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Question Number 119580 by mathocean1 last updated on 25/Oct/20

Given k ∈ N. 1) justify these relations: 3^(2k) +1≡2[8] and 3^(2k+1) +1≡4[8]. 2) Given (E): 2^n −3^m =1. n and m are unknowed. • Show that if m is even , (E) does not have solution. ■ Deduct from the first question 1) that the couple (2;1) is the only solution of (E).

$$\mathrm{Given}\:\mathrm{k}\:\in\:\mathbb{N}. \\ $$$$\left.\mathrm{1}\right)\:\mathrm{justify}\:\mathrm{these}\:\mathrm{relations}: \\ $$$$\mathrm{3}^{\mathrm{2k}} +\mathrm{1}\equiv\mathrm{2}\left[\mathrm{8}\right]\:\mathrm{and}\:\mathrm{3}^{\mathrm{2k}+\mathrm{1}} +\mathrm{1}\equiv\mathrm{4}\left[\mathrm{8}\right]. \\ $$$$\left.\mathrm{2}\right)\:\mathrm{Given}\:\left(\mathrm{E}\right):\:\mathrm{2}^{\mathrm{n}} −\mathrm{3}^{\mathrm{m}} =\mathrm{1}.\:\mathrm{n}\:\mathrm{and}\:\mathrm{m}\:\mathrm{are}\:\mathrm{unknowed}. \\ $$$$\bullet\:\mathrm{Show}\:\mathrm{that}\:\mathrm{if}\:\mathrm{m}\:\mathrm{is}\:\mathrm{even}\:,\:\left(\mathrm{E}\right)\:\mathrm{does}\:\mathrm{not}\:\mathrm{have}\: \\ $$$$\mathrm{solution}. \\ $$$$\left.\blacksquare\:\mathrm{Deduct}\:\mathrm{from}\:\mathrm{the}\:\mathrm{first}\:\mathrm{question}\:\mathrm{1}\right)\:\mathrm{that}\:\mathrm{the} \\ $$$$\mathrm{couple}\:\left(\mathrm{2};\mathrm{1}\right)\:\mathrm{is}\:\mathrm{the}\:\mathrm{only}\:\mathrm{solution}\:\mathrm{of}\:\left(\mathrm{E}\right). \\ $$

Answered by mindispower last updated on 25/Oct/20

3^(2k+1) +1≡4[8] ?

$$\mathrm{3}^{\mathrm{2}{k}+\mathrm{1}} +\mathrm{1}\equiv\mathrm{4}\left[\mathrm{8}\right]\:? \\ $$

Commented by mathocean1 last updated on 25/Oct/20

yes sir; or there is an error ?

$$\mathrm{yes}\:\mathrm{sir};\:\mathrm{or}\:\mathrm{there}\:\mathrm{is}\:\mathrm{an}\:\mathrm{error}\:? \\ $$

Answered by mindispower last updated on 25/Oct/20

3^(2k) +1=9^k +1≡1^k +1=2[8] 3^(2k+1) +1=9^k .3+1≡3.(1)^k +1≡4]8] (E)⇔2^n −(3^m +1)=0 m=2k ⇒2^n =(1+3^(2k) )≡2[8] withe 1⇒n=1 but n=1 (E)⇒2−9^k =1⇒k=0 (E) has solution (n,m)=(1,0) m=2k+1 ⇒2^n =(1+3^(2k+1) )≡4(8)...withe Quation 1 ⇒n=2, 4−3.9^k =1⇒k=0 m=2.0+1=1,n=2 solution are(n,m)∈{(1,0);(2,1)}

