Prove that $3^{2015}-2^{2015}>2016^2$ I have to prove:$$3^{2015}-2^{2015}>2016^2$$
by induction.
I wanted to use the standard factorizing formula:
$$a^{mn}−b^{mn}=(a^n−b^n)(a^{n(m−1)}+a^{n(m−2)}b^n+...+a^nb^{n(m−2)}+b^{n(m−1)}).$$
$2015=403\cdot 5$
What would be the next step?

We will prove for a more general case , i.e :
$$3^x - 2^x > (x+1)^2 \quad\quad \text{ For x > 3 }$$
For base case , $3^3 - 2^3 = 19 > 16$
Now let be true for a value $n$ , such that $3^n - 2^n > (n+1)^2$
$$\begin{align}3^n - 2^n  + 2n+3& > (n+1)^2 + 2n+3\\ 
3^n - 2^n  + 2n+3& > (n+2)^2 \end{align}$$
Now it is left to show that $3^{n+1} - 2^{n+1} >3^n - 2^n  + 2n+3 $.We show it by assuming it is true.
$$\begin{align}3^{n+1} - 2^{n+1} &>3^n - 2^n  + 2n+3 \\ 
3^{n+1} - 3^n + 2^n - 2^{n+1} &> 2n+3 \\ 
2.3^n - 2^n &> 2n+3\end{align}$$
Which follows from the fact that $3^n - 2^n > (n+1)^2 > 2n+3$.
Hence our induction is complete.
