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Question Number 5152 by 1771727373 last updated on 23/Apr/16
there is an ineterger a,b,c  can there be an interger as   a^n +b^(n+1) =c^(n+2)       (n is a interger)
$${there}\:{is}\:{an}\:{ineterger}\:{a},{b},{c} \\ $$$${can}\:{there}\:{be}\:{an}\:{interger}\:{as}\: \\ $$$${a}^{{n}} +{b}^{{n}+\mathrm{1}} ={c}^{{n}+\mathrm{2}} \:\:\:\:\:\:\left({n}\:{is}\:{a}\:{interger}\right) \\ $$$$ \\ $$
Commented by Yozzii last updated on 25/Apr/16
Let n=0⇒ 1+b=c^2   Let (a,b,c)=(0,8,3)  ⇒1+8=3^2 ⇒9=9.
$${Let}\:{n}=\mathrm{0}\Rightarrow\:\mathrm{1}+{b}={c}^{\mathrm{2}} \\ $$$${Let}\:\left({a},{b},{c}\right)=\left(\mathrm{0},\mathrm{8},\mathrm{3}\right) \\ $$$$\Rightarrow\mathrm{1}+\mathrm{8}=\mathrm{3}^{\mathrm{2}} \Rightarrow\mathrm{9}=\mathrm{9}. \\ $$
Commented by Rasheed Soomro last updated on 06/May/16
If (a,b,c)=(0,8,3) then for n=0  a^n +b^(n+1) =c^(n+2)  ⇒ 0^0 +8^1 =3^2   Does 0^0 =1?  I think 0^0  is indeterminate and  a is an integer other than 0.
$$\mathrm{If}\:\left({a},{b},{c}\right)=\left(\mathrm{0},\mathrm{8},\mathrm{3}\right)\:\mathrm{then}\:\mathrm{for}\:\mathrm{n}=\mathrm{0} \\ $$$${a}^{{n}} +{b}^{{n}+\mathrm{1}} ={c}^{{n}+\mathrm{2}} \:\Rightarrow\:\mathrm{0}^{\mathrm{0}} +\mathrm{8}^{\mathrm{1}} =\mathrm{3}^{\mathrm{2}} \\ $$$$\mathrm{Does}\:\mathrm{0}^{\mathrm{0}} =\mathrm{1}? \\ $$$$\mathrm{I}\:\mathrm{think}\:\mathrm{0}^{\mathrm{0}} \:\mathrm{is}\:\mathrm{indeterminate}\:\mathrm{and} \\ $$$$\mathrm{a}\:\mathrm{is}\:\mathrm{an}\:\mathrm{integer}\:\mathrm{other}\:\mathrm{than}\:\mathrm{0}. \\ $$
Commented by prakash jain last updated on 08/May/16
0^0  is indeteminate only when it occurs as  a limit in that case that value of limits  depends on the function, in this case  lim_(a→0)  a^0 =1
$$\mathrm{0}^{\mathrm{0}} \:\mathrm{is}\:\mathrm{indeteminate}\:\mathrm{only}\:\mathrm{when}\:\mathrm{it}\:\mathrm{occurs}\:\mathrm{as} \\ $$$$\mathrm{a}\:\mathrm{limit}\:\mathrm{in}\:\mathrm{that}\:\mathrm{case}\:\mathrm{that}\:\mathrm{value}\:\mathrm{of}\:\mathrm{limits} \\ $$$$\mathrm{depends}\:\mathrm{on}\:\mathrm{the}\:\mathrm{function},\:\mathrm{in}\:\mathrm{this}\:\mathrm{case} \\ $$$$\underset{{a}\rightarrow\mathrm{0}} {\mathrm{lim}}\:{a}^{\mathrm{0}} =\mathrm{1} \\ $$
Commented by Rasheed Soomro last updated on 08/May/16
This is knowledge for me. Thanks!
$$\mathrm{This}\:\mathrm{is}\:\mathrm{knowledge}\:\mathrm{for}\:\mathrm{me}.\:\mathrm{Thanks}! \\ $$

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