$$\mathrm{3}^{\mathrm{2}{k}} +\mathrm{1}=\mathrm{9}^{{k}} +\mathrm{1}\equiv\mathrm{1}^{{k}} +\mathrm{1}=\mathrm{2}\left[\mathrm{8}\right] \\ $$$$\left.\mathrm{3}^{\mathrm{2}{k}+\mathrm{1}} \left.+\mathrm{1}=\mathrm{9}^{{k}} .\mathrm{3}+\mathrm{1}\equiv\mathrm{3}.\left(\mathrm{1}\right)^{{k}} +\mathrm{1}\equiv\mathrm{4}\right]\mathrm{8}\right] \\ $$$$\left({E}\right)\Leftrightarrow\mathrm{2}^{{n}} −\left(\mathrm{3}^{{m}} +\mathrm{1}\right)=\mathrm{0} \\ $$$${m}=\mathrm{2}{k} \\ $$$$\Rightarrow\mathrm{2}^{{n}} =\left(\mathrm{1}+\mathrm{3}^{\mathrm{2}{k}} \right)\equiv\mathrm{2}\left[\mathrm{8}\right] \\ $$$${withe}\:\mathrm{1}\Rightarrow{n}=\mathrm{1} \\ $$$${but}\:{n}=\mathrm{1}\:\left({E}\right)\Rightarrow\mathrm{2}−\mathrm{9}^{{k}} =\mathrm{1}\Rightarrow{k}=\mathrm{0} \\ $$$$\left({E}\right)\:{has}\:{solution}\:\left({n},{m}\right)=\left(\mathrm{1},\mathrm{0}\right) \\ $$$${m}=\mathrm{2}{k}+\mathrm{1} \\ $$$$\Rightarrow\mathrm{2}^{{n}} =\left(\mathrm{1}+\mathrm{3}^{\mathrm{2}{k}+\mathrm{1}} \right)\equiv\mathrm{4}\left(\mathrm{8}\right)...{withe}\:{Quation}\:\mathrm{1} \\ $$$$\Rightarrow{n}=\mathrm{2}, \\ $$$$\mathrm{4}−\mathrm{3}.\mathrm{9}^{{k}} =\mathrm{1}\Rightarrow{k}=\mathrm{0} \\ $$$${m}=\mathrm{2}.\mathrm{0}+\mathrm{1}=\mathrm{1},{n}=\mathrm{2} \\ $$$${solution}\:{are}\left({n},{m}\right)\in\left\{\left(\mathrm{1},\mathrm{0}\right);\left(\mathrm{2},\mathrm{1}\right)\right\} \\ $$$$ \\ $$

Answered by 1549442205PVT last updated on 25/Oct/20

1a−We have 3^(2k) +1=9^k +1=(8+1)^k +1 =Σ_(m=0) ^(k) C_k ^m 8^(k−m) =(8^k +k.8^(k−1) +((k(k−))/2)8^(k−2) +...+8+1)+1≡2(mod8)(q.e.d) b−By above proof we have 3^(2k) +1=8q+2 ⇒3^(2k+1) +1=3.(3^(2k) +1)−2=3.(8q+2) −2=3.8q+4≡4(mod8)( q.e.d) 2)a−Consider the equation 2^n −3^m =1 If m is even then 2^n =3^(2k) +1(∗) (by above proof ).For n=0,1,2 we see the equation (∗) has no roots For n≥3 L.H.S (∗)is divisible by 8 while R.H.S isn′t divisible 8 since 3^(2k) +1≡2(mod8)(by above proof). Hence this case the equation has no roots (q.e.d) b−We see that (by directly checking) the (n,m)=(2,1) satisfy the equation 2^n −3^m =1(1).We prove that is unique solution of given equation.Indeed, When m is even the equation has no roots (above proof).We now consider the case m is odd.⇒m=2k+1.Then (1)⇔2^n =3^(2k+1) +1(∗∗) For n=0,1 It is clear that the equation (∗∗)has no roots For n≥2 L.H.S(∗∗)is divisible by 4 while R.H.S (∗∗)isn′t divisible (by above proof 1b).Hence this case the given has no roots which means the (n,m)=(2,1) is unique solution of the given equation (q.e.d)

