Power of $4$

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samiul_samin
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Power of $4$

Unread post by samiul_samin » Thu Mar 01, 2018 11:21 am

If $a,b,c,d$ are integers,find all of the solutions of the given equation:
$4^a+4^b+4^c=4^d$

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samiul_samin
Posts: 1004
Joined: Sat Dec 09, 2017 1:32 pm

Re: Power of $4$

Unread post by samiul_samin » Sat Mar 03, 2018 1:36 pm

I am giving two different approaches.
Approach 1
By using Infinite descent,
$4^a+4^b+4^c=4^d$
$\Rightarrow 4^{a-1}+4^{b-1}+4^{c-1}=4^{d-1}$ [Divided both sides by $4$]
$\Rightarrow 4^{a-2}+4^{b-2}+4^{c-2}=4^{d-2}$ [ Divided both sides by $4$]
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$\Rightarrow 4^{a-a}+4^{b-a}+4^{c-a}=4^{d-a}\Rightarrow 1+4^{b-a}+4^{c-a}=4^{d-a}$
Which is impossible.Because LS is not divided by $4$ and RS is obviously divided by $4$.
So there is no such $a,b,c,d$.
Approach 2
By using Extreme Principle

Let,there is a solution where the lowest values are ($a,b,c,d$).
So,$4^a+4^b+4^c=4^d$
If we divide both sides by $4$,we get $4^{a-1}+4^{b-1}+4^{c-1}=4^{d-1}$
But these values are less than ($a,b,c,d$).
A contradiction.
So,there is no such solution.
We are done.
If there is any type of mistake,please comment it.

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