Which numbers can be realized in the form $a^x - b^x$ for $a, b$ of opposite parity and $x$ even? 
How can we check whether a number can be represented in the form of $a^x - b^x$, where $a$ and $b$ are integers of opposite parity and $x$ is positive and even integer.

I thougth of using $$a^n - b^n = (a - b) (a^{n-1} + a^{n-2}b + \cdots + b^{n-2}a +b^{n-1}) .$$

If $a,b$ have different parity then $a^x - b^x$ must be odd.
And if $x$ is even then $a^x - b^x = (a^{\frac x2} - b^{\frac x2})(a^{\frac x2} + b^{\frac x2})$
So for any odd $2k-1$, the $a = k; b=k-1; x= 2$ and $2k-1= (k-(k-1))(k+(k-1)) =k^2 - (k-1)^2$
So all odd numbers and only odd numbers can be so represented.
