Exponential equation.

Question

Find all $ y \in \mathbb{Z} $ so that: $$ (1 + a)^y = 1 + a^y \;,\; a \in \mathbb{R}$$

I have tried to use the following formula: $$ a^n - b^n = (a - b)(a^{n - 1} + a^{n - 2}b + a^{n - 3}b^2 + ... + a^2b^{n - 3} + ab^{n - 2} + b^{n - 1})$$

but it didn't work.

Answer 1

HINT

$$(1+a)^y =1+ \color{blue}{\sum\limits_{k=1}^{y-1}\binom{y}{k}a^k} + a^y$$

Answer 2

If you take $a = 1\in \mathbb R$ in the equation, you get

$$2^y = 2\implies y = 1$$

Answer 3

Clearly $y=1$ is true.

Now assume there exists $y\in \mathbb{Z}$ such that $y>1$ and $(1+a)^y=1+a^y$.

Then,

$$1+a^y=\sum\limits_{n=0}^{y}{}^y{C_n}a^n\Rightarrow {}^y{C_n}=0\text { for each }0<n<y\Rightarrow {}^y{C_1}=0. $$

But ${}^y{C_1}=\dfrac{y!}{(y-1)!.1!}=y>1\ne 0$. Contradiction.