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Find the number of solutions of the following equation

$$a^{3}+2^{a+1}=a^4,\ \ 1\leq a\leq 99,\ \ a\in\mathbb{N}$$.

I tried , $$a^{3}+2^{a+1}=a^4\\ 2^{a+1}=a^4-a^{3}\\ 2^{a+1}=a^{3}(a-1)\\ (a+1)\log 2=3\log a+\log (a-1)\\ $$

This is from chapter quadratic equations.

I look for a short and simple way.

I have studied maths up to $12$th grade.

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    $\begingroup$ Hint: rearrange to get $2^{a+1} = a^3(a-1)$. By unique factorization both $a^3$ and $a-1$ have to be powers of 2. $\endgroup$ – lulu Jul 19 '15 at 13:22
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$$a^{3}+2^{a+1}=a^4$$

$$2^{a+1}=a^3(a-1)$$

Now we should use some number theory instead of taking logs. Note that now $a^3$ and $a-1$ must be powers of two, but the one is even and the other odd. So one of both should be 1. If $a^3=1$, then $a-1=0$, but 0 is not a power of two. If $a-1=1$ then $a=2$ and $a^3=8$.

We see that $a=2$ is in fact a solution, since it states $8+8=16$. This is the only natural solution.

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$$2^{a+1}=a^{4}-a^{3}=a^{3}\left(a-1\right)$$

So the only prime that divides $a^{3}\left(a-1\right)$ is $2$ so the factors $a^3$ and $a-1$ are both powers of $2$.

This leads almost directly to the conclusion that $a=2$ wich is indeed a root of the equality.

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