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Question-158843

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Question Number 158843 by mnjuly1970 last updated on 09/Nov/21
Commented by mr W last updated on 09/Nov/21
(1) not true! example: x=4, y=6 6x+11y=24+66=90 ≡0 mod(30) but x+7y=4+42=46 ≢0 mod(30)
$$\left(\mathrm{1}\right) \\ $$$${not}\:{true}! \\ $$$${example}:\:{x}=\mathrm{4},\:{y}=\mathrm{6} \\ $$$$\mathrm{6}{x}+\mathrm{11}{y}=\mathrm{24}+\mathrm{66}=\mathrm{90}\:\equiv\mathrm{0}\:{mod}\left(\mathrm{30}\right) \\ $$$${but}\:{x}+\mathrm{7}{y}=\mathrm{4}+\mathrm{42}=\mathrm{46}\:≢\mathrm{0}\:{mod}\left(\mathrm{30}\right) \\ $$
Commented by mnjuly1970 last updated on 09/Nov/21
yes you are right... thx mr W
$$\:\:{yes}\:\:{you}\:{are}\:{right}…\:{thx}\:{mr}\:{W} \\ $$
Answered by Rasheed.Sindhi last updated on 09/Nov/21
Q_2 5^9 +64=(5^3 )^3 +(4)^3 =(5^3 +4)( (5^3 )^2 −(5^3 )(4)+(4)^2 ) =(125+4)(15625−500+16) =129×15141 ∴ 5^9 +64 is composite.
$$\mathrm{Q}_{\mathrm{2}} \:\mathrm{5}^{\mathrm{9}} +\mathrm{64}=\left(\mathrm{5}^{\mathrm{3}} \right)^{\mathrm{3}} +\left(\mathrm{4}\right)^{\mathrm{3}} \\ $$$$\:\:\:\:\:=\left(\mathrm{5}^{\mathrm{3}} +\mathrm{4}\right)\left(\:\left(\mathrm{5}^{\mathrm{3}} \right)^{\mathrm{2}} −\left(\mathrm{5}^{\mathrm{3}} \right)\left(\mathrm{4}\right)+\left(\mathrm{4}\right)^{\mathrm{2}} \:\right) \\ $$$$\:\:\:=\left(\mathrm{125}+\mathrm{4}\right)\left(\mathrm{15625}−\mathrm{500}+\mathrm{16}\right) \\ $$$$\:\:\:=\mathrm{129}×\mathrm{15141} \\ $$$$\therefore\:\mathrm{5}^{\mathrm{9}} +\mathrm{64}\:{is}\:{composite}. \\ $$
Answered by Rasheed.Sindhi last updated on 09/Nov/21
Q_3 N=n^5 −5n^3 +4n =(n−2)(n−1)(n)(n+1)(n+2) (i)• n modulo 3: n=3k: Obviously 3 ∣ n⇒3 ∣ N n=3k+1: n+2=3k+3⇒3 ∣ (n+2)⇒3 ∣ N n=3k+2: n+1=3k+3⇒3 ∣ (n+1)⇒3 ∣ N ∴ 3 ∣ N...........................A (ii)• n modulo 2 n=2k: N=(2k−1)(2k−2)(2k)(2k+1)(2k+2) =8(4k^3 −5k^2 +k)⇒8 ∣N n=2k+1:N=(2k)(2k−1)(2k+1)(2k+2)(2k+3) =32k^5 +80k^4 +40k^3 −20k^2 −12k =8(4k^5 +10k^4 +5k^3 −((k(5k+3))/2) ) Can be proved that k(5k+3) is divisible by 2 in both cases:(i)k∈E or (ii)k∈O ∴ 8 ∣ N................................B (iii)• n modulo 5: n=5k⇒5 ∣ n⇒5 ∣ N n=5k+1⇒n−1=5k⇒5 ∣ n−1⇒5 ∣ N n=5k+2⇒n−2=5k⇒5 ∣ n−2⇒5 ∣ N n=5k+3⇒n+2=5(k+1)⇒5 ∣ n+2⇒5 ∣ N n=5k+4⇒n+1=5(k+1)⇒5 ∣ n+1⇒n ∣ N ∴ 5 ∣ N..............................C A & B & C⇒120 ∣ N
