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Question Number 144482 by ArielVyny last updated on 25/Jun/21
show that ∀n∈Z E(((n−1)/2))+E(((n+2)/4))+E(((n+4)/4))=n
showthat∀n∈ZE(n−12)+E(n+24)+E(n+44)=n
Answered by Olaf_Thorendsen last updated on 25/Jun/21
x = E(((n−1)/2))+E(((n+2)/4))+E(((n+4)/4)) 1st case : n = 4p x = E(((4p−1)/2))+E(((4p+2)/4))+E(((4p+4)/4)) x = E(2p−(1/2))+E(p+(1/2))+E(p+1) x = (2p−1)+(p)+(p+1) x = 4p = n (1) 2nd case : n = 4p+1 x = E(((4p)/2))+E(((4p+3)/4))+E(((4p+5)/4)) x = E(2p)+E(p+(3/4))+E(p+(5/4)) x = (2p)+(p)+(p+1) x = 4p+1 = n (2) 3rd case : n = 4p+2 x = E(((4p+1)/2))+E(((4p+4)/4))+E(((4p+6)/4)) x = E(2p+(1/2))+E(p+1)+E(p+(3/2)) x = (2p)+(p+1)+(p+1) x = 4p+2 = n (3) 4th case : n = 4p+3 x = E(((4p+2)/2))+E(((4p+5)/4))+E(((4p+7)/4)) x = E(2p+1)+E(p+(5/4))+E(p+(7/4)) x = (2p+1)+(p+1)+(p+1) x = 4p+3 = n (4) (1), (2), (3), (4) : In all cases possible x = n
x=E(n−12)+E(n+24)+E(n+44)1stcase:n=4px=E(4p−12)+E(4p+24)+E(4p+44)x=E(2p−12)+E(p+12)+E(p+1)x=(2p−1)+(p)+(p+1)x=4p=n(1)2ndcase:n=4p+1x=E(4p2)+E(4p+34)+E(4p+54)x=E(2p)+E(p+34)+E(p+54)x=(2p)+(p)+(p+1)x=4p+1=n(2)3rdcase:n=4p+2x=E(4p+12)+E(4p+44)+E(4p+64)x=E(2p+12)+E(p+1)+E(p+32)x=(2p)+(p+1)+(p+1)x=4p+2=n(3)4thcase:n=4p+3x=E(4p+22)+E(4p+54)+E(4p+74)x=E(2p+1)+E(p+54)+E(p+74)x=(2p+1)+(p+1)+(p+1)x=4p+3=n(4)(1),(2),(3),(4):Inallcasespossiblex=n
Commented by ArielVyny last updated on 25/Jun/21
thank Mr olaf
thankMrolaf

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