I was going through Calculus by Spivak when in the first chapter I encountered problem 16, which in the end recites
you should know make a good guess as to when $\left(x+y\right)^n = x^n + y^n$
quoting the results of previous exercises with a fixed $n$, which had this result iff $x = 0 \lor y = 0 \lor x = -y$
I started going on with the proof and immediately thought of using induction, but I was later discouraged about it when, seeing how I "proved" the cases with fixed $n$, I wasn't using an "induction-like" procedure, and I found myself lost (I saw Spivak proof and he uses Rolle's theorem on the function $f\left(x\right) = x^n + y^n - \left(x+y\right)^n$, but it isn't a proof that I find satisfying, as I think there's a proof that doesn't use such methods.
Here is a look into the procedure I used for $n=5$: $$\left(x+y\right)^n = x^5 + 5 x^4 y + 10 x^3 y^2 + 10 x^2 y^3 + 5 x y^4 + y^5 = x^5 + y^5$$ $$5\left(x^4y + 2x^3y^2 + 2x^2y^3 + xy^4\right) = 0$$ $$xy\left(x^3 + 2x^2y + 2xy^2 + y^3\right) = 0$$ So either $x = 0 \lor y = 0$ gives us a solution, so now we consider the part inside parenthesis $$x^3 + 2x^2y + 2xy^2 + y^3= x^3 + 3x^2y + 3xy^2 + y^3 - x^2y - xy^2 = 0$$ $$\left(x+y\right)^3 - xy\left(x + y\right) = 0$$ So either $$xy = \left(x+y\right)^2 \iff 0 = x^2 + xy + y^2 \iff x = -y $$ or $$\left(x+y\right)^3 = 0 \land \left(x+y\right) = 0 \iff x = -y$$
For a generic $n$, I went with: $$\left(x+y\right)^n = \sum_{k = 0}^n \binom{n}{k}x^{n-k}y^k = x^n + y^n$$ Unpacking the first and last term of the sum that gives us $$\sum_{k=1}^{n-1} \binom{n}{k}x^{n-k}y^k = 0$$ $$xy\left(\sum_{k=1}^{n-1} \binom{n}{k}x^{n-1-k}y^{k-1}\right) = 0$$ Which proves the condition $x = 0 \lor y = 0 \implies \left(x+y\right)^n = x^n + y^n$ Now I tried to find a way to transform the inner sum into the form $$\left(x+y\right)^{n-2} - z\left(x+y\right)$$ Where $z$ is any possible polynomial that uses $x$ or $y$ or both This way I should resolve $$\left(x+y\right)^{n-2} = \left(x+y\right) = 0$$ which yields $x = -y$ as the only result excluding $x = y = 0$
--- Update ---
Following Mark Bennet advise, I went on to prove that, with $|x| > |y|, z = y/x$ (I also tried with $z = x/y$), the sum $$\sum_{i=1}^{n-1}\binom{n}{i}z^i = 0 \iff z = -1 \lor z = 0$$ which for $z < 0$ has the same number of positive and negative terms. Due to the properties of the binomial, we also know that the $i$-th and the $\left(n-i\right)$-th term of the sum have the same coefficient $\binom{n}{i}$. So we can write it as $$\sum_{i=1}^{\left(n-1\right)/2}\binom{n}{i} \cdot z^i\left(1+z^{n-2i}\right)$$ My idea was to prove that such sum is never equal to $0$ except when $z = 0 \lor z = -1$. Do to prove such a thing, I had 3 main ideas:
- -prove that every element of the sum differs by values whose sum $\neq 0$
- -prove that the sum has only 2 real solutions, so that they can only be $z = 0 \lor z = 1$
- -prove that the sum is a parabola: this implies that it only has 2 real solutions
But again, I cannot find a way to prove such things
--- Update 2 ---
Thinking about the problem, I tried to pose it as $$\left(1+z\right)^n < 1 + z^n$$ with $z = y/x \land -1 < z < 0$ (We already proved for $z$ > 0 that the claim is false) So, listing down a couple of things: $$1 + z < 1 \land z^n < z \implies \left(1+z\right)^n < 1 +z < 1 + z^n \forall n$$ This means that $$\forall z \in \left(-1, 0\right), \left(1+z\right)^n\neq 1 + z^n$$ As this is the only left range, it means that $z$ must either be $-1 \lor 0$, which implies $y = 0 \lor x = -y$, which proves our claim