In triangle $ABC$ length of side $BC$ is $293$ (a prime). If length of side $AB$ is a perfect square, length of side $AC$ power of 2 and $AC$ twice length of $AB$, find the perimeter.
Kind of stumped here. From my understanding, if something is a 'power of 2' it is 2 to a power, with nothing else, no other prime factors. That would make $AC$ equal to $2^{2n+1} ,n\ge0$ because it is twice $AB$ which must then be $2^{2n}$ as the only prime factor in $AC$ is 2 and $AB$ is a perfect square.
Then it being a triangle probably use the Pythagorean Theorem. Thanks.
In the OP, it is shown that $AB=2^{2n}$ and $AC=2^{2n+1}$ for some integer $n\ge 0$. That argument is done well, and takes us most of the way to the solution.
Now we use the Triangle Inequality. Because $AB+BC \gt 293$, we nust have $2^{2n}+2^{2n+1}\gt 293$. That forces $n\ge 4$.
Could we have $n\ge 5$? Use the Triangle Inequality to show this is not possible. (The sum of sides $AB$ and $BC$ is too small.)
Remark: The Fundamental Theorem of Arithmetic was not used. Neither was the primality of $293$, only its rough size.