$$\mathrm{Q}_{\mathrm{3}} \:\:{N}={n}^{\mathrm{5}} −\mathrm{5}{n}^{\mathrm{3}} +\mathrm{4}{n} \\ $$$$=\left({n}−\mathrm{2}\right)\left({n}−\mathrm{1}\right)\left({n}\right)\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right) \\ $$$$\left(\mathrm{i}\right)\bullet\:\:{n}\:{modulo}\:\mathrm{3}:\: \\ $$$${n}=\mathrm{3}{k}:\:{Obviously}\:\mathrm{3}\:\mid\:{n}\Rightarrow\mathrm{3}\:\mid\:{N} \\ $$$${n}=\mathrm{3}{k}+\mathrm{1}:\:{n}+\mathrm{2}=\mathrm{3}{k}+\mathrm{3}\Rightarrow\mathrm{3}\:\mid\:\left({n}+\mathrm{2}\right)\Rightarrow\mathrm{3}\:\mid\:{N} \\ $$$${n}=\mathrm{3}{k}+\mathrm{2}:\:{n}+\mathrm{1}=\mathrm{3}{k}+\mathrm{3}\Rightarrow\mathrm{3}\:\mid\:\left({n}+\mathrm{1}\right)\Rightarrow\mathrm{3}\:\mid\:{N} \\ $$$$\therefore\:\mathrm{3}\:\mid\:{N}………………………{A} \\ $$$$\left(\mathrm{ii}\right)\bullet\:{n}\:{modulo}\:\mathrm{2} \\ $$$${n}=\mathrm{2}{k}:\:{N}=\left(\mathrm{2}{k}−\mathrm{1}\right)\left(\mathrm{2}{k}−\mathrm{2}\right)\left(\mathrm{2}{k}\right)\left(\mathrm{2}{k}+\mathrm{1}\right)\left(\mathrm{2}{k}+\mathrm{2}\right) \\ $$$$\:\:\:\:=\mathrm{8}\left(\mathrm{4}{k}^{\mathrm{3}} −\mathrm{5}{k}^{\mathrm{2}} +{k}\right)\Rightarrow\mathrm{8}\:\mid{N} \\ $$$${n}=\mathrm{2}{k}+\mathrm{1}:{N}=\left(\mathrm{2}{k}\right)\left(\mathrm{2}{k}−\mathrm{1}\right)\left(\mathrm{2}{k}+\mathrm{1}\right)\left(\mathrm{2}{k}+\mathrm{2}\right)\left(\mathrm{2}{k}+\mathrm{3}\right) \\ $$$$=\mathrm{32}{k}^{\mathrm{5}} +\mathrm{80}{k}^{\mathrm{4}} +\mathrm{40}{k}^{\mathrm{3}} −\mathrm{20}{k}^{\mathrm{2}} −\mathrm{12}{k} \\ $$$$=\mathrm{8}\left(\mathrm{4}{k}^{\mathrm{5}} +\mathrm{10}{k}^{\mathrm{4}} +\mathrm{5}{k}^{\mathrm{3}} −\frac{{k}\left(\mathrm{5}{k}+\mathrm{3}\right)}{\mathrm{2}}\:\right) \\ $$$${Can}\:{be}\:{proved}\:{that}\:{k}\left(\mathrm{5}{k}+\mathrm{3}\right)\:{is} \\ $$$$\:{divisible}\:{by}\:\mathrm{2}\:{in}\:{both}\:{cases}:\left({i}\right){k}\in\mathbb{E} \\ $$$${or}\:\left({ii}\right){k}\in\mathbb{O} \\ $$$$\therefore\:\mathrm{8}\:\mid\:{N}…………………………..{B} \\ $$$$\left(\mathrm{iii}\right)\bullet\:{n}\:{modulo}\:\mathrm{5}: \\ $$$${n}=\mathrm{5}{k}\Rightarrow\mathrm{5}\:\mid\:{n}\Rightarrow\mathrm{5}\:\mid\:{N} \\ $$$${n}=\mathrm{5}{k}+\mathrm{1}\Rightarrow{n}−\mathrm{1}=\mathrm{5}{k}\Rightarrow\mathrm{5}\:\mid\:{n}−\mathrm{1}\Rightarrow\mathrm{5}\:\mid\:{N} \\ $$$${n}=\mathrm{5}{k}+\mathrm{2}\Rightarrow{n}−\mathrm{2}=\mathrm{5}{k}\Rightarrow\mathrm{5}\:\mid\:{n}−\mathrm{2}\Rightarrow\mathrm{5}\:\mid\:{N} \\ $$$${n}=\mathrm{5}{k}+\mathrm{3}\Rightarrow{n}+\mathrm{2}=\mathrm{5}\left({k}+\mathrm{1}\right)\Rightarrow\mathrm{5}\:\mid\:{n}+\mathrm{2}\Rightarrow\mathrm{5}\:\mid\:{N} \\ $$$${n}=\mathrm{5}{k}+\mathrm{4}\Rightarrow{n}+\mathrm{1}=\mathrm{5}\left({k}+\mathrm{1}\right)\Rightarrow\mathrm{5}\:\mid\:{n}+\mathrm{1}\Rightarrow{n}\:\mid\:{N} \\ $$$$\therefore\:\mathrm{5}\:\mid\:{N}…………………………{C} \\ $$$${A}\:\&\:{B}\:\&\:{C}\Rightarrow\mathrm{120}\:\mid\:{N} \\ $$
Commented by mnjuly1970 last updated on 09/Nov/21
thanks alot sir
$${thanks}\:{alot}\:{sir} \\ $$

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