# Find $x$ such that $2^x+3^x-4^x+6^x-9^x=1$

The question:

Find values of $x$ such that $2^x+3^x-4^x+6^x-9^x=1$, $\forall x \in \mathbb R$.

Notice the numbers $4$, $6$ and $9$ can be expressed as powers of $2$ and/or $3$. Hence let $a = 2^x$ and $b=3^x$.

\begin{align} 1 & = 2^x+3^x-4^x+6^x-9^x \\ & = 2^x + 3^x - (2^2)^x + (2\cdot3)^x-(3^2)^x\\ & = 2^x + 3^x - (2^x)^2 + 2^x\cdot3^x-(3^x)^2 \\ & = a+b-a^2+ab-b^2 \\ 0 & = a^2-ab+b^2-a-b+1 \end{align}

\begin{align} 0 & = a^2-ab+b^2-a-b+1 \\ & = 2a^2-2ab+2b^2-2a-2b+2 \\ & = (a^2-2ab+b^2)+(a^2-2a+1)+(b^2-2b+1) \\ & = (a-b)^2 + (a-1)^2 + (b-1)^2 \end{align}

This is where I am stuck. I am convinced that this factorisation could help solve the question, but I don't know how. Also, once we find values for $x$, we must prove that there are no further values of $x$. Could someone complete the question?

• The $\forall x\in\Bbb{R}$ seems a bit misplaced. – Servaes Nov 30 '17 at 10:08
• You have done the hard work – lab bhattacharjee Nov 30 '17 at 10:09
• @Servaes I assume it just wants us to find all real values of $x$. – Landuros Nov 30 '17 at 10:10

If the sum of a finite number of non-negative expressions is $0$, each of them has to be zero.
In other words, when $a,b,c\geq0$
$a+b+c=0 \implies a=b=c=0$
You have done the hard part by showing that it can be written as the sum of three squares. This means that $a=b$, $a=1$ and $b=1$. What does that tell you about $x$?
• Of course! I wish I could've thought of that... After some algebraic manipulation we get $x=0$. Thanks! – Landuros Nov 30 '17 at 10:13
• @Landuros: Yes, you have shown that too. You have turned the original problem by variable substitutions and algebraic manipulation into an equivalent problem (because all of your steps are reversible and your substitutions are not restrictive either) that is true only when $a=b=1$. Therefore, $x=0$ is the only solution. – stressed out Nov 30 '17 at 10:17
A sum of squares equals zero if and only if each of the squares equals zero. So you get $a=b=1$.