$$\mathrm{1a}−\mathrm{We}\:\mathrm{have}\:\mathrm{3}^{\mathrm{2k}} +\mathrm{1}=\mathrm{9}^{\mathrm{k}} +\mathrm{1}=\left(\mathrm{8}+\mathrm{1}\right)^{\mathrm{k}} +\mathrm{1} \\ $$$$=\underset{\mathrm{m}=\mathrm{0}} {\overset{\mathrm{k}} {\Sigma}}\mathrm{C}_{\mathrm{k}} ^{\mathrm{m}} \mathrm{8}^{\mathrm{k}−\mathrm{m}} =\left(\mathrm{8}^{\mathrm{k}} +\mathrm{k}.\mathrm{8}^{\mathrm{k}−\mathrm{1}} +\frac{\mathrm{k}\left(\mathrm{k}−\right)}{\mathrm{2}}\mathrm{8}^{\mathrm{k}−\mathrm{2}} \right. \\ $$$$\left.+...+\mathrm{8}+\mathrm{1}\right)+\mathrm{1}\equiv\mathrm{2}\left(\mathrm{mod8}\right)\left(\boldsymbol{\mathrm{q}}.\boldsymbol{\mathrm{e}}.\boldsymbol{\mathrm{d}}\right) \\ $$$$\mathrm{b}−\mathrm{By}\:\mathrm{above}\:\mathrm{proof}\:\mathrm{we}\:\mathrm{have}\:\mathrm{3}^{\mathrm{2k}} +\mathrm{1}=\mathrm{8q}+\mathrm{2} \\ $$$$\Rightarrow\mathrm{3}^{\mathrm{2k}+\mathrm{1}} +\mathrm{1}=\mathrm{3}.\left(\mathrm{3}^{\mathrm{2k}} +\mathrm{1}\right)−\mathrm{2}=\mathrm{3}.\left(\mathrm{8q}+\mathrm{2}\right) \\ $$$$−\mathrm{2}=\mathrm{3}.\mathrm{8q}+\mathrm{4}\equiv\mathrm{4}\left(\mathrm{mod8}\right)\left(\:\boldsymbol{\mathrm{q}}.\boldsymbol{\mathrm{e}}.\boldsymbol{\mathrm{d}}\right) \\ $$$$\left.\mathrm{2}\right)\mathrm{a}−\mathrm{Consider}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{2}^{\mathrm{n}} −\mathrm{3}^{\mathrm{m}} =\mathrm{1} \\ $$$$\mathrm{If}\:\mathrm{m}\:\mathrm{is}\:\mathrm{even}\:\mathrm{then}\:\mathrm{2}^{\mathrm{n}} =\mathrm{3}^{\mathrm{2k}} +\mathrm{1}\left(\ast\right) \\ $$$$\left(\mathrm{by}\:\mathrm{above}\:\mathrm{proof}\:\right).\mathrm{For}\:\mathrm{n}=\mathrm{0},\mathrm{1},\mathrm{2}\:\mathrm{we}\:\mathrm{see} \\ $$$$\mathrm{the}\:\mathrm{equation}\:\left(\ast\right)\:\mathrm{has}\:\mathrm{no}\:\mathrm{roots} \\ $$$$\mathrm{For}\:\mathrm{n}\geqslant\mathrm{3}\:\mathrm{L}.\mathrm{H}.\mathrm{S}\:\left(\ast\right)\mathrm{is}\:\mathrm{divisible}\:\mathrm{by}\:\mathrm{8} \\ $$$$\mathrm{while}\:\mathrm{R}.\mathrm{H}.\mathrm{S}\:\mathrm{isn}'\mathrm{t}\:\mathrm{divisible}\:\mathrm{8}\:\mathrm{since} \\ $$$$\mathrm{3}^{\mathrm{2k}} +\mathrm{1}\equiv\mathrm{2}\left(\mathrm{mod8}\right)\left(\mathrm{by}\:\mathrm{above}\:\mathrm{proof}\right). \\ $$$$\mathrm{Hence}\:\mathrm{this}\:\mathrm{case}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{has}\:\mathrm{no}\:\mathrm{roots}\: \\ $$$$\left(\boldsymbol{\mathrm{q}}.