My challenge problem - - - Find all of the integer solutions to ...
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Take logs (I'll choose the natural log) of both sides to get:
\(\displaystyle (x^2-9)\ln(x)=0\)
Hence:
\(\displaystyle x=1,\pm3\)
However, we must also include \(\displaystyle x=-1\) since:
\(\displaystyle (-1)^{1-9}=(-1)^{-8}=(1)^{-8}=1\)
Thus, the integer solutions are:
\(\displaystyle x=\pm1,\pm3\)
\(\displaystyle (x^2-9)\ln(x)=0\)
Hence:
\(\displaystyle x=1,\pm3\)
However, we must also include \(\displaystyle x=-1\) since:
\(\displaystyle (-1)^{1-9}=(-1)^{-8}=(1)^{-8}=1\)
Thus, the integer solutions are:
\(\displaystyle x=\pm1,\pm3\)
Here is a more of a "discrete" approach.
Note that even integers cannot be a solution to \(\displaystyle f(x)=x^{x^2-9}-1=0\), as if \(\displaystyle x\) is an even integer then \(\displaystyle x^2-9\) is odd. An even integer raised to an odd integral power cannot give an odd integer (in fact it will be of the form \(\displaystyle (2k)^{\pm 1}\) for some integer \(\displaystyle k\)).
So it suffices (to find solutions) to restrict this function to a domain of odd integers, for which it is an even function, i.e. symmetric about 0. To see this, we note \(\displaystyle x^2-9=2k\) is even, so we have \(\displaystyle f(- x) = (-x)^{2k}-1 = x^{2k}-1 = f(x)\). So it further suffices to restrict the domain to positive odd integers.
Now assume \(\displaystyle x\) is a positive odd integer. We can transform the problem into solving \(\displaystyle x^{x^2}=x^9\). This right away gives that all solutions must satisfy \(\displaystyle x \in \{1,3\}\), otherwise \(\displaystyle x^{x^2} > x^{3^2} = x^9\) (here we use the fact that for \(\displaystyle x>1\) and \(\displaystyle a>b\), \(\displaystyle x^a > x^b\)).
Testing these values gives \(\displaystyle x=1,x=3\), and hence all possible solutions are \(\displaystyle \pm 1, \pm 3\).
Note that even integers cannot be a solution to \(\displaystyle f(x)=x^{x^2-9}-1=0\), as if \(\displaystyle x\) is an even integer then \(\displaystyle x^2-9\) is odd. An even integer raised to an odd integral power cannot give an odd integer (in fact it will be of the form \(\displaystyle (2k)^{\pm 1}\) for some integer \(\displaystyle k\)).
So it suffices (to find solutions) to restrict this function to a domain of odd integers, for which it is an even function, i.e. symmetric about 0. To see this, we note \(\displaystyle x^2-9=2k\) is even, so we have \(\displaystyle f(- x) = (-x)^{2k}-1 = x^{2k}-1 = f(x)\). So it further suffices to restrict the domain to positive odd integers.
Now assume \(\displaystyle x\) is a positive odd integer. We can transform the problem into solving \(\displaystyle x^{x^2}=x^9\). This right away gives that all solutions must satisfy \(\displaystyle x \in \{1,3\}\), otherwise \(\displaystyle x^{x^2} > x^{3^2} = x^9\) (here we use the fact that for \(\displaystyle x>1\) and \(\displaystyle a>b\), \(\displaystyle x^a > x^b\)).
Testing these values gives \(\displaystyle x=1,x=3\), and hence all possible solutions are \(\displaystyle \pm 1, \pm 3\).
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