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

Travis Willse
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yfusav
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2 Answers2

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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.

fleablood
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  • https://math.stackexchange.com/questions/810753/prove-that-every-odd-prime-number-can-be-written-as-a-difference-of-two-squares#comment1676488_810806 – yfusav Nov 01 '19 at 04:16
  • a and b should be natural numbers – yfusav Nov 01 '19 at 04:30
  • then how 1 can be formed – yfusav Nov 01 '19 at 04:31
  • Can you please help me in that – yfusav Nov 01 '19 at 07:00
  • $1 = 1^2 - 0^2$. – fleablood Nov 01 '19 at 15:49
  • If $a,b$ are not allowed to be $0$ then $1=a^2 -b^2=(a+b)(a-b)$ where requires $a+b=a-b =\pm 1$ which can only be solved by $a=1;b=0$ so $1$ is the only number that can not be don't. Any other odd number, except $-1$, can be done by $2k-1= k^2 - (k-1)^2$ (note $k$ can be negative). If $2k-1$ is prime then $a=k$ and $b=k-1$ is the only way but if $2k-1=m*n$ is composite that $a-b=m,a+b=n$ or $a=\frac{m+n}2$ and $b=\frac{n-m}2$ will also be a solutions. (and if $a$ and $b$ both squares $x=4$ and $a'=\sqrt a; b'=\sqrt b$ will work as well) – fleablood Nov 01 '19 at 15:59
  • "a and b should be natural numbers" Well, you should have said something about that then. – fleablood Nov 01 '19 at 16:00
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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?

Travis Willse
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