{\mathrm{2}} ,{n}=\mathrm{3}\Rightarrow\mathrm{3}=\mathrm{9},{n}=\mathrm{4}\Rightarrow\mathrm{7}=\mathrm{16} \\ $$$${n}=\mathrm{5}\Rightarrow\mathrm{25}=\mathrm{5}^{\mathrm{2}} \\ $$$$\Rightarrow{n}=\mathrm{5} \\ $$$$ \\ $$$$ \\ $$
Commented by mr W last updated on 17/Oct/21
thanks sir!
$${thanks}\:{sir}! \\ $$
Commented by mindispower last updated on 18/Oct/21
withe pleasur sir
$${withe}\:{pleasur}\:{sir}\: \\ $$

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