Let $k \geq 2$ be an integer. Let $x$ and $y$ be positive integers. Show that $x^k- y^k > 2$.

I'm a little confused by this because this is the only information given. The book I'm using doesn't specify whether or not $x$ needs to be greater than $y$, so couldn't you have $2^2 - 3^2 = -5$?

Also, I just don't know how to approach this problem.

  • $\begingroup$ Presumably $x>y$. Try factoring $x^k-y^k$. $\endgroup$ – carmichael561 Apr 3 '16 at 23:25
  • $\begingroup$ Or... let $x = y + n; n>0$ then prove $(x+n)^k - x^k >2$. $\endgroup$ – fleablood Apr 3 '16 at 23:36
  • $\begingroup$ Although I have to get irked that it doesn't stipulate x > y. If $x \le ys this is obviously not true. $\endgroup$ – fleablood Apr 3 '16 at 23:38

If there is not the condition that $x>y$ then your counterexample is a good one and immediately yields that the statement is false in general.

If we add the additional constraint that $x>y$, then we know that $x^k-y^k$ is a positive integer. (if you do not wish to take this as fact and wish to prove this statement, note that integers are closed under multiplication and $x>y\Rightarrow x^k>y^k$)

Next, we note that we can factor $x^k-y^k$.

$x^k-y^k = (x-y)(x^{k-1}+x^{k-2}y+x^{k-3}y^2+\dots+xy^{k-2}+y^{k-1})$

Since $k\geq 2$, and $x>y>0$ are both integers, we know that $x\geq 2$ as well. The above factorization then has that both parts are positive integers and furthermore that the righthand part is at least $3$

since there are at least two positive integer terms in the summation, with at least one of them a multiple of two.

Thus $x^k-y^k>2$ for all positive integers $x>y$ and $k\geq 2$.


Since $x,y\in\Bbb N$ we may as well assume $x>y\iff x\ge y+1$. But then

$$x^k-y^k\ge x^k -(x-1)^k.$$

Set $z+1= x$ then it is enough to show $(z+1)^k-z^k>2$.

$$(z+1)^k-z^k = \sum_{n=0}^{n-1}{k\choose n} z^n> {k\choose 1}$$

here the last inequality follows by positivity of $z$. But ${k\choose 1} = k$ and $k\ge 2$.

So we have

$$x^k-y^k\ge x^k-(x-1)^k > {k\choose 1} \ge 2.$$


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