Pythagoras with a twist

Without loss of generality, $a\le b$ . Then we can factor out $2^a$ : $2^a(1+2^{b-a}) = 2^c$ Here, for any $a$ and $b$ such that $b-a\ge1$ , $1+2^{b-a}$ is not multiples of 2, so we must have $a=b$ .

Knowing that $a=b$ , we have $2^a + 2^a = 2^c$ $2\cdot2^a = 2^c$ $2^{a+1} = 2^c$

Clearly there are infinitely many solutions to $a+1 = c$ , therefore there are infinitely many solutions to $2^a + 2^b = 2^c$ .

Looking at the equation in binary, $2^c$ can be written as $100...0$ , with $c$ zeros. Adding in binary, the only way to get $100...0$ is to add two identical numbers with one less zero. $1+1=10$ in binary. Thus the solution is $2^{c-1}+2^{c-1}=2^c$ .

answer: infinitely many

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$2^x+2^x=2^y$ in which $y=x+1$ ...

Because of this, as long as $a$ is equal to $b$ and $c$ is 1 more than $a$ and $b$ (not combined), we have a solution. We can turn $a$ and $b$ into one variable, $x$ , since $a=b$ . And, since $c=x+1$ gives us solutions and is a linear function with infinite solutions, (2^a+2^b=2^c/) has infinitely many whole number solutions