# Find $a,b,c,d$ such that $2^a + 2^b + 2^c = 4^d$

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?

• 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. – stuart stevenson Jun 1 at 9:49
• 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. – Dunkel Jun 1 at 9:55
• @Dunkel $a,b,c,d$ are positive whole numbers. – Landuros Jun 1 at 10:43

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 $$b=c=2d-2$$, $$a=2d-1$$.

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.

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.

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.

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

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.