Don't Overthink It

Number Theory Level 5

$\large{ 2^{a+1} - 2^a + 2 = \ 2^b - 2^c }$

Evaluate the sum of the sums of all ordered tuples of integers $(a,b,c)$ that exist as solutions to the equation above.

Details and Assumptions:

$a, b,$ and $c$ are neither necessarily distinct nor positive.
The "sum of a tuple" is the sum of its members.

Bonus: For extra fun, consider the same question but with the equation $2^{a+1}-2^a+1=2^b-2^c$ .

The answer is 14.

1 solution

Kazem Sepehrinia
Jul 10, 2015

Simplify to get $2^b-2^c=2^a+2 >0$ Results in $b>c$ , let $b-c=d \in \mathbb{N}$ and simplify: $2^{a-c}+2^{1-c}=2^d-1 \ \ \star$ If $d=1$ only possibility for $2^{a-c}+2^{1-c}=1$ is $a-c=1-c=-1$ , which gives us one answer: $(a, b, c)=(1, 3, 2)$ .

For $d>1$ RHS of $\star$ is an odd number greater than $1$ and this is possible if and only if one of $a-c$ or $1-c$ be $0$ to give a $1$ in LHS of $\star$ . So we have two cases.

Case 1) $a-c=0$ leading to $1+2^{1-c}=2^d-1 \\ 1+2^{-c}=2^{d-1}$ RHS is even, only possible when $c=0$ , so $(a, b, c)=(0, 2, 0)$ .

Case 2) $1-c=0$ leading to $2^{a-1}+1=2^d-1 \\ 2^{a-2}+1=2^{d-1}$ RHS is even, only possible when $a=2$ , so $(a, b, c)=(2, 3, 1)$ .

Ah!!i missed third ordered tuple

Aakash Khandelwal - 5 years, 11 months ago

On the LHS, The summation limits are from n=1 to (a-1). This employs a-1>=1. i.e. (a>=2) should be satisfied. This means the only valid tuple should be (a,b,c) = (2,3,1). And the other two tuples with (a < 2) should be rejected.

Abhishek Kumar - 5 years, 10 months ago

Thanks for pointing that out. I've edited the problem accordingly.

Alex Delhumeau - 5 years, 10 months ago

Don't Overthink It

The answer is 14.

1 solution

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