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Question Number 63645 by mathmax by abdo last updated on 06/Jul/19

n integr natural prove that 5 divide n^5 −n

$${n}\:{integr}\:{natural}\:{prove}\:{that}\:\mathrm{5}\:{divide}\:{n}^{\mathrm{5}} −{n} \\ $$

Commented by Prithwish sen last updated on 06/Jul/19

n=5k ⇒n^5 −n divided by 5 n=5k+1 ⇒(5k+1)^(5 ) −(5k+1)= 5m divided by 5 n=5k+2⇒(5k+2)^5 −(5k+2)=5m+30 divided by 5 n = 5k+3⇒(5k+3)^5 −(5k+3)=5m+240 divided by 5 n = 5k + 4⇒(5k+4)^5 −(5k+4)=5m+1020 divided by 5 Hence proved.

$$\mathrm{n}=\mathrm{5k}\:\Rightarrow\mathrm{n}^{\mathrm{5}} −\mathrm{n}\:\mathrm{divided}\:\mathrm{by}\:\mathrm{5} \\ $$$$\mathrm{n}=\mathrm{5k}+\mathrm{1}\:\Rightarrow\left(\mathrm{5k}+\mathrm{1}\right)^{\mathrm{5}\:} −\left(\mathrm{5k}+\mathrm{1}\right)=\:\mathrm{5m}\:\:\mathrm{divided}\:\mathrm{by}\:\mathrm{5} \\ $$$$\mathrm{n}=\mathrm{5k}+\mathrm{2}\Rightarrow\left(\mathrm{5k}+\mathrm{2}\right)^{\mathrm{5}} −\left(\mathrm{5k}+\mathrm{2}\right)=\mathrm{5m}+\mathrm{30}\:\mathrm{divided}\:\mathrm{by}\:\mathrm{5} \\ $$$$\mathrm{n}\:=\:\mathrm{5k}+\mathrm{3}\Rightarrow\left(\mathrm{5k}+\mathrm{3}\right)^{\mathrm{5}} −\left(\mathrm{5k}+\mathrm{3}\right)=\mathrm{5m}+\mathrm{240}\:\mathrm{divided}\:\mathrm{by}\:\mathrm{5} \\ $$$$\mathrm{n}\:=\:\mathrm{5k}\:+\:\mathrm{4}\Rightarrow\left(\mathrm{5k}+\mathrm{4}\right)^{\mathrm{5}} −\left(\mathrm{5k}+\mathrm{4}\right)=\mathrm{5m}+\mathrm{1020}\:\mathrm{divided}\:\mathrm{by}\:\mathrm{5} \\ $$$$\mathrm{Hence}\:\mathrm{proved}. \\ $$

Commented by Prithwish sen last updated on 06/Jul/19

Another method n^5 −n=n(n−1)(n+1)(n^2 +1) for n=5k , n is divisible by 5 for n=5k+1,(n−1) is divisible by 5 for n=5k+2, (n^2 +1) is divosible by 5 for n = 5k+3, (n^2 +1) is divisible by 5 for n = 5k+4, (n+1) is divisibleby 5 ∴ n^5 −n is divisible by 5 . Hence proved.

$$\mathrm{Another}\:\mathrm{method} \\ $$$$\mathrm{n}^{\mathrm{5}} −\mathrm{n}=\mathrm{n}\left(\mathrm{n}−\mathrm{1}\right)\left(\mathrm{n}+\mathrm{1}\right)\left(\mathrm{n}^{\mathrm{2}} +\mathrm{1}\right) \\ $$$$\mathrm{for}\:\mathrm{n}=\mathrm{5k}\:,\:\mathrm{n}\:\mathrm{is}\:\mathrm{divisible}\:\mathrm{by}\:\mathrm{5} \\ $$$$\mathrm{for}\:\mathrm{n}=\mathrm{5k}+\mathrm{1},\left(\mathrm{n}−\mathrm{1}\right)\:\mathrm{is}\:\mathrm{divisible}\:\mathrm{by}\:\mathrm{5}\: \\ $$$$\mathrm{for}\:\mathrm{n}=\mathrm{5k}+\mathrm{2},\:\left(\mathrm{n}^{\mathrm{2}} +\mathrm{1}\right)\:\mathrm{is}\:\mathrm{divosible}\:\mathrm{by}\:\mathrm{5} \\ $$$$\mathrm{for}\:\mathrm{n}\:=\:\mathrm{5k}+\mathrm{3},\:\left(\mathrm{n}^{\mathrm{2}} +\mathrm{1}\right)\:\mathrm{is}\:\mathrm{divisible}\:\mathrm{by}\:\mathrm{5} \\ $$$$\mathrm{for}\:\mathrm{n}\:=\:\mathrm{5k}+\mathrm{4},\:\left(\mathrm{n}+\mathrm{1}\right)\:\mathrm{is}\:\mathrm{divisibleby}\:\mathrm{5} \\ $$$$\therefore\:\boldsymbol{\mathrm{n}}^{\mathrm{5}} −\boldsymbol{\mathrm{n}}\:\boldsymbol{\mathrm{is}}\:\boldsymbol{\mathrm{divisible}}\:\boldsymbol{\mathrm{by}}\:\mathrm{5}\:.\:\boldsymbol{\mathrm{H}}\mathrm{ence}\:\mathrm{proved}. \\ $$$$ \\ $$$$ \\ $$

Commented by mathmax by abdo last updated on 06/Jul/19

thank you sir.

$${thank}\:{you}\:{sir}. \\ $$

Commented by Prithwish sen last updated on 07/Jul/19

welcome

$$\mathrm{welcome} \\ $$

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