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.

Are you looking for the $y\in \mathbb{Z}$ such that $(1+a)^y = 1 + a^y$ holds for all $a\in\mathbb{R}$, or that it holds for some $a\in\mathbb{R}$? — Daniel Fischer, Mar 19 '15 at 17:10

AgentS · Answer 1 · 2015-03-19T15:55:08.797

1

HINT

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

edited Mar 19 '15 at 15:55

answered Mar 19 '15 at 15:10

AgentS

12,195

Thats the reason it was only an hint and the extended binomial thm works for negative exponents right ? – AgentS Mar 19 '15 at 15:11
I thought we can use it to conclude that there are no solutions for $y\ge 2$ because that middle term is nonzero for $y\ge 2$. Isn't it simple ? – AgentS Mar 19 '15 at 15:20
I misread ... the problem asks to find all $y$ for which the expression is valid – Mark Viola Mar 19 '15 at 15:44
Ahh I see... I read it as finding integer $y$ values that work for all $a\in \mathbb{R}$ – AgentS Mar 19 '15 at 15:47
Yes, you're correct – Mark Viola Mar 19 '15 at 15:50

score 0 · Answer 2 · answered Mar 19 '15 at 16:26

0

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

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

answered Mar 19 '15 at 16:26

themaker

2,451

score 0 · Answer 3 · answered Mar 19 '15 at 17:05

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.

Exponential equation.

3 Answers3