<s>[INST] Finding the value of $n\in\mathbb{Z}$ in the equation $16^n-25\times 12^{n-1}+9^n=0$ If $n$ is satisfy the equation $16^n-25\times 12^{n-1}+9^n=0$. Find the value of $n$.
Here is my work:
$$16^n-2\times12\times 12^{n-1}+9^n=12^{n-1}$$
$$(4^n-3^n)^2=12^{n-1}$$
For $n\in\mathbb{Z}^+\cup\{0\}$ we can see the LHS is an odd number and RHS is odd only for $n=1$. But for $n\in\mathbb{Z}^-$ we have fractions in both sides of the equation. I don't know how to check if there are answers for negative values of $n$.
 [/INST] The original expression $16^n-25\times 12^{n-1}+9^n=0$ when taken $\dfrac{1}{12}$ out common reduces to $\dfrac{1}{12}(4\cdot 3^n-3 \cdot 4^n)(3^{n+1}-4^{n+1})=0$
This results in 2 cases:
Case 1:
$(3^{n+1}-4^{n+1})=0 \implies 3^{n+1}=4^{n+1}$
This is one possible when both have power $0$
So $n+1=0 \implies n=-1$
Case 2:
$ (4\cdot 3^n-3 \cdot 4^n)=0 \implies 4 \cdot 3^n=3 \cdot 4^n $
If you try to see deep you will notice that when $n=1$ only then the equation is satisfied.
So $n=-1,1$
Edit: In case you failed here is other solution-
Take $a=4^n, b=3^n$ and yay there you get quadratic $12(a-b)^2=ab$. Just solve this and get your answer
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