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Question Number 99495 by bobhans last updated on 21/Jun/20

solve for x,y ∈ N 7^y +2 = 3^x

$$\mathrm{solve}\:\mathrm{for}\:\mathrm{x},\mathrm{y}\:\in\:\mathbb{N}\: \\ $$$$\mathrm{7}^{\mathrm{y}} +\mathrm{2}\:=\:\mathrm{3}^{\mathrm{x}} \: \\ $$

Commented by PRITHWISH SEN 2 last updated on 21/Jun/20

x=1,y=0

$$\mathrm{x}=\mathrm{1},\mathrm{y}=\mathrm{0} \\ $$

Commented by floor(10²Eta[1]) last updated on 21/Jun/20

go to michael penn channel on yt he made a video for that 4 days ago

$$ \\ $$$${go}\:{to}\:{michael}\:{penn}\:{channel}\:{on}\:{yt} \\ $$$${he}\:{made}\:{a}\:{video}\:{for}\:{that}\:\mathrm{4}\:{days}\:{ago} \\ $$

Answered by bobhans last updated on 21/Jun/20

(x,y) = { (((1,0))),(((2,1))) :}

$$\left(\mathrm{x},\mathrm{y}\right)\:=\:\begin{cases}{\left(\mathrm{1},\mathrm{0}\right)}\\{\left(\mathrm{2},\mathrm{1}\right)}\end{cases} \\ $$

Answered by 1549442205 last updated on 21/Jun/20

It is easy to check that for (x;y)∈{(1;0);(2;1)} are roots of given equation.Also, for y=3 ∄x∈N satisfying the given equation. Now we prove that for y>3,y∈N the given equation has no solutions.Indeed, putting y=k+3(k∈N^∗ )we get 7^(3+k) +2=3^x ⇒x>3 and we have 243.7^k −3^x =−2.This equation has no solutions because LHS always is divisible by 9 while RHF isn′t divisible by 9.Hence, the given equation has only roots (x;y)∈{(1;0);(2;1)}

$$\mathrm{It}\:\mathrm{is}\:\mathrm{easy}\:\mathrm{to}\:\mathrm{check}\:\mathrm{that}\:\mathrm{for}\:\left(\mathrm{x};\mathrm{y}\right)\in\left\{\left(\mathrm{1};\mathrm{0}\right);\left(\mathrm{2};\mathrm{1}\right)\right\} \\ $$$$\mathrm{are}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{given}\:\mathrm{equation}.\mathrm{Also},\:\mathrm{for}\:\mathrm{y}=\mathrm{3} \\ $$$$\nexists\mathrm{x}\in\mathbb{N}\:\mathrm{satisfying}\:\mathrm{the}\:\mathrm{given}\:\mathrm{equation}. \\ $$$$\mathrm{Now}\:\mathrm{we}\:\mathrm{prove}\:\mathrm{that}\:\mathrm{for}\:\mathrm{y}>\mathrm{3},\mathrm{y}\in\mathbb{N}\:\mathrm{the}\:\mathrm{given} \\ $$$$\mathrm{equation}\:\mathrm{has}\:\mathrm{no}\:\mathrm{solutions}.\mathrm{Indeed}, \\ $$$$\mathrm{putting}\:\mathrm{y}=\mathrm{k}+\mathrm{3}\left(\mathrm{k}\in\mathbb{N}^{\ast} \right)\mathrm{we}\:\mathrm{get} \\ $$$$\mathrm{7}^{\mathrm{3}+\mathrm{k}} +\mathrm{2}=\mathrm{3}^{\mathrm{x}} \Rightarrow\mathrm{x}>\mathrm{3}\:\mathrm{and}\:\mathrm{we}\:\mathrm{have}\: \\ $$$$\mathrm{243}.\mathrm{7}^{\mathrm{k}} −\mathrm{3}^{\mathrm{x}} =−\mathrm{2}.\mathrm{This}\:\mathrm{equation}\:\mathrm{has}\:\mathrm{no} \\ $$$$\mathrm{solutions}\:\mathrm{because}\:\mathrm{LHS}\:\mathrm{always}\:\mathrm{is}\:\mathrm{divisible}\:\mathrm{by}\:\mathrm{9} \\ $$$$\mathrm{while}\:\mathrm{RHF}\:\mathrm{isn}'\mathrm{t}\:\mathrm{divisible}\:\mathrm{by}\:\mathrm{9}.\mathrm{Hence}, \\ $$$$\mathrm{the}\:\mathrm{given}\:\mathrm{equation}\:\mathrm{has}\:\mathrm{only}\:\mathrm{roots} \\ $$$$\left(\boldsymbol{\mathrm{x}};\boldsymbol{\mathrm{y}}\right)\in\left\{\left(\mathrm{1};\mathrm{0}\right);\left(\mathrm{2};\mathrm{1}\right)\right\} \\ $$

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