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Let $a,b,c,d$ be whole numbers that satisfy

$$2^a + 2^b + 2^c = 4^d$$

What values of $(a,b,c,d)$ would make this equation true?


Here is my work so far.

Without loss of generality, assume $a\ge b\ge c$. Then one trivial solution by inspection is $(1,0,0,1)$. Playing around, I also found a solution at $(3,2,2,2)$. Then I checked $a=5,b=4,c=4$ and found that it also worked.

It seems that there is a family of solutions at $(2n-1,2n-2,2n-2,n)$. I can prove this easily:

\begin{align} LHS&=2^{2n-1}+2^{2n-2}+2^{2n-2}\\ &=2^{2n-1}+2^{2n-1}\\ &=2^{2n}\\ &=4^n\\ &=RHS \end{align}

Is this the only solution? If it is, how do I go about proving it?

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  • $\begingroup$ Have another look at your trivial solution. There are also more trivial solutions with two $0$'s. I would try a proof by contradiction for cases outside of the one I have put and your own. I think these are the only cases. $\endgroup$ – stuart stevenson Jun 1 at 9:49
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    $\begingroup$ Any restrictions on $a,b,c,d$? Can they be negative? If we choose $a=-1, b=-2, c=-2,d=0$, the equation is satisfied. $\endgroup$ – Dunkel Jun 1 at 9:55
  • $\begingroup$ @Dunkel $a,b,c,d$ are positive whole numbers. $\endgroup$ – Landuros Jun 1 at 10:43
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Obviously, $d>0$. WLOG, assume that $a\ge b\ge c$

$2^c(2^{a-c}+2^{b-c}+1)=2^{2d}$

$2^{a-c}+2^{b-c}+1\ge 3$. Therefore, $b-c$ must be $0$ as otherwise $2^{a-c}+2^{b-c}+1$ will be an odd number larger than $1$.

$2^{a-c}+2^{b-c}+1=2^{a-c}+2=2(2^{a-c-1}+1)$ and hence $2^{a-c-1}+1$ must be even. $a-c=1$.

We have $2^c\times 4=2^{2d}$ and hence $b=c=2d-2$, $a=2d-1$.

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If we write the numbers in binary system we get $$1\underbrace{0\ldots0}_a+1\underbrace{0\ldots0}_b+1\underbrace{0\ldots0}_c=1\underbrace{0\ldots0}_{2d}$$ This is possible only if $b=c$ to get $1+1=10$ and $a=b+1$ to make the $1$ dissapear with the carry.

In addition, $a$ must be odd.

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I'd write this as $$\frac1{2^r}+\frac1{2^s}+\frac1{2^t}=1$$ where $r=2d-a$, $s=2d-b$ and $t=2d-c$ are integers. Each of $r$, $s$ and $t$ must be positive, and they can't be all $\ge2$, since then the LHS is $\le 3/4$. So one of them, say $r$, equals $1$. In that case $$\frac1{2^s}+\frac1{2^t}=\frac12.$$ The only possibility here is $s=t=2$. So the solutions for $(r,s,t)$ are $(1,2,2)$, $(2,1,2)$, $(2,2,1)$.

What about $(a,b,c)$? If $(r,s,t)=(1,2,2)$ then $(a,b,c)=(c+1,c,c)$ and $2^a+2^b+2^c=2^{c+2}$. Then $c=2d-2$ must be even, and $(a,b,c) =(2d-1,2d-2,2d-2)$ etc.

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WLOG $a\ge b\ge c$. If $b\neq c$ then consider $\pmod {2^{c+1}}$ we get a contradiction (why?). Then observe $a=b+1$ by considering $\pmod{2^{b+2}}$. Then $a=b+1=c+1$, so $2^b=4^d\implies b$ is even. Hence the solution you found are the only solutions.

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$$2^a+2^b+2^c=2^c(2^{a-c}+2^{b-c}+1)$$ Since the second factor must be even, we see that $a>b=c$, and then it becomes $$2^c(2^{a-c}+2)=2^{c+1}(2^{a-c-1}+1)$$ And now we see that $a=c+1$.

Now, $$2^a+2^b+2^c=4\cdot2^c$$ so $c$ is even.

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We have $$ \eqalign{ & \left( {a,b,c,d} \right)\quad \Rightarrow \quad 2^{\,a} + 2^{\,b} + 2^{\,c} = 4^{\,d} \quad \Rightarrow \cr & \Rightarrow \quad 2^{\,2n} \left( {2^{\,a} + 2^{\,b} + 2^{\,c} } \right) = 2^{\,2n} 4^{\,d} \quad \Rightarrow \cr & \Rightarrow \quad 2^{\,a + 2n} + 2^{\,b + 2n} + 2^{\,c + 2n} = 4^{\,d + n} \quad \Rightarrow \cr & \Rightarrow \quad \left( {a + 2n,b + 2n,c + 2n,d + n} \right) \cr} $$

So the problem shifts to find the minimal quadruples.

In binary notation it is clear how it shall be $$ \eqalign{ & 2^{\,a} \quad \quad \quad \quad = 0,0, \cdots ,1,0,0 \cr & 2^{\,b} \quad \quad \quad \quad = 0,0, \cdots ,1,0,0 \cr & 2^{\,a} + 2^{\,b} \quad \quad = 0,0, \cdots ,0,1,0 \cr & 2^{\,c} \quad \quad \quad \quad = 0,0, \cdots ,0,1,0 \cr & 2^{\,a} + 2^{\,b} + 2^{\,c} = 0,0, \cdots ,0,0,1 \cr & 4^{\,d} = 2^{\,2d} \quad \;\; = \underbrace {0,0, \cdots ,0,0,1}_{0,1, \cdots ,\;2d - 2,\;2d - 1,\;2d} \cr} $$

So
- $2^k$ has only one $1$, and same has $4^d$;
- in summing the $2$'s, to reach to have only one $1$, we cannot leave any $1$ behind;
- so the (only) one of $2^b$ must be just below that of $2^a$ - ... and we must end on bit $2d$.

The conclusion is clear.

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