Adding monomials of different degree. Can you prove that x^m + x^n can never equal x^k, where k is some rational number, and m is not equal to n.
I know we've all been doing it since middle school, but is there a mathematical way of proving it?
 A: First, note that there will be trivial solutions; indeed, if $m\leq n$ then there will be $m$ of them. We can eliminate these by dividing out $x^m$, so the only case we really need concern ourselves with is $1+x^n = x^k$.
But this certainly has solutions. For example, $x^2=1+x$ includes the golden ratio $\phi=\frac{1}{2}(1+\sqrt{5})$ as an irrational solution. A better question is what kinds of solutions one gets: must they always be irrational?
A: $x=0$ is trivial and works. If $x \not = 0$, necessarily $k>max(n,m)$ so let's suppose $m<n<k$. Then you're trying to solve $x^k - x^n - x^m = 0 \Leftrightarrow x^m(x^{k-m}-x^{n-m}-1)=0 \Leftrightarrow (x^{k-m}-x^{n-m}-1)=0$ which is a perfectly valid expression that has roots and everything depending on the parameters.
A: So here's how I did it. I wrote down the equality, and calculated the Wronskian for the expression, assuming x^m, x^n and x^k to be all functions. The Wronskian comes out to be non-zero if k is any value other than either m or n. Hence proving that the three terms are linearly independent, and hence, no equality exists for k not equal to m or n.
Glad to see that there are easier ways as well, and also learnt to watch out for trivial solutions after reading the other responses. 
