Which numbers can be realized in the form $a^x - b^x$ for $a, b$ of opposite parity and $x$ even?

Question

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

Answer 1

Hint For $x = 2 m$, we have $$a^x - b^x = (a^m)^2 - (b^m)^2 ,$$ that is, any difference of even powers is a difference of squares, so we may as well restrict our attention to $x = 2$.

Now, the factorization of a difference of squares is $$a^2 - b^2 = (a + b) (a - b) ,$$ which specializes the factorization formula in the question statement.

Additional hint Since $a$ and $b$ have opposite parity, $a - b$ and $a + b$ are both odd, hence so is $a^2 - b^2$. Are all odd numbers expressible this way?

Answer 2

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.