# Solving for $(a,\,b,\,c)$ with Powers of 2

• MHB
• anemone
In summary, solving for $(a,\,b,\,c)$ with powers of 2 serves to simplify and express equations in terms of powers of 2, making them easier to solve and providing insight into the relationship between variables. The basic rules involve using properties of exponents, and this technique has various applications in fields such as computer science, finance, and physics. To apply this method, look for terms or variables involving powers of 2 in the equation. However, there are limitations and restrictions, such as the variables being real numbers and the exponents being integers. Some equations may not be suitable for this method, such as those with negative or fractional exponents. Careful consideration is necessary before using powers of 2 to solve
anemone
Gold Member
MHB
POTW Director
Find all positive integers $(a,\,b,\,c)$ such that $ab-c,\,bc-a,\,ca-b$ are all powers of 2.

Let $a=b$. Then $a(c-1)$ and $a^2-c=2^n$ are powers of 2, hence, $a=2^k,\,c-1=2^m$. Considering equality $2^{2k}=a^2=1+(c-1)+(a^2-c)=1+2^m+2^n$ modulo 4, we get $(m, \,n)=(0,\,1)$, $k=1$. This leads to solutions $(2,\,2,\,2)$ and $(2,\,2,\,3)$.

Now, assume that $a<b<c$. Denote $bc-a=2^A,\,ac-b=2^B$, then $2^A-2^B=(c+1)(b-a)>0$ hence $A>B$ and $2^B$ divides $(c+1)(b-a)$. Also, $2^B$ divides $2^A+2^B=(c-1)(a+b)$. Either $c-1$ or $c+1$ is not divisible by 4, hence either $2(b-a)$ or $2(b+1)$ is divisible by $2^B$, in both cases we have $a+b\ge 2^{B-1}$. So, $a(b+1)\le ac=2^B+b\le 2(a+b)+b$, i.e. $ab\le a+3b<4b$. Hence, $a<4$.

If $a=2$, then $2^C=2b-c,\,2^B=2c-b>1$, hence $B>0$. We get $b=\dfrac{(2^{C+1}+2^B)}{3},\,c=\dfrac{(2^{B+1}+2^C)}{3}$. If $C\ge 1$ then both $b$ and $c$ are even, hence $bc-2$ is not divisible by 4, in this case, $bc-2\ge 3\cdot 4-2=10$ cannot be a power of 2.

If $C=0$, then by modulo 3 we see that $B$ is even and $bc-2=\dfrac{(2+2^B)(2^{B+1}+1)-18}{9}=\dfrac{2^{2B+1}+5\cdot2^B-16}{9}$ must be a power of 2. This is not possible if $B\ge 6$, since 32 does not divide numerator $2^{2B+1}+5\cdot 2^B-16$ and it is greater than $16\cdot 9$. For $B=2$, we get $b=a=2$ and for $B=4$, we get $b=6$ and $c=11$.

If $a=3$, then $2^C=3b-c,\,2^B=3c-b\ge 2c+1>4$, hence $B\ge 3$. We get $b=\dfrac{(3\cdot 2^{C}+2^B)}{8},\,c=\dfrac{(3\cdot 2^{B}+2^C)}{8}$. We get $B>C$ (from $b<c$) and since $B\ge 3$, we get that $C\ge 3$ as well and $2^A=bc-3=(3\cdot 2^{C-3}+2^{B-3})(3\cdot 2^{B-3}+2^{C-3})-3$. If $B>C>3$, this is odd and greater than 1, hence they are not power of 2.

If $C=3$ we get $2^A=2^{B-3}(10+3\cdot2^{B-3})$. This is a power of 2 if and only if $B-3=1 \implies B=4$, $b=5,\,c=7$.

Hence, $(a,\,b,\,c)=(2,\, 2,\, 2),\,(2,\, 2,\, 3),\,(2,\, 6,\, 11),\,(3,\, 5,\, 7)$ and their permutations.

## What is "Solving for $(a,\,b,\,c)$ with Powers of 2"?

"Solving for $(a,\,b,\,c)$ with Powers of 2" is a mathematical concept that involves finding the values of the variables $a$, $b$, and $c$ in an equation that contains powers of 2.

## Why is it important to solve for $(a,\,b,\,c)$ with Powers of 2?

Solving for $(a,\,b,\,c)$ with Powers of 2 is important because it allows us to simplify complex equations and make them easier to understand and work with.

## What are some common strategies for solving for $(a,\,b,\,c)$ with Powers of 2?

Some common strategies for solving for $(a,\,b,\,c)$ with Powers of 2 include factoring, using logarithms, and using the laws of exponents.

## What are some real-world applications of solving for $(a,\,b,\,c)$ with Powers of 2?

Solving for $(a,\,b,\,c)$ with Powers of 2 is used in various fields such as computer science, physics, and finance. For example, it is used in computer algorithms and in calculating compound interest.

## Are there any limitations to solving for $(a,\,b,\,c)$ with Powers of 2?

While solving for $(a,\,b,\,c)$ with Powers of 2 can be useful, it is not always applicable to every equation or problem. Some equations may require different methods of solving, and there may be cases where solving for $(a,\,b,\,c)$ with Powers of 2 is not possible.

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