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Question Number 9485 by Joel575 last updated on 10/Dec/16

a_n = a_(n−1) ^2 + a_(n−2) ^2 If a_1 =1 and a_2 =1, what is the remainder of a_(2016) when divided by 10 ?

$${a}_{{n}} \:=\:{a}_{{n}−\mathrm{1}} ^{\mathrm{2}} \:+\:{a}_{{n}−\mathrm{2}} ^{\mathrm{2}} \\ $$$$\mathrm{If}\:{a}_{\mathrm{1}} =\mathrm{1}\:\mathrm{and}\:{a}_{\mathrm{2}} =\mathrm{1},\:\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{remainder}\:\mathrm{of}\:{a}_{\mathrm{2016}} \:\mathrm{when}\:\mathrm{divided}\:\mathrm{by}\:\mathrm{10}\:? \\ $$

Commented by sou1618 last updated on 10/Dec/16

(mod 10) a_3 =1^2 +1^2 =2 a_4 =1^2 +2^2 =5 a_5 =2^2 +5^2 =29≡9 a_6 =5^2 +29^2 ≡5^2 +9^2 ≡5+1=6 a_7 ≡9^2 +6^2 ≡1+6=7 a_8 ≡6^2 +7^2 ≡6+9≡5 a_9 ≡7^2 +5^2 ≡9+5≡4 a_(10) ≡5^2 +4^2 ≡5+6≡1 a_(11) ≡4^2 +1^2 ≡6+1=7 a_(12) ≡1^2 +7^2 ≡1+9≡0 a_(13) ≡7^2 +0^2 ≡9+0=9 a_(14) ≡0^2 +9^2 ≡0+1=1 a_(15) ≡9^2 +1^2 ≡1+1=2 a_2 ≡a_(14) ,a_3 ≡a_(15) a_(16) =a_(14) ^2 +a_(15) ^2 ≡a_2 ^2 +a_3 ^2 ≡a_4 a_(17) =a_(15) ^2 +a_(16) ^2 ≡a_3 ^2 +a_4 ^2 ≡a_5 .... (k=1,2,3...10,11,12 / n=0,1,2,3.....) a_(k+12n) ≡a_k a_(2016) =a_(12+12×167) ≡a_(12) =0 (∵2016=12×168) a_(2016) ≡0 (mod 10)

$$\left({mod}\:\mathrm{10}\right) \\ $$$${a}_{\mathrm{3}} =\mathrm{1}^{\mathrm{2}} +\mathrm{1}^{\mathrm{2}} =\mathrm{2} \\ $$$${a}_{\mathrm{4}} =\mathrm{1}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} =\mathrm{5} \\ $$$${a}_{\mathrm{5}} =\mathrm{2}^{\mathrm{2}} +\mathrm{5}^{\mathrm{2}} =\mathrm{29}\equiv\mathrm{9} \\ $$$${a}_{\mathrm{6}} =\mathrm{5}^{\mathrm{2}} +\mathrm{29}^{\mathrm{2}} \equiv\mathrm{5}^{\mathrm{2}} +\mathrm{9}^{\mathrm{2}} \equiv\mathrm{5}+\mathrm{1}=\mathrm{6} \\ $$$${a}_{\mathrm{7}} \equiv\mathrm{9}^{\mathrm{2}} +\mathrm{6}^{\mathrm{2}} \equiv\mathrm{1}+\mathrm{6}=\mathrm{7} \\ $$$${a}_{\mathrm{8}} \equiv\mathrm{6}^{\mathrm{2}} +\mathrm{7}^{\mathrm{2}} \equiv\mathrm{6}+\mathrm{9}\equiv\mathrm{5} \\ $$$${a}_{\mathrm{9}} \equiv\mathrm{7}^{\mathrm{2}} +\mathrm{5}^{\mathrm{2}} \equiv\mathrm{9}+\mathrm{5}\equiv\mathrm{4} \\ $$$${a}_{\mathrm{10}} \equiv\mathrm{5}^{\mathrm{2}} +\mathrm{4}^{\mathrm{2}} \equiv\mathrm{5}+\mathrm{6}\equiv\mathrm{1} \\ $$$${a}_{\mathrm{11}} \equiv\mathrm{4}^{\mathrm{2}} +\mathrm{1}^{\mathrm{2}} \equiv\mathrm{6}+\mathrm{1}=\mathrm{7} \\ $$$${a}_{\mathrm{12}} \equiv\mathrm{1}^{\mathrm{2}} +\mathrm{7}^{\mathrm{2}} \equiv\mathrm{1}+\mathrm{9}\equiv\mathrm{0} \\ $$$${a}_{\mathrm{13}} \equiv\mathrm{7}^{\mathrm{2}} +\mathrm{0}^{\mathrm{2}} \equiv\mathrm{9}+\mathrm{0}=\mathrm{9} \\ $$$${a}_{\mathrm{14}} \equiv\mathrm{0}^{\mathrm{2}} +\mathrm{9}^{\mathrm{2}} \equiv\mathrm{0}+\mathrm{1}=\mathrm{1} \\ $$$${a}_{\mathrm{15}} \equiv\mathrm{9}^{\mathrm{2}} +\mathrm{1}^{\mathrm{2}} \equiv\mathrm{1}+\mathrm{1}=\mathrm{2} \\ $$$$ \\ $$$${a}_{\mathrm{2}} \equiv{a}_{\mathrm{14}} ,{a}_{\mathrm{3}} \equiv{a}_{\mathrm{15}} \\ $$$${a}_{\mathrm{16}} ={a}_{\mathrm{14}} ^{\mathrm{2}} +{a}_{\mathrm{15}} ^{\mathrm{2}} \equiv{a}_{\mathrm{2}} ^{\mathrm{2}} +{a}_{\mathrm{3}} ^{\mathrm{2}} \equiv{a}_{\mathrm{4}} \\ $$$${a}_{\mathrm{17}} ={a}_{\mathrm{15}} ^{\mathrm{2}} +{a}_{\mathrm{16}} ^{\mathrm{2}} \equiv{a}_{\mathrm{3}} ^{\mathrm{2}} +{a}_{\mathrm{4}} ^{\mathrm{2}} \equiv{a}_{\mathrm{5}} \\ $$$$.... \\ $$$$\left({k}=\mathrm{1},\mathrm{2},\mathrm{3}...\mathrm{10},\mathrm{11},\mathrm{12}\:\:/\:{n}=\mathrm{0},\mathrm{1},\mathrm{2},\mathrm{3}.....\right) \\ $$$${a}_{{k}+\mathrm{12}{n}} \equiv{a}_{{k}} \\ $$$${a}_{\mathrm{2016}} ={a}_{\mathrm{12}+\mathrm{12}×\mathrm{167}} \equiv{a}_{\mathrm{12}} =\mathrm{0}\:\:\left(\because\mathrm{2016}=\mathrm{12}×\mathrm{168}\right) \\ $$$$ \\ $$$${a}_{\mathrm{2016}} \equiv\mathrm{0}\:\left({mod}\:\mathrm{10}\right) \\ $$$$ \\ $$

Commented by Joel575 last updated on 11/Dec/16

thank you very much

$${thank}\:{you}\:{very}\:{much} \\ $$

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