\boldsymbol{\mathrm{e}}.\boldsymbol{\mathrm{d}}\right) \\ $$$$\mathrm{b}−\mathrm{We}\:\mathrm{see}\:\mathrm{that}\:\left(\mathrm{by}\:\mathrm{directly}\:\mathrm{checking}\right) \\ $$$$\mathrm{the}\:\left(\mathrm{n},\mathrm{m}\right)=\left(\mathrm{2},\mathrm{1}\right)\:\mathrm{satisfy}\:\mathrm{the}\:\mathrm{equation} \\ $$$$\mathrm{2}^{\mathrm{n}} −\mathrm{3}^{\mathrm{m}} =\mathrm{1}\left(\mathrm{1}\right).\mathrm{We}\:\mathrm{prove}\:\mathrm{that}\:\mathrm{is}\:\mathrm{unique} \\ $$$$\mathrm{solution}\:\mathrm{of}\:\mathrm{given}\:\mathrm{equation}.\mathrm{Indeed}, \\ $$$$\mathrm{When}\:\mathrm{m}\:\mathrm{is}\:\mathrm{even}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{has}\:\mathrm{no} \\ $$$$\mathrm{roots}\:\left(\mathrm{above}\:\mathrm{proof}\right).\mathrm{We}\:\mathrm{now}\:\mathrm{consider} \\ $$$$\mathrm{the}\:\mathrm{case}\:\mathrm{m}\:\mathrm{is}\:\mathrm{odd}.\Rightarrow\mathrm{m}=\mathrm{2k}+\mathrm{1}.\mathrm{Then} \\ $$$$\left(\mathrm{1}\right)\Leftrightarrow\mathrm{2}^{\mathrm{n}} =\mathrm{3}^{\mathrm{2k}+\mathrm{1}} +\mathrm{1}\left(\ast\ast\right) \\ $$$$\mathrm{For}\:\mathrm{n}=\mathrm{0},\mathrm{1}\:\mathrm{It}\:\mathrm{is}\:\mathrm{clear}\:\mathrm{that}\:\mathrm{the}\:\mathrm{equation} \\ $$$$\left(\ast\ast\right)\mathrm{has}\:\mathrm{no}\:\mathrm{roots} \\ $$$$\mathrm{For}\:\mathrm{n}\geqslant\mathrm{2}\:\mathrm{L}.\mathrm{H}.\mathrm{S}\left(\ast\ast\right)\mathrm{is}\:\mathrm{divisible}\:\mathrm{by}\:\mathrm{4} \\ $$$$\mathrm{while}\:\mathrm{R}.\mathrm{H}.\mathrm{S}\:\left(\ast\ast\right)\mathrm{isn}'\mathrm{t}\:\mathrm{divisible}\:\left(\mathrm{by}\right. \\ $$$$\left.\mathrm{above}\:\mathrm{proof}\:\mathrm{1b}\right).\mathrm{Hence}\:\mathrm{this}\:\mathrm{case}\:\mathrm{the} \\ $$$$\mathrm{given}\:\mathrm{has}\:\mathrm{no}\:\mathrm{roots}\:\mathrm{which}\:\mathrm{means}\:\mathrm{the} \\ $$$$\left(\mathrm{n},\mathrm{m}\right)=\left(\mathrm{2},\mathrm{1}\right)\:\mathrm{is}\:\mathrm{unique}\:\mathrm{solution}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{given}\:\mathrm{equation}\:\left(\boldsymbol{\mathrm{q}}.\boldsymbol{\mathrm{e}}.\boldsymbol{\mathrm{d}}\right) \\ $$

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