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I have been boggled by this question for a while as well. Prove that $$(2a+b)(2b+a)=2^c$$

Is impossible.

I know that if a and b do exist then they must be even. I am trying to use this fact to contradict the statement. I haven also tried rewriting a and b as products of powers of twos and a odd factor

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    Sorry proofwriters, Wolfram Alpha says that using negative numbers, when $a=5, b=-2$, then the result is $8$. The link is here: (https://www.wolframalpha.com/input/?i=log+base+2+(25%2B(-2))(2-2%2B5)). Is the problem restricted to natural numbers only? – Toby Mak Jul 01 '17 at 07:06
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    It seems that when $2b + a = 1$, the proof fails. @Dwo Lineard, is this case excluded? – Toby Mak Jul 01 '17 at 07:07

2 Answers2

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You have $2a+b=2^r$ and $a+2b=2^s$. Then $a+b=\frac13(2^r+2^s)$ so $r+s$ has to be odd. If $a$ and $b$ have to be positive, and say $r>s$ then $a+b\ge\frac13(2^{s+1}+2^s)=2^s=a+2b$. This is looking unlikely...

Angina Seng
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Presumably $a,b,c$ are positive integers. Without loss of generality, suppose $a\le b$. The given condition implies that both $2a+b$ and $a+2b$ are powers of $2$. Hence $a+2b=2^k(2a+b)$ for some nonnegative integer $k$. Yet this is impossible because $a+2b>2^0(2a+b)$ and $a+2b<2^k(2a+b)$ for every $k\ge1$. Hence the given condition cannot be satisfied.

user1551
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