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Suppose $Ae^{iax} + Be^{ibx} = Ce^{icx}$ for some nonzero constants $A, B, C, a, b, c$, and for all $x$. Prove that $a = b = c$ and $A + B = C$.

When given questions like this, my first instinct is to plug in the givens and try to see if I reach a true conclusion like $1=1$ or something. Is this the wrong way to approach such problems?

Note: I am not asking for someone to solve this, I am merely asking for advice on my approach to solving such problems where I'm asked to "prove" something.

For example, this is what I did for this problem, yet it feels unfulfilled, almost as if I have not guaranteed the proof. If my technique is not wrong here, then it is my way of thinking of when a mathematical thing is "fully proven".

$$Ae^{iax}+Be^{ibx}=Ce^{icx}$$ $$Ae^{iax}+Be^{iax}=Ce^{iax}$$ $$e^{iax}(A+B)=Ce^{iax}$$ $$e^{iax}(C)=Ce^{iax}$$ $$Ce^{iax}=Ce^{iax}$$

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    It's the wrong way to *prove* it. But it's a good way to brainstorm to see if something comes to you. But no, a single true example is of no use in proving something. – 2017-02-13

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No, it is not a proof. You have assumed exactly what you're trying to prove. You used in your proof that $a=b=c$ and $A+B=C$, but that is exactly what you're trying to show.

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Doing what you do does not constitute a proof of course, but sometimes it may create some insight as to what might be happening, which could give you some inspiration for how to prove what you need to prove.

For your specific example, for example, you do gain some insight as to how the $a$, $b$, and $c$ relate to each other, and same for the $A$, $B$, and $C$.

So, in general, I think it is a good idea, sure!

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It's an excellent way to get an intuition for what's going on, and sometimes there's a way to get a proof out of it. The problem that may occur is that some of your steps may lose information - for example, you could be multiplying by zero, resulting in a true equation even if the equation you began with was false.

But the thing is, that's the only issue. That means that if your steps are reversible - that is, if they're equally correct running backwards as forwards - then you can get a proof by just doing that. For example, here's a proof of the example you gave:

$Ce^{iax} = Ce^{iax}$

$e^{iax}(C) = Ce^{iax}$

$e^{iax}(A + B) = Ce^{iax}$

$Ae^{iax} + Be^{iax} = Ce^{iax}$

$Ae^{iax} + Be^{ibx} = Ce^{icx}$

What this proof does is start from something true and use only valid steps to reach a conclusion - that makes it a genuine proof. Unfortunately, it's a proof of $Ae^{iax} + Be^{ibx} = Ce^{icx}$ given the statements $A + B = C$ and $a = b = c$, which isn't what you wanted. But this sort of approach may give you insight into the correct proof of the desired result.

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    " You didn't say what your givens were, but I assume they included and .) " why would you assume that? – 2017-02-13
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    @fleablood Because OP used those in the deduction given; it's the natural way to make the leap from $Ae^{ax} + Be^{bx} = Ce^{cx}$ to $Ae^{ax} + Be^{ax} = Ce^{ax}$ (for that you need $a=b=c$) and from $e^{ax}(A+B)=Ce^{ax}$ to $e^{ax}(C) = Ce^{ax}$ (which calls for $A + B = C$). – 2017-02-13
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    How is $Be^{ax}=Ce^{ax}\implies Be^{bx}=Ce^{cx} $ either valid or reversible. – 2017-02-13
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    @fleablood As I've now stated twice, assuming that OP has the given $a=b=c $, that step is both valid and reversible. Since the given derivation was just an example, it is reasonable to suppose that the givens are sufficient to make the steps valid. – 2017-02-13
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    But the op absolutely was not given a=b=c as that is something that is being asked to be proven. If we assume it, the question is utterly trivial. You are making an equation with six variables, and you are simply throwing two of them away with utterly no justification at all. – 2017-02-14
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    And you are clearly ignoring the I in the exponent. This statement is simply not true without it. $e+2e^3=5e^{\ln (e+2e^3)/5} $ but the clearly fails. – 2017-02-14
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    @fleablood I've edited my answer to address the $i$ issue and make clearer the idea I was going for - I gather I didn't express my point as clearly as I'd hoped. You're correct that I'd missed the $i$, but that isn't important - I was addressing the technique of proof, not the problem itself, since OP explicitly asked about the former rather than the latter. – 2017-02-14